Section 6-2: Volumes of Rotation with Solids of Known...

Volumes of Rotation with Solids of Known Cross Sections In this lesson we are going to learn how to find the volume of a solid which is swept out by a curve revolving about an axis. There are three main methods, and one minor method, of doing this and we will first look at the minor method, finding the volume of a solid by slicing. If we can take a cross section of a volume, and find the area of that cross-section, then I can use calculus and integrals to "add up" all the areas of all the cross-sections. This method works with solids of any shape as long as you know a formula for the area of the cross section. Usually we see rectangles, squares, triangles, semicircles, and trapezoids. Our general guidelines are: 1. Sketch a cross section of the solid. It will be perpendicular to either the x- or y-axis. The axis it is perpendicular to gives the variable you will be using in the integral. If you do not have a 3-D mind, don’t despair! If you concentrate on the shape that your problem describes to you, you will still be able to do the problem even if you can’t visualize or draw the region. 2. Find the area of the cross section, A (x). 3. Find the limits of integration along the axis that the cross section is perpendicular to. The problem will tell you if you have a section that is perpendicular to the x- or y-axis.

4. Integrate the area formula throughout the interval: ( )

b

a

V A x dx= ∫

That is all there is to it Of course you still have to find the area of that cross section, but with practice you will be able to do it. Just remember to read the problem carefully to find out what shape you are talking about, and then find the area of that shape. Almost always, the difference between two given curves or the height of a single curve will give you the edge of a cube, the diameter or radius of a circle, or the base of a triangle that you need in order to find the area of the cross section.

Volumes of Rotation with Solids of Known Cross Sections To review, our general guidelines are: 1. Sketch a cross section of the solid. 2. Find the area of the cross section, A (x). 3. Find the limits of integration along the axis that the cross section is perpendicular to.

4. Integrate the area formula throughout the interval: ( )

b

a

V A x dx= ∫

Almost always, the difference between two given curves or the height of a single curve will give you the edge of a cube, the diameter or radius of a circle, or the base of a triangle. *** I find it almost impossible to draw these cross-sections unless I "lift" them out of the coordinate axes. If your 3-dimensional mind is better than mind, go for the 3-D picture. If not, follow my lead and just look at the isolated cross-section. Let's look at a few examples: A: We have a solid between planes perpendicular to the x-axis at x = -1 and x = 1. The cross sections are perpendicular to the x-axis. Note: they have already given us some information. Our limits of integration come from the x-axis, from x = -1 to x = 1. a) The cross sections will be circles whose diameters stretch from the

curve 2

11

yx

−=

+ to the curve

2

11

yx

=+

2 2 2

1 1 21 1 1

Dx x x

− −= − =

+ + +

2

1 12 1

R Dx

−= =

+

The diameter of the circle is the difference between the two curves between which it stretches.

The radius of the circle is half the diameter.

2

222

111

A Rxx

ππ π −

= = = ++

Find the area of the circular cross section.

1

211

V dxx

π

−

=+∫

211

1tan

4 4 2V x π π ππ π−

−

− = = − =

Finally write the volume integral.

This is easily integrated. Evaluate. Recall that tan-1 1 means that we want the angle whose tangent is 1.

b) This time we have vertical squares whose bases run from the curve

2

11

yx

−=

+ to the curve

2

11

yx

=+

2 2 2

1 1 21 1 1

Ex x x

− −= − =

+ + + The edge of the square is the

difference between the two curves between which it stretches.

2

222

2 411

A Exx

−= = = ++

Find the area of the square cross-section.

1

21

41

V dxx−

=+∫

11

14 tan 4 2

4 4V x π π π−

−

− = = − =


Again, this is easily integrated. Evaluate.

B. We have a solid with a base bounded by the circle x2 + y

2 = 4. We are

going to find the volume of four solids, each with a different cross section. Notice, though, that the base is the same in each case, is always perpendicular to the x-axis, and in each case the distance from one side of the cross section base is double the distance from the x-axis to the outside of the circle:

22 4y x= − . The edges of the circle along the x-axis go from x = -2 to x = 2. a) Here my cross sections are squares.

22 4E x= − The edge of the square is double the distance from the x-axis to the outside of the circle.

( )2

2 2 22 4 4 4A E x x = = − = − Find the area of the square cross-section.

( )2

2

2

4 4V x dx−

= −∫

23

2

1 1284 43 3

V x x−

= − =


This is easily integrated. Evaluate.

b) Here my cross sections are equilateral triangles. This one is a bit trickier because you have to know your geometry relationships of equilateral triangles.

22 4b x= −

22 4 x−22 4 x−

24 x−

23 4 x−

22 4 x−22 4 x−

24 x−

23 4 x−

23 4h x= −

The base of the triangle is double the distance from the x-axis to the outside of the circle.

Draw an equilateral triangle with all sides equal to the b we just found. Drop an altitude from the top vertex. You now have a 30-60-90 triangle, so you can find the length of the altitude. Remember that the altitude will bisect the base.

2 21 1 2 4 3 42 2

A bh x x = = − −

( )23 4A x= −

Find the area of the cross-section.

( )2

2

2

3 4V x dx−

= −∫

23

2

1 32 33 4 18.4753 3

V x x−

= − = ≈



c) Here my cross sections are semicircles.

24r x= − The radius of the each semicircle is the distance from the x-axis to the outside of the circle. We don’t double it this time – it is the diameter that goes all the way across, not the radius.

22 21 1 4

2 2A r xπ π = = −

( )242

A xπ= −

Find the area of the half-circle cross-section. Be careful here, to take half the area of the circle – we are given semi-circles!

( )2

2

2

42

V x dxπ

−

= −∫

23

2

1 164 16.7552 3 3

V x xπ π

−

= − = ≈



d) This time my cross sections are isosceles right triangles. Again, this one is a bit trickier because it depends on some geometry knowledge. Don’t worry, most problems use rectangles, squares, and circles or semicircles.

22 4b x= −

22 4 x− 22 4 x−24 x−

24 x−

22 4 x− 22 4 x−24 x−

24 x−

24h x= −

The base of the triangle is double the distance from the x-axis to the outside of the circle.

Draw an right isosceles triangle with the base equal to the b we just found. Drop an altitude from the top vertex. You now have a 45-45-90 triangle, so you can find the length of the altitude. Remember that the altitude will bisect the base.

2 21 1 2 4 42 2

A bh x x = = − −

24A x= −


( )2

2

2

4V x dx−

= −∫

23

2

1 324 10.6663 3

V x x−

= − = ≈



Volumes of Rotation with the Disk Method Now we continue to find volumes of rotation, this time with the disk method. This will be a real test of your ability to see in three dimensions. If you have one of those great see-in-3-D minds, this will seem easier for you than it does for two dimensional people like me! The premise behind this method is that when we rotate a curve around an axis it will form a solid which is in the shape of a cylinder, although it will not be a "nice" cylinder with straight sides. We will use the method of limits to find the volume of an infinite number of small cylinders and add them up, akin to what we did with rectangles and trapezoids when finding the area under a curve. Here is a picture (best I could do!) to illustrate what we are doing:

y = x

dx

R

In this graph we have the curve y = √ x which is revolving around the x-axis. It is sweeping out an urn-shaped figure whose volume we want to determine. I have drawn a thin cylinder within the urn whose radius is equal to the distance from the x-axis to the y-value of the curve and whose height is denoted as dx. The cylinder is standing on its edge vertically. The volume of any cylinder is V = π r

2 h so the volume of this cylinder will be

V = π r

2 h This is the formula for the volume of a cylinder.

= π (√x)2 dx I put in the values for the radius and the height of the

cylinder in our picture. = π x dx. Simplify.

However, there is more than that one cylinder sandwiched into this urn. In fact there are an infinite number of them and they are also infinitely thin; so thin that dx will approach 0. As you have probably guessed by now, to find the total of all these small volumes and get the volume of the urn, we will be taking the limit of the sum of these volumes, which is the integral of our volume formula. The trick is to make sure to get the proper volume formula and make the proper substitutions.

Here are the steps to take as you set up your integrals to find volumes of rotation by the disk method:

1) Draw the region for which you want to find the volume. The sketch does not have to be fancy, or even three dimensional, but you do have to be able to see the curve and the axis about which it will rotate. 2) Draw a thin rectangle perpendicular to the axis of rotation and across the region. It should look like the one above. Be sure that it only extends across the region for which you want the volume, from the axis of rotation to the curve. 3) Determine the function which describes the radius. This function will always need to be in terms of the variable about whose axis you are rotating (or about a line parallel to the axis). Hence, in the sketch above we are rotating about the x-axis so the radius was expressed in terms of x. 4) Determine the limits of integration. These will come from along the axis of rotation; from one end of the "urn" to the other.

5) Write your volume integral: 2b

a

V R dxπ= ∫

6) Integrate to find the volume. Let me reiterate the process. The most important thing is to sketch the curve and show the axis of rotation. Then you draw the rectangle through the region between the axis of rotation and the curve. That rectangle represents your radius, you need an expression for it (usually just the curve, unless the axis of rotation is not a coordinate axis). Then write your integral and evaluate to get your volume. Not too terrible, but as usual it will take practice.

Volumes of Rotation with the Disk Method - examples To review our disk method: 1 – draw the region 2 – draw in the rectangle perpendicular to the axis of rotation and extending through the region. 3 – determine the function that describes the radius, which will be the function that describes the height of the rectangle. 4 – determine the limits of integration. They always come from along the axis of rotation.

5 – write the volume integral: 2b

a

V R dxπ= ∫

6 – integrate and evaluate the antiderivative.

A: We are going to revolve the region bounded by the graphs of 8ln yxy

= ,

y = e, and x = 0 about the y-axis. The region resembles a quarter circle bounded on the right and bottom by the curve, on the left by the y-axis, and on the top by the horizontal line. Notice that we are going to rotate about the y-axis, so our radius and limits of integration will both be in terms of y. You can imagine the "bowl" we will have when the region rotates around the y-axis.

Draw the rectangle through the region, horizontally (perpendicular to the axis of rotation). I want to see the region and rectangle on your paper. Next we have to define the radius of the cylinder. In this case it is going to be

8ln yy

the x-value along the curve at any y.

e

1

R

The limits of integration will be along the y-axis from y = 1 to y = e. Be sure to start at the lowest value. As y moves from the value of 1 to the value of e we will get the area of all the circles which are piled up on top of each other.

Now we can set up the integral and simplify. 2

1 1

8ln ln8e ey yV dy dy

y yπ π

= =

∫ ∫

Now integrate and evaluate the antiderivative.

u = ln y du = 1/y dy.

To find the integral here we will need to use substitution:

2 218 8 4

2V u du u uπ π π = = =

∫ Make the substitution and take the antiderivative. ** Leave the limits of integration off until you put the x’s back in, or change them to match the u’s!

( )2

14 ln

eV yπ = Reverse the substitution (unless you have

already changed the limits of integration to fit u) and get ready to evaluate.

= 4π [ (ln e)2 - (ln 1)

2 ]

= 4π [12 - 0

2 ]

= 4π

Finish the evaluation.

You can see that our integration skills from the last couple of chapters will still be greatly needed! On the test I will let you use your calculator to evaluate the integrals once you have them set up, but for now I want you to practice your integration skills on the HW. You have to be very good at integration!! B: The solid is bounded by the curves y = x3, y = 0, and x = 2 and is formed by a rotation about the x-axis. First we have to draw the region:

2

y = x^3

R

I have drawn the rectangle perpendicular to the x-axis, the axis of rotation. The line x = 2 forms the right-hand border and the y-axis forms the left-hand border so the limits of integration will be x = 0 to 2. The radius is x

3.

( )2 2

23 6

0 0

V x dx x dxπ π= =∫ ∫

We can now set up the integral.

I simplified by squaring the radius and moving the constant π out in front.

227 7

00

17 7

V x xππ = = ( ) 128128 0 57.446

7 7V π π

= − = ≈

This integral can be directly evaluated without substitution.

C: The region is confined to the first quadrant and bounded by the curves y = 1, y = tan x, and the y-axis. We will be rotating about the line y = 1. The diagram comes first:

Notice how the radius only extends to the curve y = tan x; it does not go all the way down to the x-axis. The formula y = tan x gives the value of y - its height above the x-axis. Our radius lies between this curve and the line y = 1, so the expression for the radius will be 1 - tan x. Our limits of integration have to be in terms of x, so we are moving from x = 0 to x = π/4 (the value at which tan x = 1; the point of intersection of the curves y = tan x and y = 1). Look here at a simplified version of our rectangle and curve to see how I found the length of our rectangle:

y = 1

π/

y = tan

R

( )4

2

0

1 tanV x dxπ

π= −∫

Now we will set up the integral. To further simplify this so that we can integrate the long way would require that we square the binomial with FOIL and then use some trig identities to rewrite the integrand so that we can find the antiderivative. Ugh – in this case, let’s not do all that work!

0.3068 0.964V π≈ ≈

Evaluate the integral with your calculator. Make sure that you take any decimal answers out correct to three places.

I expect to see the graphs with your rectangles drawn. Don't skip that step - it will become increasingly important as the module continues. Show all your steps and work. If you keep organized you should be fine. Again let me say that on the test you will be able to use your calculator to evaluate most of the integrals, once you have the problem set up. But for the homework, I want you to practice integration, unless I specify otherwise.

tan x

1 R = 1 – tan x

More Examples with the Disk Method Here are some more examples that use the disk method: D: Our region is bound on the top by the curve y = √x, on the left by the vertical line x = 4, on the left by the vertical line x = 1, and on the bottom by the x-axis.

The rectangle is drawn perpendicular to the axis of rotation, the x-axis.

y = √x => R = x1/2

x = 1 to x = 4

Radius = the length of the rectangle = y-value of the curve in terms of x because we are rotating about the x-axis and have to use dx The limits of integration come from along the x-axis.

( )4 4

21/2

1 1

V x dx x dxπ π= =∫ ∫ Write the volume integral and simplify.

42

1

12

V xπ =

( ) 1516 1 7.5 23.56192 2

V π π π= − = = ≈

Find the antiderivative and evaluate.

E: Our region is bound on the right by the curve y = x2, on the left by the y-axis,

below by the origin, and on the top by the horizontal line y = 4. This time we are rotating about the y-axis, and so our rectangle is horizontal. Since that rectangle intersects the y-axis, we will be using dy in the integral. That means that the limits and the integrand must be in terms of y, not x.

The rectangle is drawn perpendicular to the axis of rotation, the y-axis.

y = x2

x = y

1/2 = R

y = 0 to y = 4

Radius = the length of the rectangle = x-value of the curve in terms of y because we are rotating about the y-axis and have to use dy The limits of integration come from along the y-axis.

( )4 4

21/2

0 0

V y dy y dyπ π= =∫ ∫ Write the volume integral and simplify.

42

0

12

V yπ =

( )16 0 8 25.13272

V π π= − = ≈


F: Our region is bound on the right by the curve y = x2/3

, on the left by the y-axis, below by the origin, and on the top by the horizontal line y = 1.

The rectangle is drawn perpendicular to the axis of rotation, the y-axis.

y = x2/3

x = y

3/2 = R

y = 0 to y = 1


( )1 1

23/2 3

0 0

V y dy y dyπ π= =∫ ∫ Write the volume integral and simplify.

( )1

4

0

1 1 04 4 4

V y π ππ = = − =


G: We have to find the volume of the solid generated by the region bounded by the graphs of y = x, y = 0, y = 4, and x = 6. We are going to revolve the region about the line x = 6.

The graph is above. Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is horizontal). This means that we will be using dy in the integral.

First you have to graph the region. It is a trapezoidal region, with a bottom base along the x-axis, top base along the line y = 4, right side along the vertical line x = 6, and left side slanted along the line y = x, from the origin to the point (4, 4).

R = 6 - x = 6 - y The length of the rectangle is from the vertical line x = 6 to the slant line x = y. We have to subtract them to get the radius. y = 0 to y = 4


( )4

2

0

6V y dyπ= −∫ Write the volume integral and simplify.

( )43

06

3V yπ− = −

[ ] 2088 216 217.8173 3

V π π−= − = ≈

Find the antiderivative and evaluate. I used substitution here (u = 6 - y so - du = dy) but I could have also expanded the ( ) with FOIL and then integrated.

y = x

y = 4

x = 6

dy

6

x

H: We have to find the volume of the solid generated by the region bounded by the graphs of y = 1/x, y = 0, x = 1, and x = 4. We are going to revolve the region about the x-axis.

The region is drawn above. We drew the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is vertical). This means that we will be using dx in the integral.

First you have to graph the region.

y = R = 1/x The length of the rectangle is from the x-axis to the curve y = 1/x. We have to put the equation in terms of x. x = 1 to x = 4


24 42

1 1

1V dx x dxx

π π − = = ∫ ∫

41

1V xπ − = −

1 314 4

V π π = − − =

Write the volume integral and simplify.

1 4

R

I: We have to find the volume of the solid generated by the region bounded by the graphs of y = e

-x^2, y = 0, x = 0, and x = 2. We are going to revolve the region about the

x-axis.

The graph is shown, with the rectangle drawn through the middle of the region, perpendicular to the axis of rotation (so it is vertical). This means that we will be using dx in the integral.

First you have to graph the region. It is a bell-shaped region, with a bottom base along the x-axis, top curving base along the curve y = e

-x^2,

right side along the vertical line x = 2, and left side along the y-axis.

y = R = e-x^2

The length of the rectangle is from the x-axis to the curve y = e

-x^2. We keep the

equation in terms of x. x = 0 to x = 2


( )2 2

2^2 2 ^2

0 0

x xV e dx x dxπ π− −= =∫ ∫

V ≈ 1.9686

Write the Volume integral and evaluate. We have to use the calculator to evaluate this integral.

2

R

AP Calculus Mrs. Jo Brooks

1

Volumes of Rotation with the Washer Method In this part of lesson 7-3 we will look at how to find the volume swept out by a region which does not border on the axis of revolution. What happens when the region does not border on the axis of revolution is that there will be a hole in our solid and a cross-section of the solid will look like a washer instead of a disk. Let's look at a diagram:

R ra

b

The rectangular area on the top is going to be rotated about the x-axis (that is what that little arrowed-circle means on the far right). The rectangle does not touch the x-axis, thus when it rotates it will leave a hole. The shaded portion shows the solid and the white portion is the hole. I have drawn a rectangle across the region, just as we did in the last lesson.

21

b

a

V R dxπ= ∫

If you look at the line R, it shows the radius the outer shell. This integral will give us volume of the solid that extends all the way down to the x-axis (using the disk method from before).

22

b

a

V r dxπ= ∫

Now look at the line r; it shows the radius of the hole. This integral will give us the volume of the inner, smaller solid (the volume of the hole).

2 2b b

a a

V R dx r dxπ π= −∫ ∫

2 2b b

a a

V R dx r dxπ π= −∫ ∫

2 2b b

a a

V R dx r dxπ

= − ∫ ∫

( )2 2b

a

V R r dxπ= −∫

Thus, to get the volume of the shaded washer we only have to subtract these two integrals. Move π the out in front of each integral. Factor the π out of the expression.

This is our final formula for the volume of rotation when the solid is in the shape of a washer and where R = outer radius, r = inner radius, and a and b are the limits of integration.


2

Thus, the method of volume of revolution using washers is just a variation on the theme of the disk method. As you do these, remember that the formula calls for the difference of the squares of the two radii, not the square of the difference of the two radii. In other words, we have to square each radius first and then subtract; not the other way around. Please be careful to not make the algebraic error of thinking that R

2 - r

2 is the same as (R - r)

2. They are

not!!! Think FOIL and remember that (R - r)2 = (R - r)(R - r) = R

2 - 2Rr + r

2

So, to wrap it up, the steps we take with the washer method are basically the same as with the disk method: 1) Draw the region for which you want to find the volume. The sketch does not have to be fancy, or even three dimensional, but you do have to be able to see the curve and the axis or line about which it will rotate. 2) Draw a thin rectangle perpendicular to the axis of rotation and across the region. It should look like the one above. Be sure that it only extends across the region for which you want the volume. Draw in arrows for the two radii. 3) Determine the functions which describe the outer radius, R, and the inner radius, r. These functions will always need to be in terms of the variable about whose axis you are rotating (or about a line parallel to the axis). Hence, in the sketch above we are rotating about the x-axis so the radii will be expressed in terms of x. 4) Determine the limits of integration. These will come from along the axis of rotation; from one end of the washer to the other. They are determined by the points of intersection of your boundary curves.

5) Write your volume integral: ( )2 2

b

a

V R r dxπ= −∫ Simplify the expression if

possible. 6) Integrate to find the volume.


1

Volumes of Rotation with the Washer Method To review, the formula for volumes of rotation with the washer method is:

Washer Method

( )2b

r

a

V R r dx= −∫

Where R = outer radius r = inner radius

a, b are the limits of integration from along the x-axis and the axis of rotation is the x-axis (or parallel to it)

** If you are revolving about the y-axis or a line parallel to it, everything will be in terms of y.

And the steps that we follow are: 1) Draw the region for which you want to find the volume. 2) Draw a thin rectangle perpendicular to the axis of rotation and across the region. Draw in arrows for the two radii. 3) Determine the functions which describe the outer radius, R, and the inner radius, r. 4) Determine the limits of integration.

5) Write your volume integral: ( )2b

r

a

V R r dx= −∫ Simplify the expression if possible.

6) Integrate to find the volume. Now let’s look at three examples.


2

A: Our region is bounded by the curves y = 2 √x, y = 2, x = 0 and will be rotated about the x-axis. The first task is to draw the region:

r R

y = 2

y = 2 x

Since we are rotating about the x-axis, everything will be in terms of x and our rectangle is perpendicular to the x-axis. The radii are always measured from the axis of rotation so our outer radius, R, will go from the x-axis to the horizontal line and will be equal to the height of the upper line y = 2. The inner radius, r, will extend from the x-axis to the inner curve and be equal to the height of our curve y = 2 √x. The limits of integration will move along from x = 0 to x = 1 (which is where y = 2 intersects y = 2 √x). Now we can set up our integral and solve for the volume.

( )( )1 22

0

2 2V x dxπ= −∫

( )1

0

4 4V x dxπ= −∫

This is our integral.

I squared each radius and simplified.

12

04 2V x xπ = −

This integral can be easily integrated.

= π [ (4 - 2) - 0 ] = 2π

Evaluate the antiderivative.

As you can see, the integral was not difficult on this problem. The trick is getting the problem set up properly in the first place.


3

B: Here we have a region bounded by the circle, x2 + y

2 = 25, and the line x = 4 being

rotated around the y-axis. First we will draw the region:

r

R

x = -4 x = 4

3

-3x2 + y2 = 25

Since we are rotating about the y-axis, the rectangle is perpendicular to the y-axis and all functions will be in terms of y. Notice how the arrows I draw for the inner and outer radii always extend from the axis of rotation to the curve which bounds the region. This is important to remember. Our outer radius, R, will be the curve in terms of y: x = √(25 - y

2 ). This is the distance from the

y-axis to the outer curve. The inner radius, r, will be the line x = 4, the distance 4 from the y-axis to the vertical line. Our limits of integration are from y = -3 to y = 3; these are the points were the line x = 4 meets the curve x = √(25 - y

2 ).

Now we can set up the integral:

( ) ( )3 2 22

3

25 4V y dyπ−

= − − ∫

( )3

2

3

25 16V y dyπ−

= − −∫

( ) ( )3 3

2 2

3 0

9 2 9V y dy y dyπ π−

= − = −∫ ∫


I squared each radius. When I simplified I used symmetry to change the limits and just double the integral.

33

0

12 93

V y yπ = −

This integral is can be easily integrated.

= π [ (27 - 9) - 0] = 36π



4


5

C: In this problem our region is confined to the first quadrant and is bounded by the curves y = x

2, the x-axis, and x = 1. We will be rotating about the line x = -1, which is parallel to the y-

axis. Thus the rectangle and all functions will be in terms of y. First we draw the region:

1-1-2

y = x2

Rr

We will have to be more careful this time when we find our radii because the axis of rotation is not a coordinate axis. The outer radius, R, will be the distance from the line x = 1 to the axis of rotation: x = 1 + 1 = 2. The inner radius will be the distance from the curve to the axis of rotation (and we have to remember to put the curve in terms of y): x = 1 + √y . For each of the radii I had to add 1 in order to get the total distance from the curve to the axis of rotation. Tricky, eh? You have to be careful and don't skip the step of drawing the rectangle through the region and the arrows for R and r. The limits of integration will be y = 0 to y = 1 (which is where the curves x = √y and x = 1 meet). Now to set up the integral:

( ) ( )( )1 22

0

2 1V y dyπ= − +∫

( )( )1

0

4 1 2V y y dyπ= − + +∫

( )1

1/2

0

3 2V y y dyπ= − −∫


I squared each radius and simplified. I also wrote the radical with exponents.

13/2 2

0

4 133 2

V y y yπ = − −

The integral is can be easily integrated.

= π (3 - 4/3 - 1/2 - 0 ) = 7π/6



6

Again, like the disk method, this should not be too bad but you will have to be very careful about what you are doing when you choose your radii and limits of integration. I expect to see your graphs on your papers. The graphs do not have to be fancy. Just the general shape of the region with your rectangle and radii marked and big enough so that you can read them.

More Examples with the Washer Method D: The region is a crescent moon shape, bounded on the top by the curve y = x

2

and on the bottom by the curve y = x3. We are going to rotate the region about the x-

axis.

The endpoints of our region are from x = 0 to x = 1, the points of intersection of the two curves. When we revolve about the x-axis, we see that there will be an open space between the axis of rotation (the x-axis) and the region, so we have to use washers. The rectangle is perpendicular to the axis of rotation, so it is vertical.

R = x2

r = x

3

x = 0 to x = 1

Outer Radius = the distance from the axis of rotation to the top of the rectangle Inner Radius = the distance from the axis of rotation to the bottom of the rectangle The limits of integration come from along the x-axis.

( ) ( )1

2 22 3

0

V x x dxπ = − ∫

14 6

0

V x x dxπ = − ∫

Write the Volume integral and simplify.

15 7

0

1 15 7

V x xπ = −

V = 2π/35


E: We are going to find the volume of the solid generated by the region bounded by the graphs of y = √x , y = 0, and x = 4 and we are going to do the problem four times, once revolving about the x - axis, once around the y-axis, around the line x = 4, and once about the line x = 6. We will be using both disks and washers in this section, so you will have to decide which fits the situation at hand. Remember that you use disks when the region lies completely along the axis of rotation. If there is a gap between the region and the axis of rotation then you have to use washers. E-1) This time we revolve about the x-axis.

y=√x

4

R

First you have to graph the region. Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is vertical). This means that we will be using dx in the integral. Since the region bounded by the heavier lines is completely along the axis of rotation, we can use disks, the first method we learned.

R = √x x = 0 to x = 4

Radius = the distance from the axis of rotation to the top of the rectangle The limits of integration come from along the x-axis.

( )4 42

0 0

V x dx x dxπ π= =∫ ∫ Write the Volume integral and simplify.

42

0

1 82

V xπ π = =


E-2) This time we revolve about the y-axis.

First you have to graph the region. Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is horizontal). This means that we will be using dy in the integral. This time there is a gap between the end of the rectangle and the axis of rotation, so we will be using washers.

R = 4 y = √x => r = y

2

y = 0 to y = 2

Outer Radius = the distance from the axis of rotation to the far end of the rectangle; in this problem that distance is always 4 Inner Radius = the distance from the axis of rotation to the near end of the rectangle; in this problem that distance is always the curve, which has to be in terms of y. The limits of integration come from along the y-axis. The intersection of the vertical line and the curve is (4, 2).

( ) ( )2

22 2

0

4V y dyπ = − ∫

24

0

16V y dyπ = − ∫

Write the Volume integral and simplify.

25

0

1165

V y xπ = −

32 12831 05 5

V π π = − − =


E-3) This time we revolve about the line x = 4.

y=√x

4

R

First you have to graph the region. The right side of the region lies completely along the axis of rotation in this case, so we will be able to use disks. Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is horizontal). This means that we will be using dy in the integral.

R = 4 - √x = 4 - y2

we have to subtract the distance up to the curve y = x2 from the distance to the vertical line x = 4 to get the distance from the axis of rotation to the left end of the rectangle y = 0 to y = 2

Radius = the distance from the axis of rotation to the far end of the rectangle. We have to do some subtracting here and also use the equation in terms of y. The limits of integration come from along the y-axis.

( )2

22

0

4V y dyπ= −∫

( )2

2 4

0

16 8V y y dyπ= − +∫

Write the Volume integral and simplify. Remember to use FOIL when expanding the parenthesis squared.

23 5

0

8 1163 5

V y y yπ = − +

64 32323 5

V π = − + 256 17.066 53.616515

V π π= ≈ ≈

Find the antiderivative and evaluate. You can give your answer in either of these forms.

E-4) This time we revolve about the line x = 6.

y=√x

4

R

6

r

First you have to graph the region. The right side of the region no longer lies completely along the axis of rotation in this case, so we have to use washers. Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is horizontal). This means that we will be using dy in the integral.

R = 6 - √x = 6 - y2

we have to subtract the distance up to the curve y = x

2 from the

distance to the vertical line x = 6 to get the distance from the axis of rotation to the left end of the rectangle r = 6 - 4 = 2 y = 0 to y = 2

Outer Radius = the distance from the axis of rotation to the far end of the rectangle Inner Radius = the distance from the axis of rotation to the near end of the rectangle; in this problem that distance is always 2. The limits of integration come from along the y-axis.

( )2

22 2

0

6 2V y dyπ = − − ∫

22 4

0

36 12 4V y y dyπ = − + − ∫

22 4

0

32 12V y y dyπ = − + ∫

Write the Volume integral and simplify. Remember to use FOIL when expanding the parenthesis squared.

23 5

0

132 45

V y y yπ = − +

3264 325

V π = − +

192 38.4 120.6375

V π π= = ≈

Find the antiderivative and evaluate. Again, any form of the answer is fine.

F: We have to find the volume of the solid generated by the region bounded by the graphs of y = x

2 and y = 4x - x

2. We are going to do the problem twice, once revolving

about the x - axis and once about the line y = 6.

F-1) This time we revolve about the x-axis.

The vertical rectangle, perpendicular to the axis of rotation means that we will be using dx in the integral.

First you have to graph the region. The point of intersection of the two curves, (2, 4)

R = 4x - x2

r = x

2

x = 0 to x = 2


( ) ( )2

2 22 2

0

4V x x x dxπ = − − ∫

22 3

0

16 8V x x dxπ = − ∫

Write the Volume integral and simplify. I skipped the algebra steps here. Remember to use FOIL on the first term. The two x4 terms canceled out.

23 4

0

16 23

V x xπ = −

32 210 33.51033 3

V π π= = ≈

Find the antiderivative and evaluate. Give your answer in whichever form you prefer.

R r

24y x x= −

2y x=

F-2) This time we revolve about the line y = 6.

The graph is the same as for part a with the exception of the new axis of rotation.

First you have to graph the region. curves, (2, 4)

R = 6 - x2

we have to subtract the distance up to the curve y = x2 from the distance to the horizontal line y = 6 to get the distance from the axis of rotation to the bottom of the rectangle r = 6 - (4x - x

2 ) = 6 - 4x + x

2

subtraction was required here too x = 0 to x = 2

Outer Radius = the distance from the axis of rotation to the bottom of the rectangle (the end farthest from the axis of rotation). Inner Radius = the distance from the axis of rotation to the top of the rectangle (the end closest to the axis of rotation. The limits of integration come from along the x-axis.

( ) ( )2

2 22 2

0

6 6 4V x x x dxπ = − − − + ∫

23 2

0

8 5 6V x x x dxπ = − + ∫

Write the Volume integral and simplify. Remember to use FOIL when expanding the parentheses squared. I left out the algebra steps here. I was able to factor out an 8 from all terms.

24 3 2

0

1 58 34 3

V x x xπ = − +

64 121 67.02063 3

V π π= = ≈


R r 24y x= −

2y x=

6

G: We have to find the volume of the solid generated by the region bounded by the graphs of y = 1/x and y = 0, x = 1, and x = 4. We are going to revolve the region about the line y = 4.

Draw the rectangle through the middle of the region, perpendicular to the axis of rotation (so it is vertical). This means that we will be using dx in the integral.

First you have to graph the region. It is a trapezoid-shaped region, with a bottom base along the x-axis, top base along the curve y = 1/x, and vertical sides at x = 1 and 4. We have to use washers because there is a gap between the region (where the rectangle is) and the axis of rotation.

R = 4 - 0 = 4 this is just the distance from the x-axis to the axis of rotation since the region sits on the x-axis r = 4 - 1/x You had to subtract here. x = 1 to x = 4


( )24

2

1

14 4V dxx

π = − −

∫

4 42

21 1

8 1 8V dx x dxx x x

π π − = − = − ∫ ∫

Write the Volume integral and simplify. Remember that the antiderivative of 8/x is a natural log, but when the exponent is not 1, it must be rewritten with negative exponents first.

41

18ln 2V x xπ − = +

38ln 4 10.3403 32.4854

V π π = − ≈ ≈

Find the antiderivative and evaluate. Any form of the answer is fine.

4

1 4

4

1 4

r R


1

Finding Volumes of Rotation with the Shell Method The Shell method of finding volumes of revolution is an alternative to the washer method. Usually, the two methods can be used interchangeably. The times that this method of shells may be the better choice is where it is inconvenient to do the squaring involved in the R

2 - r

2 formula

of washers, or when it is difficult or impossible to express a function implicitly in terms of y when revolving about the y-axis. In the previous methods we were finding the volumes of successive cylindrical disks or washers and then "adding them up". In this method we will instead find the surface area of cylinders with successive radii and add them up. What we are doing, in essence, is finding the surface area of the outer layer of the cylinder, then adding it to the surface areas of an infinite number of successive layers. This will cause us to do a couple of things differently as we set up our problem. First look at a diagram of a volume formed and see the parts which we will be concerned with:

rh

dx

y = f(x)

r = x (distance from axis of rotation)h = f(x)dx = thickness of shell

The curve f(x) is being rotated about the y-axis. We will be concerned with the surface area of the cylinder, so the rectangle will be drawn parallel to the axis of rotation, rather than perpendicular to it. The thickness of the shell is our dx and corresponds to the thickness of the rectangle. This means that all of our functions and limits of integration will be in terms of x. Always draw that rectangle in so that you will have a reference to remind you which variable you will be using. Notice that with this method the variable in the integral is not the same as the axis about which we are rotating. We will also need the radius of the shell (the distance from the axis of rotation), which is usually just the independent variable. The height of the shell, which is the height of the rectangle at any point, is usually the function. After you go over this section, be sure to look at the video for this method. It will help you visualize what is happening and give you some practice. You may remember that the formula for the surface area of a cylinder comes from the area of a rectangle. If we take a cylindrical shell, open it up, and flatten it out, we get a rectangle whose length is the same as the circumference of the cylinder/circle and whose width is the height of the cylinder:


2

length = Circumference = 2 π r

h

h

r

Surface Area = 2 π r h

This is the formula, then, that we will use in our integral, where a and b are the limits of

integration. 2b

a

V r h dxπ= ∫

So, to summarize, the steps involved in finding the volume of rotation by the shell method are: 1) Draw the region for which you want to find the volume. The sketch does not have to be fancy, or even three dimensional, but you do have to be able to see the curve and the axis about which it will rotate. 2) Draw a thin rectangle parallel to the axis of rotation and across the region. It should look like the one above. Be sure that it only extends across the region for which you want the volume. 3) Determine the function which describes the radius. This function will always need to be in terms of the variable whose axis is perpendicular to the one about which you are rotating. Hence, in the sketch above we are rotating about the y-axis so the radius was expressed in terms of x. We will be using the variable which matches the dx or dy which is the thickness of the rectangle. 4) Determine the function which describes the height of the cylinder and rectangle. Be careful to use the correct variable. 5) Determine the limits of integration. These will come from along the axis which is perpendicular to the axis of rotation; from the center of the cylinder to the outside.


a

V r h dxπ= ∫

7) Integrate to find the volume.


1

Volumes of Rotation with the Shell Method – Examples A-C To summarize, the steps involved in finding the volume of rotation by the shell method are: 1) Draw the region for which you want to find the volume. 2) Draw a thin rectangle parallel to the axis of rotation and across the region. 3) Determine the function which describes the radius. This function will always need to be in terms of the variable whose axis is perpendicular to the one about which you are rotating. 4) Determine the function which describes the height of the cylinder and rectangle. 5) Determine the limits of integration.


a

V r h dxπ= ∫

7) Integrate to find the volume. Now let’s look at a few examples. A: We are going to rotate the curve y = x about the x-axis. First we draw the curve with its rectangle parallel to the axis of rotation, the x-axis:

dyr

2

y = x

r = yh = 2 - x = 2 - y Notice how I extended with a

dotted line the rectangle all the way to the y-axis. I write the dy there and this reminds me that everything will be integrated with respect to y. The radius is the distance from the axis of rotation to the rectangle, which is just the value y on the y-axis. The height of the rectangle is often simply the function itself, but if you look at the rectangle, its height is the x-value of 2 minus the x-value of the curve. Our limits of integration will be along the axis which is perpendicular to the axis of rotation, the y-axis (remember that everything in this problem has to be in terms of y). At x = 2, y = 2, so the limits of integration will be y = 0 to 2.


2

Now we are ready to write out our integral and solve.

( )2

0

2 2V y y dyπ= −∫ This is our integral with values put in for r and h.

( )2

2

0

2 2V y y dyπ= −∫ Simplify this by moving the constants out front and multiplying through by the y.

22 3

0

123

V y yπ = −

We can now integrate easily, using the power rule.

8 82 4 03 3

V π π = − − =

Now we evaluate between 0 and 2.

B: We are rotating the curve y = 1/(x

2) this time. Again, we start with the graph and the

rectangle parallel to the axis of rotation, which in this problem is the y-axis:

1/2 2dxr h

r = xh = x -2

y = 1/x2

The distance from the axis of rotation to the rectangle is simply the x-value at the rectangle. The height of the rectangle is the height of the curve. The limits of integration are the values along the x-axis from x = 1/2 to 2. Now, to set up the integral:


3

( )2

2

1/2

2V x x dxπ −= ∫ This is our integral with values put in for r and h.

21

1/2

2V x dxπ −= ∫ Simplify this by moving the constants out front and multiplying the variables.

[ ]2

1/22 lnV xπ= We can now integrate easily.

= 2π[ ln 2 - ln 1/2 ] = 2π[ln 2 + ln 2 ] = 2π(2 ln 2) = 4πln 2 = π ln 16

Now we evaluate between 1/2 and 2. I have to use the laws of exponents/logs to simplify our answer. On the AP exam you are allowed to stop at the first line, but in this module I want you to practice your rules of manipulation. You do not have to do that last simplification.

C: We will do this problem twice, rotating about a different axis each time. C-1: The first time we are rotating the curve x = y - y

3 about the x-axis. First we need the

graph and the rectangle parallel to the axis of rotation, the x-axis. Notice how I keep repeating that these rectangles are parallel to the axis of rotation? Since the main difference between shells and our previous methods is the orientation of the rectangle I want to make sure you are aware of where you need to draw the rectangle.

1

x = y - y3rdy

h

r = yh = y - y3

The radius is the y-value at the rectangle. The height is the distance from the y-axis to the curve. All of this is in terms of y; our width of the rectangle is dy. The limits of integration will be from the y-value of 0 to 1. Set up the integral:


4

( )1

3

0

2V y y y dyπ= −∫ This is our integral with values put in for r and h.

( )1

2 4

0

2V y y dyπ= −∫ Simplify this by moving the constants out front and multiplying the variables.

13 5

0

1 123 5

V y yπ = −

We can now integrate easily.

= 2π [ (1/3 - 1/5 ) - 0 ] = 2π (2/15) = (4/15) π

Now we evaluate between 0 and 1. Simplify. Again on the AP exam you are not required to simplify completely. I like you to practice!

C-2: This time we will set up to rotate about the line y = 1:

1

x = y - y3r

dyh

r = 1 - yh = y - y3

The radius here is the distance from the axis of rotation (the line y = 1) and the rectangle. This will have to be 1 minus the y-value of the rectangle: r = 1-y. The height of the rectangle is again the curve. The limits of rotation will be along the y-axis from 0 to 1. Set up the integral:


5

( )( )1

3

0

2 1V y y y dyπ= − −∫ This is our integral with values put in for r and h.

( )1

3 2 4

0

2V y y y y dyπ= − − +∫

Simplify this by moving the constants out front and using FOIL to multiply out the integrand. Substitution will not work here so multiplying everything out is our only option.

12 4 3 5

0

1 1 1 122 4 3 5

V y y y yπ = − − +

We can now integrate easily with the power rule.

1 1 1 122 4 3 5

V π = − − + 7 7260 30

V π π = =

Now we evaluate between 0 and 1. On the AP exam you are allowed to stop at the first line, but in this chapter I want you to practice your rules of manipulation and those horrid fractions that many of you dread!

That's it. Not too difficult, but you have to remember what the orientation of the rectangle is which variable you need to be setting up your integral in terms of. I have more of a problem visualizing these shells and the stacked surface areas than I do with the disks and washers, but if I just concentrate on the process and don't worry about my mind's picture I can do these without too much problem. So don't worry if your 3-D picture is harder to come up with - you can still do it. On the test I will not generally require you to use any one particular method when you have to find the volume formed by a revolution about an axis and the same is true about the HW problems. Occasionally you will be told which method to use, so make sure to read the directions.


1

The Shell Method – Examples D-G Here are some more examples. D: We are told to use the shell method to find the volume of revolution. We are rotating the curve y = √x about the y-axis. The graph is below and the rectangle vertical, parallel to the axis of rotation.

The radius is the distance from the axis of rotation to the rectangle, the x-value at the rectangle: r = x The height of the rectangle is the height of the curve: h = √x The limits of integration are the values along the x-axis from x = 0 to 4.

Now, to set up the integral: 2b

a

V r h dxπ= ∫

4

0

2V x x dxπ= ∫ This is our integral with values put in for r and h.

43/2

0

2V x dxπ= ∫ Simplify this by moving the constants out front and multiplying the variables.

445/2 5/2

00

2 425 5

V x xπ π = =


[ ]4 12832 05 5

V π π= − =

Evaluate between 0 and 4.


2

E: We are told to use the shell method to find the volume of revolution. We are rotating a tear-drop shaped region bounded by the curve y = x

2 and the curve y = 4x - x

2 about the y-

axis. The region stretches up diagonally from the origin, between the two curves. The rectangle is vertical, parallel to the axis of rotation.

y = 4x-x2

y = x2

(2, 4)

y = 4x-x2

y = x2

y = 4x-x2

y = x2

(2, 4)

The radius is the distance from the axis of rotation to the rectangle. r = x The height of the rectangle is the height of the top curve minus the height of the bottom curve: h = [ 4x - x

2 ] - [ x

2 ] = 4x - 2 x

2

To find the limits of integration we need to know the point of intersection of the two curves (you can use the calculator to find these if you want to): 4x - x

2 = x

2

2 x2 - 4x = 2x (x - 2) = 0

x = 0 and 2 Now, to set up the integral:

( )2

2

0

2 4 2V x x x dxπ= −∫ This is our integral with values put in for r and h.

( )2

2 3

0

2 4 2V x x dxπ= −∫ Simplify this by moving the constants out front and multiplying the variables.

23 4

0

4 123 2

V x xπ = −

We can integrate easily.

32 162 8 03 3

V π π = − − =


F: We are again told to use the shell method to find the volume of revolution. We are rotating a triangular shaped region bounded by the curve y = 4x - x

2, the y-axis, and the

horizontal line y = 4 about the y-axis. The region lies long the y-axis has "vertices" at the origin and vertex of the parabola, (2, 4). The rectangle is vertical, parallel to the axis of rotation.


3

4

4x-x2

The distance from the axis of rotation to the rectangle is simply the x-value at the rectangle. r = x The height of the rectangle is the height of the horizontal line minus the height of the curve. h = 4 – 4x – x2 The limits of integration are the values along the x-axis from x = 0 to 2. Now, to set up the integral:

( )2

2

0

2 4 4V x x x dxπ= − −∫ This is our integral with values put in for r and h.

( )2

2 3

0

2 4 4V x x x dxπ= − −∫ Simplify this by moving the constants out front and multiplying the variables.

22 3 4

0

4 12 23 4

V x x xπ = − −


32 82 8 4 03 3

V π π = − − − =



4

G: This is an ugly problem that cannot be done with shells if we want to use a single integral because of the shape of the region. I would much rather do this one with disks (In fact, for a point of quick EC, do this problem with disks and send it to me in a Message with the heading “shells example G”). However, we are asked to use shells, so I will do it so you can see how careful you have to be about what the region looks like when you set up these problems. We are going to rotate the trapezoidal region bounded by the curve y = 1/x , the x-axis, and the vertical lines x = 1 and x = 2 about the x-axis.

The rectangle is horizontal, parallel to the axis of rotation. Look carefully at this region – I have split it into two parts with a dotted line. If you place the rectangle in the bottom of the region, it is first bounded on each end by the vertical lines and its length is a constant distance of 2 - 1 = 1. However, if you move it up into the top part of the region (above the dotted line), the rectangle is now bounded on the left by the vertical line x = 1 but on the right by the curve. Since the boundaries of the rectangle change, we have to have two regions, two integrals! In the bottom sub-region, the radius is the distance from the axis of rotation to the rectangle, which is just y. The height of the rectangle is the difference between the two vertical lines: 2 - 1 = 1. The limits of integration will go from y = 0 to y = 1/2. In the top sub-region, the radius is again the distance from the axis of rotation to the rectangle, which is just y. The height of the rectangle is the difference between the vertical line x = 1 and the curve. The curve is y = 1/x, so in terms of y, we have x = 1/y and h = 1/y – 1. Our limits of integration will be along the axis which is perpendicular to the axis of rotation, the y-axis (remember that everything in this problem has to be in terms of y), so the limits of integration will be y = 1/2 to 1. Now we are ready to write out our integral and solve.


5

( )1/2 1

0 1/2

12 1 2 1V y dy y dyy

π π

= + −

∫ ∫ We have two integrals for this problem.

( )1/2 1

0 1/2

2 2 1V y dy y dyπ π= + −∫ ∫ Simplify this by moving the constants out front and multiplying through by the y.

1/2 12 2

0 1/2

1 12 22 2

V y y yπ π = + −

We can now integrate easily, using the power rule.

1 1 1 10 2 14 2 2 8

V π π = − + − − −

1 1 3 1 1 124 2 8 4 4 2

V π π π π π = + − = + =

Now we evaluate between 0 and 2. Be careful when you evaluate the second integral! You will not get 0 at the lower limit of integration!


1

The Shell Method – Examples H-J Here are a few more examples. H: We are told to use the shell method to find the volume of revolution. We are rotating the domed region bounded by the curve y = 4x - x

2 and the x-axis about the vertical line x = 5.

It has bottom vertices at (0,0) and (4, 0). The rectangle is vertical, parallel to the axis of rotation. This is one example of a problem with cannot easily be done with any other method; shells is the preferred way to find the volume. The distance from the axis of rotation to the rectangle is the difference between the vertical line, 5, and the x-value at the rectangle. r = 5 – x The height of the rectangle is the height of the curve. h = 4x - x

2

The limits of integration are the values along the x-axis from x = 0 to 4. Now, to set up the integral:

( )( )4

2

0

2 5 4V x x x dxπ= − −∫ This is our integral with values put in for r and h.

( )4

2 3

0

2 20 9V x x x dxπ= − +∫ Simplify this by moving the constants out front and multiplying the binomials using FOIL.

42 3 4

0

12 10 34

V x x xπ = − +


( )2 160 192 64 0 64V π π= − + − = Now we evaluate between 0 and 4.


2

I: We are going to do this problem three times, each time with a different axis of rotation. We are going to rotate the triangular region bounded by the curve y = x

3 , the x-axis, and the

vertical line x = 2. a) First we rotate about the x-axis. Since the region lies along the x-axis, the disk method will be the easiest to use, although you can use shells (the integral would be

( )8

3

0

2 2V y y dyπ= −∫ ).

The rectangle is vertical, perpendicular to the axis of rotation. The radius is the distance from the axis of rotation to the top of the rectangle, which is just the height of the curve, in terms of x, R = x

3.

The limits of integration will go from x = 0 to x = 2. Now we are ready to write out our integral and solve.

( )2

23

0

V x dxπ= ∫ This is the integral.

26

0

1287

V x dxπ π= =∫ Simplify this by moving the constants out front and squaring the x3. I am not going to go through the process of integration; I think you can do it by yourself.


3

b) This time we rotate about the y-axis, so we can use either shells or the washer method.

Shells is probably easier here. and the integr would be 2

3

0

2V x x dxπ= ∫ .

For shells the rectangle is parallel to the axis of rotation. Since the rectangle intersects the x-axis, everything is in terms of x. The radius stretches from the axis of rotation to the rectangle: r = x The height is the from the x-axis to the curve: h = y = x3 The limits of integration are from x = 0 to x = 2

23

0

2V x x dxπ= ∫ This is the integral.

24

0

6425

V x dxπ π= =∫ Simplify. I am not going to go through the process of integration; I think you can do it by yourself.

But we could have used the washer method as well:

For washers, the rectangle will be horizontal. That means we will be using the variable y for everything. The outer radius is measured from the axis of rotation to the far end of the rectangle, at the vertical line, 2. R = 2


4

The inner radius is measured from the axis of rotation to the near end of the rectangle, at the curve, y = x3, so 1/33x y r y= = = The limits of integration will be from y = 0 to y = 8. Now we are ready to write out our integral and solve.

( ) ( )8

22 1/3

0

2V y dyπ = − ∫ This is the integral.

82/3

0

6445

V y dyπ π = − = ∫

Simplify. I am not going to go through the process of integration; I think you can do it by yourself.

c) This time we rotate about the vertical line x = 4, so again we can use either shells or the washer method. I think the shell method is probably easier here.

For shells, the rectangle will be vertical and everything will be in terms of x. The radius will be the difference between the distance to the axis of rotation and the x-value of the rectangle: 4 - x. The height of the rectangle is the height of the curve, y = h = x3. The limits of integration will be from x = 0 to x = 2. Now we are ready to write out our integral and solve.

( )2

3

0

2 4V x x dxπ= −∫ This is the integral.

( )2

3 5

0

962 45

V x x dxπ π= − =∫ Simplify. I am not going to go through the process of integration.

J: On this one we are going to graph the region, set up the integral, and then use the calculator to evaluate and find the volume.


5

The region that we are rotating is a half moon shape bounded on the top by the curve

( ) ( )2 23 2 6y x x= − − and the bottom by the x-axis. We are revolving about the y-axis, so

shells is the only practical way to do this problem.

For shells, the rectangle will be vertical. The radius will be the distance from the axis of rotation to the rectangle, x. r = x

The height of the rectangle is the height of the curve. ( ) ( )2 23 2 6h x x= − −

The limits of integration will be from x = 2 to x = 6. Now we are ready to write out our integral and solve.

( ) ( )6

2 23

2

2 2 6 187.249V x x x dxπ= − − ≈∫ This is the integral and the answer. The AP exam requires that you give your answers to three decimal places.

Homework Examples with Cross Sections - #1 #1: Find a formula for the area of the cross sections of the solid that are perpendicular to the x-axis. For this problem we do not have to integrate and actually find the volume, but if we had to, we would integrate that area formula between x = -1 and 1. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross sections perpendicular to the x-axis between these planes run from the semicircle

21y x= − − to the semicircle 21y x= − . a) Here my cross sections are circular disks with diameters in the xy-plane.

21r x= − The radius of the each circle is the distance from the origin out to one of the semicircles. Or if you prefer, you can think of it as half the distance between the two curves we were given.

( )22 21A r xπ π= = −

( )21A xπ= −

Find the area of the circular cross-section.

b) Here my cross sections are squares with bases in the xy-plane.

22 1E x= − The edge of the square is double the distance from the x-axis to the outside of the circle.

( )22 22 1A E x= = −

( )24 1A x= −


c) Here my cross sections are squares with diagonals in the xy-plane. (The length of a square's diagonal is 2 times the length of its sides.)

This one is a bit tougher because the diagonal is 2 times the length of its sides. Look at the equation 2d E= . Thus to find the length of the edge, which we need to find area, we will have to divide the diagonal by 2 .

22 1d x= − 22 1

2xE −

=

The diagonal is the distance between the two curves. The edge of the square is equal to the length of the diagonal divided by 2 .

22

2 2 12

xA E −

= =

( )22 1A x= −


d) Here my cross sections are equilateral triangles with bases is in the xy-plane.

22 1b x= −

23 1h x= −

The base of the triangle is the distance between the two curves.

Draw an equilateral triangle with all sides equal to the b we just found. Drop an altitude from the top vertex. You now have a 30-60-90 triangle, so you can find the length of the altitude. Remember that the altitude will bisect the base.

( )( )2 21 1 2 1 3 12 2

A bh x x= = − −

( )23 1A x= −


22 1 x−22 1 x−

21 x−

23 1 x−

Homework Examples with the Disk Method - #15, 27, 29 #15: Our region is drawn for us. It is bound on the top by the horizontal line y = 1, on

the left by the y-axis, and to the right by the curve tan4

x yπ =

.

The book has drawn in the rectangle, perpendicular to the axis of rotation.

tan4

R x yπ = =

y = 0 to y = 1

Radius = the length of the rectangle = x-value of the curve in terms of y because we are rotating about the y-axis and have to use dy. The limits of integration come from along the y-axis.

21

0

tan4

V y dyππ = ∫

12

0

tan4

V y dyππ = ∫

12

0

sec 14

V y dyππ = − ∫

Write the volume integral and simplify. Use a trig identity to rewrite the integrand.

1

0

4 tan4

V y yπππ

= −

4 tan 1 0

4V ππ

π = − −

4 1 4V π ππ

= − = −

Find the antiderivative and evaluate. Don't forget to use substitution for the angle: u = (π/4)y du = π/4 dy 4/ π du = dy

#27: This time we are rotating about the line 2y = , and so our rectangle is vertical. Our region is in the first quadrant, bound on the top by the curve 2y = , on the left by the y-axis, and below by the curve y = sec x tan x. The rectangle is drawn perpendicular to the axis of rotation, the horizontal line 2y = .

2 sec tanR x x= −

Radius = the length of the rectangle To find the length we take the distance up to the horizontal line and then subtract the distance below the curve.

Intersection: 2 sec tanx x=

0.7854x ≈

The limits of integration come from along the x-axis. To find the right one we have to set the equations equal to each other and solve for x. Use your calculator.

( )0.7854 2

0

2 sec tanV x x dxπ= −∫

2.301V ≈

Write the volume integral and evaluate with your calculator.

R

2y =

sec tany x x=

#29: Our region is enclosed by the curve 25x y= and the lines x = 0, y = -1, and y = 1. We will rotate about the y-axis, so the rectangle is drawn perpendicular to that axis of rotation.

25x y= y = -1 to y = 1 or we can double the region from y = 0 to 1


( )1 2

2

0

2 5V y dyπ= ∫

Write the volume integral and simplify. I decided to find the volume of the top region and double it.

14

0

10V y dyπ= ∫

15

0

1105

V yπ =

( )2 1 0 2V π π= − =


25x y=1

-1


1

Homework Examples with the Shell Method #21, 33, 41 #21: Find the volume of revolution. We are rotating the region bounded by the curve y = x and the line y = 1 about the x-axis. The graph is below.

I could use the washer method here, but this time I will use shells. That means that the rectangle will be horizontal, parallel to the axis of rotation. The radius is the distance from the axis of rotation to the rectangle, the y-value at the rectangle: r = y The height of the rectangle is the distance from the y-axis to the line y = x: h = y The limits of integration are the values along the y-axis from y = 0 to 1.


a

V r h dxπ= ∫

1

0

2 ( )( )V y y dyπ= ∫ This is our integral with values put in for r and h.

12

0

2V y dyπ= ∫ Simplify this by moving the constants out front and multiplying the variables.

13

0

123

V yπ =

1 22 03 3

V π π = − =


y = x 1

r

h


2

#33: Find the volume of revolution of the region generated by rotating the region bounded by the curve y = x

2 , the line x = 2, and the x-axis about the y-axis.

Again, I could have used a horizontal rectangle and washers, but I am going to use shells. The rectangle is vertical, parallel to the axis of rotation. The radius is the distance from the axis of rotation to the rectangle. r = x The height of the rectangle is the distance from the x-axis to the height of the curve h = y = x

2

To find the limits of integration come from along the x-axis.


a

V r h dxπ= ∫

2

2

0

2 ( )( )V x x dxπ= ∫

This is our integral with values put in for r and h.

23

0

2V x dxπ= ∫ Simplify this by moving the constants out front and multiplying the variables.

( )2

4

0

1 12 16 0 84 2

V xπ π π = = − =


2 r

h

2y x=


3

#41: Find the volume of revolution of the region generated by rotating the region bounded by the curve y x= , the line x = 4, and the x-axis about the y-axis.

I could have use washers here too, but I will use shells. The rectangle is vertical, parallel to the axis of rotation. The radius is the distance from the axis of rotation to the rectangle. r = x The height of the rectangle is the distance from the x-axis to the height of the curve h y x= = To find the limits of integration come from along the x-axis. Now, to set up the integral:

4

0

2 ( )( )V x x dxπ= ∫

This is our integral with values put in for r and h.

43/ 2

0

2V x dxπ= ∫

Simplify this by moving the constants out front and multiplying the variables.

45/ 2

0

2 12825 5

V xπ π = =


4

y x=

h

r


1

Homework Examples with the Washer Method #23, 25, 31 #23: Find the volume of the solid generated by revolving the region bounded by the curves y = x2 + 1 and y = x + 3 about the x-axis.

The region is not touching the axis of rotation so disks can't be used. Here I will use washers.

R = x + 3 2 1r x= +

Intersection (you can use your calculator to find these if you prefer): x2 + 1 = x + 3 x2 - x - 2 = (x - 2)(x + 1) = 0 x = -1, 2

R = distance from the x-axis to the top of the rectangle, which is touching the line r = distance from the x-axis to the bottom of the rectangle, which is touching the parabola The limits of integration come from the x-axis.

( ) ( )2

22 2

1

3 1V x x dxπ−

= + − + ∫

( ) ( )2

2 4 2

1

6 9 2 1V x x x x dxπ−

= + + − + + ∫

24 2

1

6 8V x x x dxπ−

= − − + + ∫

Write the volume integral and simplify.

25 3 2

1

1 1 3 85 3

V x x x xπ−

= − − + +

32 8 12 165 3

1 1 3 85 3

V π

− − + + = − + + −

117 23.4 73.513

5V π π= = ≈

Find the antiderivative and evaluate. You can give the answer in any of these forms.

R r


2

#25: Find the volume of the solid generated by revolving the region bounded by the curves y

= sec x , 2y = , and 4 4

xπ π− ≤ ≤ about the x-axis.

The region is not touching the axis of rotation so disks can't be used. Here I will use washers.

2R = r = sec x Limits of integration:

4x π= ±

R = distance from the x-axis to the top of the rectangle, which is touching the line r = distance from the x-axis to the bottom of the rectangle, which is touching the curve The limits of integration were given to us.

( ) ( )/ 4 2 2

/ 4

2 secV x dxπ

π

π−

= − ∫

/ 42

/ 4

2 secV x dxπ

π

π−

= − ∫

/42

0

2 2 secV x dxπ

π = − ∫

Write the volume integral and simplify. I am going to change the limits and use symmetry.

[ ] /4

02 2 tanV x x ππ= −

( )2 1 02

V ππ = − −

2 2 3.586V π π= − ≈

Find the antiderivative and evaluate. Either form of the answer is fine.

R r


3

#31: Find the volume of the solid generated by revolving the region enclosed by the triangle with vertices (1, 0), (2, 1), and (1, 1) about the y-axis.

There is a gap between the axis of rotation and the rectangle in the region, so we have to use washers. We also have to find the equation of the oblique line and the left vertical line in order to express our R and r. Everything will have to be in terms of y in order to use washers.

Outer radius is line between (2, 1) and (1, 0) m = 1 y - 0 = x - 1 x = y + 1 R = y + 1 Inner radius is vertical line: r = 1 Limits of integration: y = 0 to y = 1

R = distance from the y-axis to the far right side of the rectangle, which is touching the line r = distance from the r-axis to the vertical line The limits of integration come from the y-axis.

( ) ( )1

2 2

0

1 1V y dyπ = + − ∫

12

0

2V y y dyπ = + ∫

Write the volume integral and simplify. Don't forget to use FOIL when expanding the binomial squared.

13 2

0

13

V y yπ = +

( )1 41 0 4.18873 3

V π π = + − = ≈

Find the antiderivative and evaluate. Either form of the answer is fine.

(1, 1) (2, 1)

(1, 0) R r

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