M68 Volumes of Cones and Spheres - EkVolRev1

8/11/2019 M68 Volumes of Cones and Spheres - EkVolRev1

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Volumes of cones and spheres

For any object whose cross-sections are congruent, the volume

of the object is(area of cross section) × (height of object)

where the height is measured perpendicular to the plane of thecross section. For example:

Rectangular cross sectionsVolume = (length × width) × height

A cylinder is an object with a circular base—the cross sectionsare congruent circles.

1. Find the volume of a cylinder with radius r and height h.

2. A cone is an object whose cross sections are circles thatare not congruent. However, they are similar . Actually, all circles are similar.

Imagine approximating this cone with cylinders of equalheight. The radius of each cylinder is the radius of thecone’s cross section, at the height of the cylinder’s base:

Problems with a Point: May 8, 2002 c EDC 2002



Volumes of cones and spheres: Hints 1

Hints

Hint to problem 2b. To find the radius of a cylinder, con-sider that the cross section of a cone—through the vertex—is atriangle. Suppose you cut the top of the cone off, right wherethe bottom of the cylinder would be. How does its cross sectioncompare to the cross section of the whole cone? (You only needconsider half the cross section.)

Hint to problem 4. The cross section of the sphere is a circle.If you impose a coordinate grid, with the origin at the center of the sphere, you can write a simple equation for the circle. Now,consider the second cylinder from the bottom. The edge of thebottom base touches the circle. What are the coordinates wherethey touch? Which coordinate gives the radius of the cylinder?




Volumes of cones and spheres: Answers 1

Answers

1. πr2h.

2. (a) The height of each cylinder is h

10 .

(b) i. The radius of the top cylinder is r10 .ii. The radius of the cylinder is 2r

10, or r

5.

iii. The radius of cylinder i is ir

10 .(c) The total volume is 0.385πr2h.(d) i. The radius ith cylinder is ir

100 .ii. The total volume is 338350

1003 πr2h, or 0.338350πr2h.

(e) The total volume is πr2h (n+1)(2n+1)6n2 .

(f ) As n goes to infinity, the volume of the cylinders ap-proaches 1

3πr2h. See the solutions for an explanation

of why this is the volume of the cone.

3. The volume of the pyramid is 13s2h.

4. The volume of the sphere is 43πr3.




Volumes of cones and spheres: Solutions 1

Solutions

1. Since the area of the base is πr2, the volume is πr2h.

2. (a) The height of each cylinder is h

10 .

(b) i. To find the radius of the top cylinder, consider theright triangle formed by the altitude and radiusshown in the diagram. Connecting the two verticesthat aren’t already touching gives a line that goesthrough the bottom edge of each cylinder:

The triangle formed by the top cylinder is similarto the big triangle, so you can set up a proportion:

h

10

h =

rtop

r

So the radius of the top cylinder, rtop, is r

10 .ii. The triangle formed by the radius of the next cylin-

der and the portion of the altitude from that line

up to the top of the cone is also similar to the bigtriangle. Since the height this time is 2h

10, the radius

of the cylinder will be 2r10 , or r

5 .iii. For cylinder i, the corresponding small triangle will

have height ih

10 , so the radius of cylinder i is ir

10 .(c) The volume of the ith cylinder is π( ir

10)2( h

10), so the

volume of all 10 cylinders is π( r

10)2( h

10) + π(2r10)2( h10) +

· · · + π(10r10

)2( h

10), or π( r

2h

103 )10

i=1 i2. Since

10i=1 i

2 is16(10)(11)(21), the total volume becomes 3.85πr2h.

(d) i. Working as before, a triangle can be drawn similar

to the large triangle formed by the radius of the baseand the altitude. For the ith cylinder, the height of the triangle is ih

100 , so the base of the triangle (andtherefore the radius of the ith cylinder) is ir

100.

ii. The volume of the ith cylinder is π( ir

100)2( h

100), so the

total volume of all 100 is π( r

100)2( h

100)+π( 2r100)2( h

100)+

· · · + π(100r100

)2( h

100), or π( r

2h

1003 )100

i=1 i2. Since

100i=1 i

2





is 16(100)(101)(201), the total volume becomes

3383501003 πr

2h, or 0.338350πr2h.(e) For n cylinders, the radius of the ith cylinder would

be ri

n, and the height would be h

n. The volume of

the ith cylinder is π(ri

n )2

(h

n) and the total volume isn

i=1πr

2hi

2

n3 , or πr2h

n3

n

i=1 i2. Using the formula for the

sum gives πr2h

n3 (16

)n(n + 1)(2n + 1), or πr2h (n+1)(2n+1)6n2 .

(f) Look at the fraction in the last version of the formula

given above, (n+1)(2n+1)6n2 . (The rest doesn’t depend on

n, so we can ignore it for the moment.) By makingnumerical approximations for larger and larger n, astudent will probably conjecture (correctly) that thisapproaches 13 . Algebraically this can be found by firstexpanding the numerator and then dividing each term

by the denominator:(n + 1)(2n + 1)

6n2 =

2n2 + 3n + 1

6n2

= 1

3 +

1

2n +

1

6n2

As n goes to infinity, the last two terms go to 0, leavingonly 1

3 . The volume of the cylinders approaches 13πr2h.

By increasing the value of n, you are making more andmore divisions. The cylinders come closer to approx-imating the cone. When there are an infinite numberof cylinders, they exactly fit the cone, so the volumeof an infinite number of cylinders is the same as thevolume of the cone.

3. As with the cone, the first thing to do is find the dimensionsof each of the n sections—in this case, a rectangular prism.Similar triangles are formed: Note that the base is s

2, not s.

The side length of the ith prism is si

n, so the volume of the

ith prism iss2i2

n2

h

n

.





The volume of all n prisms is

ni=1

s2h

n3 i2 =

s2h

n3

ni=1

i2

= s2

hn3

( 16

)n(n + 1)(2n + 1)

= s2h

2n3 + 3n2 + n

6n3

Again, the fraction becomes 13 + 1

2n + 1

6n2 , which becomes 13

as n goes to infinity. The volume of the pyramid is 13s2h.

4. Divide the hemisphere into n cylinders, as you did with thecone. The radius of the ith cylinder is a little more difficultto calculate, here. A cross section of the hemisphere—

perpendicular to the base—will be a semicircle. Impose acoordinate grid with the center of the sphere at the origin:

The equation for the circle is x2 + y2 = r2. If the baseof the cylinder touches the circle at (x, y), the radius willbe x. The distance from the base (x-axis) to the bottom

of the cylinder is y , so it will probably be easier to numberthe cylinders from the bottom up, this time, and start with0 at the bottom. Then the ith cylinder touches the circleat (x, ir

n). From the equation for the circle, you can find

that x = r2 − ( ir

n)2, or r

n

√ n2 − i2.

Remember that the n cylinders are numbered starting at 0,so they go from 0 to n−1. The volume of the hemisphere,then, is Notice that since adding 02

doesn’t change the sum,

n−1

i=0 i

2

is the same as

n−1

i=1 i

2, so theusual formula—with n − 1 insteadof n—will work.

n−1

i=0

π

r

n

√ n2 − i2

2 r

n

=

n−1

i=0

π

r2

n2

(n2 − i2)

r

n

= πr3

n3

n−1i=0

(n2 − i2)

= πr3

n3

n2

n−1i=0

1 −n−1i=0

i2

= πr3

n3

n2(n) − 1

6(n− 1)n(2n− 1)





= πr3 − πr3

2n3 − 3n2 + n

6n3

= πr3 − πr3

1

3 − 1

2n +

1

6n2

As n goes to infinity, only 13πr3 is left in the second term,

so the volume of the cylinders becomes 23πr

3. This is thevolume of the hemisphere , so double it to get the volumeof the sphere: 43πr

3.


M68 Volumes of Cones and Spheres - EkVolRev1

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