Calculus with Algebra and Trigonometry II Lecture 16 ...frooney/M217_17_Volb.pdf · Calculus with...

Calculus with Algebra and Trigonometry IILecture 16

Volume of solids of revolution: shells

Mar 19, 2015

Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 1 / 19

Finding volumes using shells

To find the volume generated when the area between the graph of afuncion y = f (x) is rotated about the y axis.

A representative slice of the area when rotated about the x axis generatesa cylindrical shell with a height equal to the length of the slice and aradius equal to the distance to the axis of rotation.

The volume is the sum of all these cylinders

Volume =

∫ b

a2π(radius)(height) dx =

∫ b

a2πx(f (x)) dx


Example 1

Find the volume generated when the region, bounded byy = x(x − 2)(x − 4); 0 ≤ x ≤ 2 and the x axis, is rotated about the y axis

Use the shell method. The radius is x and the height is x(x − 2)(x − 4)

Volume = 2π

∫ 2

0x2(x2 − 6x + 8) dx = 2π

[x5

5− 3x4

2+

8x3

3

]20

=112π

15


Example 2

Find the volume generated when the region bounded by y = 1√1+x2

, x = 1

and the coordinate axes is rotated about the y axis.

Again use shells. The radius is x and the height is 1√1+x2

.

Volume = 2π

∫ 1

0

x√1 + x2

dx


To evaluate the integral use u substitution. Let

u = 1 + x2 ⇒ du = 2x dx ⇒ dx =du

2x

x = 0 ⇒ u = 1 x = 1 ⇒ u = 2

Volume = 2π

∫ 1

0

x√1 + x2

dx

= 2π

∫ 2

1

x√u

(du

2x

)= 2π

∫ 2

1

1

2u−1/2 du

= 2π[u1/2

]21

= 2π(√

2− 1)


Example 3

Find the volume generated when the region bounded byy = sin x ; 0 ≤ x ≤ π and the x axis is rotated about the y axis.

Again use shells. The radius is x and the height is sin x .

Volume = 2π

∫ π

0x sin x dx


To evaluate the integral use integration by parts. Let

f (x) = x g ′(x) = sin x ⇒ f ′(x) = 0 g(x) = − cos x

Volume = 2π

∫ π

0x sin x dx

= 2π

([x(− cos x)]π0 −

∫ π

0(− cos x) dx

)= 2π

(π +

∫ π

0cos x dx

)= 2π (π + [sin x ]π0 ) = 2π2


Example 4

Find the volume generated when the region bounded by y = x2 andy = x + 2 rotated about the line x = −2.

First the intersection points

x2 = x + 2 ⇒ x2 − x − 2 = 0 ⇒ x = 2,−1

The radius is x + 2 and the height is x + 2− x2.Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 8 / 19

The volume is given by

Volume = 2π

∫ 2

−1(x + 2)(x + 2− x2) dx

= 2π

∫ 2

−1(4 + 4x − x2 − x3) dx

= 2π

[4x + 2x2 − x3

3− x4

4

]2−1

= 2π

((8 + 8− 8

3− 4

)−(−4 + 2 +

8

3− 1

4

))=

107π

6


Example 5

Find the volume generated when the region bounded by y = x ln x , x = e,and the x axis, is rotated about the line x = 1.

Use shells: radius = x − 1 and the height = x ln x


Volume = 2π

∫ e

1(x − 1)x ln x dx

Use integration by parts

f (x) = ln x g ′(x) = x(x − 1) ⇒ f ′(x) =1

xg(x) =

x3

3− x2

2

Volume = 2π

∫ e

1(x − 1)x ln x dx

=

[(x3

3− x2

2

)ln x

]e1

−∫ e

1

(x3

3− x2

2

)1

xdx

=e3

3− e2

2−[x3

9− x2

4

]e1

=2e3

9− e2

4− 5

36


Example 6

Find the volume generated when the region between the parabolsx = 3y2 − 2, x = y2 and the x axis, is rotated about the x axis


Slice horizontally and use shells. The radius = y and the height = 2− 2y2.

To find the upper limit for the y integration we need to find theintersection point

3y2 − 2 = y2 ⇒ y = 1

Volume = 2π

∫ 1

0y(2− 2y2) dy

= 2π

[y2 − y2

2

]10

= π


Example 7

Find the volume of a torus. A torus is obtained when the circle(x − R)2 + y2 = r2 is rotated about the y axis

Using shells the radius i= x , the height = 2√r2 − (x − R)2, and

R − r ≤ x ≤ R + r .

Volume = 2π

∫ R+r

R−rx(2√r2 − (x − R)2) dx


Firstly let u = x − R then

Volume = 4π

∫ r

−r(u + R)

√r2 − u2 du

= 4π

(∫ r

−ru√

r2 − u2 du +

∫ r

−rR√r2 − u2 du

)The first integral is zero. Let u = −v then∫ r

−ru√r2 − u2 du =

∫ −rr

(−v)√r2 − v2 (−dv) =

∫ −rr

v√

r2 − v2 dv

and ∫ r

−r

√r2 − u2 du = area of a semicicle of radius r =

πr2

2

thusVolume = 2πRr2


Example 8

Find the volume remaining when a cylindrical hole with radius 3 is drilledthrough a sphere of radius 5.

The volume will be the same as rotating the reginn between the line x = 3and the circle x2 + y2 = 25 about the y axis.


Using the shell method radius = x and height = 2√

25− x2

Volume = 2π

∫ 5

32x√

25− x2 dx

Use u substitution. Let

u = 25− x2 ⇒ du = −2x dx ⇒ dx = −du

2x

x = 3 ⇒ u = 16 x = 5 ⇒ u = 0

Volume = 2π

∫ 0

162x√u

(−du

2x

)= 2π

∫ 16

0

√u du

= 2π

[2

3u3/2

]160

=256π

3


Volumes with the cross sections as a function of x

Suppose we have a body whose cross section at x is a known function of x,A(x) then its volume will be

Volume =

∫ b

aA(x) dx

For example to find the volume of a pyramid with height, h, and its base asquare of side length b.

A vertical section though the pyramid will be an isosceles triangle withheight h and base b. We can represent it as the region between the twolines y = ± b

2hx with 0 ≤ x ≤ h. So the cross section at position x is a

square of side length bxh .


The volume is then

Volume =

∫ h

0

(bx

h

)2

dx

=b2

h2

∫ h

0x2 dx =

b2

h2

[x3

3

]h0

=1

3b2h


Calculus with Algebra and Trigonometry II Lecture 16 ...frooney/M217_17_Volb.pdf · Calculus with...

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