6.2 Volumes

The idea behind computing volumes is the same as that behind computing area.

Area Approximate a 2-dimensional region by rectangles and sum their areas.

Volume Approximate a 3-dimensional region by cylinders1 and sum their volumes.

Suppose that we take the purple solid pictured. We imagine slicingthrough the solid perpendicular to the x-axis at a distance x∗ alongsaid axis. Suppose that we are able to compute the area A(x∗) ofthe resulting cross-sectional slice.The green cylinder of width ∆x has

Volume = A(x∗)∆x

and approximates part of the original solid. We should thereforebe able to approximate the original volume using a Riemann sumof the volumes of many thin cylinders.

Definition. Suppose that a solid region has cross-sectional area function A(x) whenever a ≤ x ≤ b.Then its volume is the limit of a Riemann sum

V = limn→∞

n

∑i=1

A(x∗i )∆x =∫ b

aA(x)dx

Volumes of Revolution Most of our examples will be columes obtained by rotating a curve y =f (x) around the x-axis for x in some interval [a, b]. It follows that the radius of rotation is precisely thevalue of the function. The pictures should convince you that the cross-sectional area function is

A(x) = πr2 = πy2 = π[

f (x)]2

y

xa x

f (x)

b

y = f (x)

whence the volume is

V =∫ b

aA(x)dx = π

∫ ba( f (x))2 dx

1A cylinder does not have to be round! It is merely a solid formed by taking a curve and moving it in some direction.Therefore a cube could be described as as square-based cylinder!

1

The motivating example involved rotating the curve y = 4x − x2 around the x-axis between x = 0and x = 3. Below are three approximations with 4, 10 and 30 cylinders.

V ≈ 30.9089π V ≈ 30.6457π V ≈ 30.6050πSince A(x) = πy2 = π(4x− x2)2, the exact volume is

V =∫ 3

0A(x)dx =

∫ 30

π(4x− x2

)2dx = π

∫ 30

x4 − 8x3 + 16x2 dx

= π

(15· 35 − 8

4· 34 + 16

3· 33)=

1535

π = 30.6π

Example Find the volume enclosed when the curve y = ex is rotated around the x-axis from x = −1to x = 1.

The cross-sectional area is

A(x) = πy2 = π(ex)2 = πe2x

The volume of revolution is therefore

V = π∫ 1−1

e2x dx =π

2e2x∣∣∣1−1

=π

2(e2 − e−2) ≈ 11.394

−2

−1

1

2

y

−1 1x

x

ex

2

Volume of a right-circular cone We can recover several famous expressions for the volumes ofsolids using this approach, For example, a cone of height h and base radius r can be formed byrotating the line

y =rh

x

around the x-axis for 0 ≤ x ≤ h. Its volume is therefore

V = π∫ h

0

r2

h2x2 dx =

πr2

h2· 1

3x3∣∣∣h

x=0=

13

πr2h

y

xx

rh x

h

r

Volume of a sphere A sphere of radius r can be formed by rotating the curve

y =√

r2 − x2

around the x-axis for −r ≤ x ≤ r. Its volume is therefore

V = π∫ r−r

r2 − x2 dx = 2π∫ r

0r2 − x2 dx = 2π

(r2x− 1

3x3) ∣∣∣∣r

x=0

= 2π(

r3 − 13

r3)=

43

πr3

y

xx

√r2 − x2

r−r

3

Volumes with annular cross-sections

More complicated examples involve rotating the region between two curves f (x) ≥ g(x) around thex-axis: the cross-section is an annulus (washer) of area

A(x) = π(outer radius)2 − π(inner radius)2

= π(r2out − r2in) = π( f (x)2 − g(x)2)

The volume is therefore

V = π∫ b

af (x)2 − g(x)2 dx

y

xx

f (x)− g(x)

g(x)

a b

y = f (x)

y = g(x)

Example Find the volume of the solid given by rotating the region between y = x and y =√

xaround the x-axis

Since√

x ≥ x for 0 ≤ x ≤ 1 we have

V = π∫ 1

0(√

x)2 − x2 dx = π∫ 1

0x− x2 dx = π

(12− 1

3

)=

π

6

y

xx

√x− x

x

1

y =√

x y = x

4

Revolutions around other axes For even greater complexity, we can revolve curves around the y-axis, or a completely different line: in each case you should find the radius (or radii) of revolution andapply one of the methods above.

Example Find the volume of the solid obtained by rotatingthe curve x = 2− y2 around the line x = 1 between y = ±1.The two curves meet when

y2 = 2− 1 = 1 =⇒ y = ±1

The radius of revolution is marked on the picture

r = 2− y2 − 1 = 1− y2 −1

0

1y

1 2x

ry

x = 2 − y2

and so

V =∫ 1−1

πr2 dy = π∫ 1−1(1− y2)2 dy = 2π

∫ 10

1− 2y2 + y4 dy = 1615

π

More general volumes We can use this construction to compute the volume of any object, providedwe know its cross-sectional areas. For example, the region below has square horizontal cross-sectionsof side-length

x = 1− 14

y2 for 0 ≤ y ≤ 2

Its volume is

V =∫ 2

0A(y)dy =

∫ 20

(1− 1

4y2)2

dy

=∫ 2

01− 1

2y2 +

116

y4 dy

= y− 16

y3 +180

y5∣∣∣∣20=

1615

Suggested problems

1. Let D be the region enclosed by the curves y = x2, x = 1, and y = 0.

(a) Rotate D around the x-axis. What is the resulting volume?

(b) Rotate D around the y-axis. What is the volume now?

2. Rotate the region bounded by the curves y = 4− x2 and y = 5− 2x2 around the line y = 2.What is the volume?

3. A torus (doughnut) is obtained by rotating the region within the circle (x−R)2 + y2 = r2 aroundthe y-axis. Assuming r ≤ R, find the volume of the torus.

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Volumes

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