6.2 Volumesndonalds/math2b/notes/6-2.pdf · 6.2 Volumes The idea behind computing volumes is the same as that behind computing area. ... Volumes of Revolution Most of our examples

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.

5

Volumes

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