Volume: The Disk Method

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We have seen that integration is used to compute the area of two-dimensional regions bounded by curves. Integrals are also used to find the volume of three-dimensional regions (or solids). Once again, the slice-and-sum method is the key to solving these problems.

General Slicing Method

Consider a solid object that extends in the x-direction from x = a to x = b. Imagine cutting through the solid, perpendicular to the x-axis at a particular point x, and suppose the area of the cross section created by the cut is given by a known integrable function A.

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To find the volume of this solid, we first divide [a, b] into n subintervals of length Δx = (b – a)/n. The endpoints of the subintervals are the grid points x0 = a, x1, x2, … , xn = b. We now make cuts through the solid perpendicular to the x-axis at each grid point, which produces n slices of thickness Δx.

Imagine cutting a loaf of bread to create n slices of equal width.

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On each subinterval, an arbitrary point xk* is identified. The kth slice through the

solid has a Δx, and we take A(xk*) as a representative cross-sectional area of the slice.

Therefore, the volume of the kth slice is approximately A(xk*)Δx.

Summing the volumes of the slices, the approximate volume of the solid is:

V A xk* x.

k1

n

4

As the number of slices increases (n ∞) and the thickness of each slice goes to zero (Δx 0), the exact volume V is obtained in terms of a definite integral:

V limn

A xk* x A(x)dx.

a

b

k1

n

5

General Slicing Method: Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is

V A(x)dx.a

b

Notes: A(x) is the cross-sectional area of a slice and dx represents its thickness. Summing (integrating) the volumes of the slices A(x) dx gives the volume of the solid.

Example 1: Volume of a “parabolic hemisphere”A solid has a base that is bounded by the curves y = x2 and y = 2 – x2 in the xy-plane. Cross sections through the solid perpendicular to the x-axis are semicircular disks. Find the volume of the solid.

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Since a typical cross section perpendicular to the x-axis is a semicircular disk, the area of a cross section is ½ πr2, where r is the radius of the cross section. The key observation is that this radius is one-half of the distance between the upper bounding curve y = 2 – x2 and the lower bounding curve y = x2. So the radius at the point x is

r 1

22 x 2 x 2 1 x 2.

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This means the area of the semicircular cross section at the point x is

A(x) 1

2r2

2

1 x 2 2.

The intersection points of the two bounding curves satisfy 2 – x2 = x2, which has solutions x = ± 1. Therefore, the cross sections lie between x = -1 and x = 1. Integrating the cross-sectional areas, the volume of the solid is

V A(x)dx 1

1

€

=2

1 − x 2( )

2dx

−1

1

∫

2

1 2x 2 x 4 dx 1

1

815

.8

The Disk Method

We now consider a specific type of solid known as a solid of revolution. Suppose f is a continuous function with f(x) ≥ 0 on an interval [a, b]. Let R be the region bounded by the graph of f, the x-axis, and the lines x = a and x = b.

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Now revolve R around the x-axis. As R revolves once around the x-axis, it sweeps out a three-dimensional solid of revolution.

The goal is to find the volume of the solid, and it may be done using the general slicing method.

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With a solid of revolution, the cross-sectional area function has a special form since all cross sections perpendicular to the x-axis are circular disks with radius f(x).

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Therefore, the cross section at the point x, where a ≤ x ≤ b, has area

A(x) (radius)2 f (x)2 .

By the general slicing method, the volume of the solid is

V A(x)dx f (x)2dx.a

b

a

b

Since each slice through the solid is a circular disk, the resulting method is called the disk method.

Disk Method About the x-Axis: Let f be continuous with f(x) ≥ 0 on the interval [a, b]. If the region R bounded by the graph of f, the x-axis, and the lines x = a and x = b is revolved about the x-axis, the volume of the resulting solid of revolution is

V f (x)2dx.a

b

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Example 2: Let R be the region bounded by the curve f(x) = (x + 1)2, the x-axis, and the lines x = 0 and x = 2. Find the volume of the solid of revolution obtained by revolving R about the x-axis.

When the region R is revolved about the x-axis, it generates a solid of revolution. A cross section perpendicular to the x-axis at the point 0 ≤ x ≤ 2 is a circular disk of radius f(x). Therefore, a typical cross section has area

A(x) f (x)2 x 1 2 2

.

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Integrating these cross-sectional areas between x = 0 and x = 2 gives the volume of the solid

V A(x)dx x 1 2 2

dx0

2

0

2

x 1 4dx

0

2

u5

51

3

242

5.

Substitute for A(x).

Simplify.

Let u = x + 1 and evaluate.

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Washer Method: A slight variation on the disk method enables us to compute the volume of more exotic solids of revolution. Suppose R is the region bounded by the graphs of f and g between x = a and x = b, where f(x) ≥ g(x) ≥ 0.

If R is revolved about the x-axis to generate a solid of revolution, the resulting solid generally has a hole through it.

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Once again we apply the general slicing method. In this case, a cross section through the solid perpendicular to the x-axis is a circular washer with an outer radius of R = f(x) and a hole with a radius of r = g(x), where a ≤ x ≤ b. The area of the cross section is the area of the entire disk minus the area of the hole, or

A(x) R2 r2 f (x)2 g(x)2 The general slicing method gives the area of the solid.

Washer Method About the x-Axis: Let f and g be continuous functions with f(x) ≥ g(x), and the lines x = a and x = b. When R is revolved about the x-axis, the volume of the resulting solid of revolution is

V f (x)2 g(x)2 dx.a

b

16

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Example 3: The region R is bounded by the graphs of f(x) = √x and g(x) = x2 between x = 0 and x = 1. What is the volume of the solid resulting when R is revolved about the x-axis?

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The region R is bounded by the graphs of f and g with f(x) ≥ g(x) on [0, 1], so the washer method is applicable. The area of a typical cross section at the point x is

€

A(x) = π f (x)2 − g(x)2( ) = π x( )

2

− x 2( )

2 ⎛ ⎝ ⎜ ⎞

⎠ ⎟= π x − x 4

( ).

Therefore, the volume of the solid is

V x x 4 dx0

1

x 2

2x 5

5

0

1

310

.

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Revolving About the y-Axis:

Everything you learned about revolving regions about the x-axis applies to revolving regions about the y-axis. Consider a region R bounded by the curve x = p(y) on the right, the curve x = q(y) on the left, and the horizontal lines y = c and y = d.

To find the volume of the solid generated when R is revolved about the y-axis, we use the general slicing method - now with respect to the y-axis. The area of a typical cross section is A(y) = (p(y)2 - q(y)2), where c ≤ x ≤ d. As before, integrating these cross-sectional areas of the solid gives the volume.

Disk and Washer Methods About the y-Axis

Let p and q be continuous functions with p(y) ≥ q(y) ≥ 0 on [c, d]. Let R be the region bounded by x = p(y), x = q(y), and the lines y = c and y = d. When R is revolved about the y-axis, the volume of the resulting solid of revolution is given by

V p(y)2 q(y)2 dy.c

d

Note: If q(y) = 0, the disk method results:

V p(y)2dy.c

d

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Example 4: Let R be the region in the first quadrant bounded by the graphs of x = y3 and x = 4y. Which is greater, the volume of the solid generated when R is revolved about the x-axis or the y-axis?

Solving y3 = 4y - or, equivalently, y(y2 - 4) = 0 - we find that the bounding curves of R intersect at the points (0, 0) and (8, 2). When the region R is revolved about the y-axis, it generates a funnel with a curved inner surface.

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Washer-shaped sections perpendicular to the y-axis extend from y = 0 to y = 2. The outer radius of the cross section at the point y is determined by the line x = p(y) = 4y. The inner radius of the cross section at the point y is determined by the curve x = q(y) = y3. Applying the washer method, the volume of this solid is

V p(y)2 q(y)2 dy0

2

16y 2 y 6 dy0

2

16

3y 3

y 7

7

0

2

512

2176.60.

Washer method

Substitute for p and q.

Fundamental Theorem

Evaluate.

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When the region R is revolved about the x-axis, it generates a funnel.

Vertical slices through the solid between x = 0 and x = 8 produce washers. The outer radius of the washer at the point x is determined by the curve x = y3, or y = f(x) = x1/3. The inner radius is determined by x = 4y, or y = g(x) = x/4. The volume of the resulting solid is

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V f (x)2 g(x)2 dx0

8

x2

3 x 2

16

dx

0

8

3

5x

53 x 3

48

0

8

128

1526.81.

Washer method

Substitute for f and g.

Fundamental Theorem

Evaluate.

Example 5: Find the volume V of the “wedding band” obtained by rotating the region between the graphs of f(x) = x2 + 2 and g(x) = 4 - x2 about the horizontal line y = -3.

Points of intersection: f(x) = g(x) x2 + 2 = 4 - x2 x2 = 1 x = ±1

The figure A shows that g(x) ≥ f(x) for -1 ≤ x ≤ 1.

Since we want to revolve around y = -3, we must determine how the radii are affected. Figure B shows when we rotate about y = -3, the line segment AB generates a washer whose outer and inner radii are both 3 units larger:

Router = g(x) - (-3) = (4 - x2) + 3 = 7 - x2

Rinner = f(x) - (-3) = (x2 + 2) + 3 = x2 + 5

The volume of revolution is equal to the integral of the area of this washer:

V (about y 3) Router2 Rinner

2 dx g(x) 3 2 f (x) 3 2 dx 1

1

1

1

We obtain Router by subtracting y = -3 from y = g(x) because vertical distance is the difference of the y-coordinates. Similarly, we subtract -3 from f(x) to obtain Rinner.

7 x 2 2 x 2 5 2 dx

1

1

49 14x 2 x 4 x 4 10x 2 25 1

1

dx

24 24x 2 dx 24x 8x 3 1

1

1

1

32

Example 6: Find the volume obtained by rotating the graphs of f(x) = 9 - x2 and y = 12 for 0 ≤ x ≤ 3 about the line y = 12 and then, y = 15.

Are the cross sections disks or washers???

Figure B shows the line segment AB rotated about y = 12 generates a disk of radius

R = length of line segment AB = 12 - f(x) = 12 - (9 - x2) = 3 + x2

Note: The length of line segment AB is 12 - f(x) rather than f(x) - 12 since the line y = 12 lies above the graph of f(x).

The volume when we rotate about y = 12 is

V R2dx 3 x 2 2dx 9 6x 2 x 4 dx

0

3

0

3

0

3

9x 2x 3 1

5x 5

0

3

648

5

Figure C shows that line segment AB rotated about y = 15 generates a washer. The outer radius of this washer is the distance from B to the line y = 15:

Router = 15 - f(x) = 15 - (9 - x2) = 6 + x2

The inner radius is Rinner = 3, so the volume of revolution about y = 15 is

V Router2 Rinner

2 dx 6 x 2 2 32 dx

0

3

0

3

36 12x 2 x 4 9 0

3

dx

27x 4x 3 1

5x 5

0

3

1,188

5

Example 7: Find the volume of the solid obtained by rotating the region under the graph of f(x) = 9 - x2 for 0 ≤ x ≤ 3 about the vertical axis x = -2.

The figure shows that the line segment AB sweeps out a horizontal washer when rotated about the vertical line x = -2. We are going to integrate with respect to y, so we need the inner and outer radii of this washer as functions of y. Solving for x in y = 9 - x2, we obtain

x 2 9 y

x 9 y since x 0

Therefore,

Router 9 y 2, Rinner 2

Router2 Rinner

2 9 y 2 2 22 9 y 4 9 y 4 4 9 y 4 9 y

The region extends from y = 0 to y = 9 along the y-axis, so

V Router2 Rinner

2 0

9

dy 9 y 4 9 y dy0

9

9y 1

2y 2

8

39 y

32

0

9

225

2