11.Prove that if f and g are continuous on [a, b] and g is nonnegative, then there is a number c in (a, b) for which

Z b Z b

f(x)g(x) dx = f(c) g(x) dx.

a a

(Hint: If m and M are the absolute minimum and absolute maximum of f on [a, b], then mg(x) ≤ f(x)g(x) ≤ Mg(x). By the Order Property,

R

By the Intermediate Value Theorem, there is some c such that

Z b

If g(x) dx = 0, then g(x) 6≡0 on [a, b] and all integrals are zero.)

a

Remark 20 The number f(c) is called the weighted average of f on [a, b] with respect to the weight function g.

5.7Volumes of Revolution

One simple application of the Riemann integral is to define the volume of a solid.

Theorem 5.7.1 Suppose that a solid is bounded by the planes with equations x = a and x = b. Let the cross-sectional area perpendicular to the x-axis at x be given by a continuous function A(x). Then the volume V of the solid is given by

Z b

V = A(x) dx.

a

Proof. Let P = {a = x0 < x1 < x2 < · · · < xn = b} be a partition of [a, b]. For each i = 1, 2, 3, · · · , n, let

Vi = volume of the solid between the planes with equations x = xi−1 and x = xi,

mi = absolute minimum of A(x) on [xi−1, xi], Mi = absolute maximum of A(x) on [xi−1, xi],

xi = xi − xi−1.

Then

mi xi ≤ Vi ≤ Mi xi, mi ≤ Vxi i ≤ Mi.

Since A(x) is continuous, there exists some ci such that xi−1 ≤ ci ≤ Mi and

mi ≤ A(ci) = Vxi i ≤ Mi Vi = A(ci)Δxi

n

X

V = A(ci)Δxi.

i=1

It follows that for each partition P of [a, b] there exists a Riemann sum that equals the volume. Hence, by definition,

Then the volume V obtained by rotating R about the x-axis is given by

Z b

V = π(f(x))2dx.

a

Proof. Clearly, the volume of the rotated solid is between the planes with equations x = a and x = b. The cross-sectional area at x is the circle generated by the line segment joining (x, 0) and (x, f(x)) and has area A(x) = π(f(x))2. Since f is continuous, A(x) is a continuous function of x. Then by Theorem 5.7.1, the volume V is given by

Z b

V = π(f(x))2 dx.

a

Theorem 5.7.3 Let f and R be defined as in Theorem 5.7.2. Assume that f(x) > 0 for all x [a, b], either a ≥ 0 or b ≤ 0, so that [a, b] does not contain 0. Then the volume V generated by rotating the region R about the y-axis is given by

Z b

V = (2πxf(x)) dx.

a

Proof. The line segment joining (x, 0) and (x, f(x)) generates a cylinder whose area is A(x) = 2πxf(x). We can see this if we cut the cylinder vertically at (−x, 0) and flattening it out. By Theorem 5.7.1, we get

Z b

V = 2πxf(x) dx.

a

Theorem 5.7.4 Let f and g be continuous on [a, b] and suppose that f(x) > g(x) > 0 for all x on [a, b]. Let R be the region bounded by the curves x = a, x = b, y = f(x) and y = g(x).

(i) The volume generated by rotating R about the x-axis is given by

Z b

π[(f(x))2 − (g(x))2] dx.

a

(ii)If we assume R does not cross the y-axis, then the volume generated by rotating R about the y-axis is given by

Z b

V = 2πx[f(x) − g(x)]dx.

a

(iii)If, in part (ii), R does not cross the line x = c, then the volume generated by rotating R about the line x = c is given by

Z b

V = 2π|c − x|[f(x) − g(x)]dx.

a

Proof. We leave the proof as an exercise.

Remark 21 There are other various horizontal or vertical axes of rotation that can be considered. The basic principles given in these theorems can be used. Rotations about oblique lines will be considered later.

Example 5.7.1 Suppose that a pyramid is 16 units tall and has a square base with edge length of 5 units. Find the volume of V of the pyramid.

graph

We let the y-axis go through the center of the pyramid and perpendicular to the base. At height y, let the cross-sectional area perpendicular to the y-axis be A(y). If s(y) is the side of the square A(y), then using similar triangles, we get

A(y) = 25625 (16 − y)2.

Then the volume of the pyramid is given by

Example 5.7.2 Consider the region R bounded by y = sin x, y = 0, x = 0 and x = π. Find the volume generated when R rotated about

= π2 .

2

graph

(ii) By Theorem 5.7.3, the volume V is given by (integrating by parts)

Z π

V = 2πx sin x dx ; (u = x, dv = sin x dx)

0

=2π[−x cos x + sin x]π0

=2π[π]

=2π2.

graph

(iii) In this case, the volume V is given by

Z π

V = π(sin x + 2)2 dx

0

Zπ

=π[sin2 x + 4 sin x + 4] dx

graph

(iv) In this case,

Z π

V = π[12 − (1 − sin x)2] dx.

0

graph

Z π

V = π[1 − 1 + 2 sin x − sin2x]dx

= π(8 − π). 2

(v)

Z π

V = (2π(π − x) sin x] dx

0

Zπ

=2π [π sin x − x sin x] dx

0

=2π[−π cos x + x cos x − sin x]π0

=2π[2π − π]

=2π2.

graph

(vi)

Z π

V = 2π(2π − x) sin x dx

0

=2π[−2π cos x + x cos x − sin x]π0

=2π[4π − π]

=6π2.

graph

Example 5.7.3 Consider the region R bounded by the circle (x−4)2 +y2 = 4. Compute the volume V generated when R is rotated around

(i) y = 0 (ii) x = 0 (iii) x = 2

graph

(i) Since the area crosses the x-axis, it is su cient to rotate the top half to get the required solid.

Z 6 Z 6

V = πy2 dx = π [4 − (x − 4)2] dx

dx = 2 cos tdt

Zπ/2

=4π (4 + 2 sin t)(2 cos t)(2 cos t)dt

−π/2

Zπ/2

=4π (16 cos2 t + 8 cos2 t sin t) dx

=4π[8(π)]

=32π2

(iii) In this case,

Z 6

V = 2π(x − 2)2y dx

2

Z 6

p

= 4π (x − 2) 4 − (x − 4)2 dx ; x − 4 = 2 sin t

2

dx = 2 cos tdt

Zπ/2

=4π (2 + 2 sin t)(2 cos t)(2 cos t)dt

−π/2

Zπ/2

=4π (8 cos2 t + 8 cos2 t sin t)dt

=4π[4π]

=16π2

Exercises 5.7

1.Consider the region R bounded by y = x and y = x2. Find the volume generated when R is rotated around the line with equation

3.Consider the region R bounded by y = ex, x = 0, x = ln 2, y = 0. Find the volume generated when R is rotated about the line with equation

6.Consider the region R bounded by y = x, y = x3. Find the volume generated when R is rotated about the line with equation

7.Consider the region R bounded by y = x2, y = 8 − x2. Find the volume generated when R is rotated about the line with equation

8.Consider the region R bounded by y = sinh x, y = 0, x = 0, x = 2. Find the volume generated when R is rotated about the line with equation

10. Compute the volume of a cone with height h and radius r.