Calculus II Uiz 1

1132 Calculus II uiz 1

Names:

Date: 2/6/09

Question: 1 2 3 4 Total Points: 3 2 2 3 10 Score:

You are allowed to work in groups of no more than ﬁve (5) people. This quiz is open book and open notes. Each group only has to hand in one (1) quiz, so please make sure it is written in a neat and concise manner.

1. (3 points) A hole of radius r is bored through the center of a sphere of radius R > r. Find the volume of the remaining portion of the sphere.

√ √ 2 2 2 2 Solution: The line y = r intersects the semicircle√ y = R − x when r = R − x implies√ that r2 = R2 − x2, so x2 = R2 − r2 and x = ± R2 − r2. Rotating the region (bounded by y = R2 − x2 and y = r) about the x-axis gives √ Z R2−r2 2 p 2 2 2 √ π R − x − r dx. − R2−r2 Then since the integrand is an even function, √ √ Z R2−r2 2 Z R2−r2 p 2 2 2 2 2 2 V = √ π R − x − r dx = 2π R − x − r dx. − R2−r2 0 Then √ √ 2 2 2 2 R −r Z R −r 1 2 2 2 2 2 3 V = 2π R − x − r dx = 2π (R − r )x − x 0 3 0 1 = 2π (R2 − r2)3/2 − (R2 − r2)3/2 3 2 4 = 2π · (R2 − r2)3/2 = (R2 − r2)3/2. 3 3

4 3 This answer should make sense to you, since taking the limit as r → 0 yields V → 3 πR . 2. (2 points) Newton’s Law of Gravitation states that two bodies with masses m1 and m2 attract each other with a force m m F = G 1 2 , r2 where r is the distance between the bodies and G is the gravitational constant. If one of the bodies is ﬁxed, ﬁnd the work needed to move the other from r = a to r = b.

Solution: We have b Z b Z b m m −1 1 1 W = F (r) dr = G 1 2 dr = Gm m = Gm m − . 2 1 2 1 2 a a r r a b a

Page 2 3. (2 points) If f(0) = g(0) = 0 and f 00 and g00 are continuous, show that Z a Z a f(x)g00(x) dx = f(a)g0(a) − f 0(a)g(a) + f 00(x)g(x) dx. 0 0 That is, evaluate the integral on the left until you end at the expression on the right.

Solution: Suppose f(0) = g(0) = 0 and let u = f(x) and dv = g00(x) dx. Then du = f 0(x) dx and v = g0(x). So our integral becomes

Z a a Z a Z a 00 0 0 0 0 0 0 f(x)g (x) dx = f(x)g (x) − f (x)g (x) dx = f(a)g (a) − f (x)g (x) dx. 0 0 0 0 Now we use integration by parts again on the integral on the right. Let U = f 0(x) and dV = g0(x) dx. Then dU = f 00(x) dx and V = g(x). Then

Z a a Z a Z a 0 0 0 00 0 00 f (x)g (x) dx = f (x)g(x) − f (x)g(x) dx = f (a)g(a) − f (x)g(x) dx. 0 0 0 0 Combining the two lines gives Z a Z a f(x)g00(x) dx = f(a)g0(a) − f 0(a)g(a) + f 00(x)g(x) dx. 0 0

Page 3 4. (3 points) Use the method of cylindrical shells to ﬁnd the volume generated by rotating the region bounded by y = ex, y = e−x, and x = 1 about the y-axis.

Solution: The volume of the solid is given by

Z 1 Z 1 Z 1 Z 1 V = 2π(ex − e−x) dx = 2π (xex − xe−x) dx = 2π xex dx − xe−x dx . 0 0 0 0 We must use integration by parts on both integrals. For the ﬁrst integral, let u = x and dv = ex. Then du = dx and v = ex. So Z Z xex dx = xex − ex dx = xex − ex.

A similar process on the other integral yields Z xe−x dx = −xe−x − e−x.

Z 1 Z 1 1 2 4π x −x x x −x V = 2π xe dx − xe dx = 2π((xe − e ) − (−xe − x)) = 2π − 0 = . 0 0 0 e e

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