y x arctan x arctan x x dx 2 ln by first expressing the integral as an iterated integral. f t dt dz dy 1 2 y x x t 2 f t dt F 2 Gm d 1 d 1 sr 2 d 2

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1 CHALLENGE PROBLEMS CHALLENGE PROBLEMS: CHAPTER A Click here for answers. S Click here for solutions.. If denotes the greatest integer in, evaluate the integral where R, y 3, y 5.. Evaluate the integral where ma, y means the larger of the numbers and. 3. Find the average value of the function f cost dt on the interval [, ]. 4. If a, b, and c are constant vectors, r is the position vector i y j z k, and E is given by the inequalities a r, b r, c r, show that yyy E 5. The double integral y is an improper integral and could be defined as the y d dy y limit of double integrals over the rectangle, t, t as t l. But if we epand the integrand as a geometric series, we can epress the integral as the sum of an infinite series. Show that y y yy R y y a rb rc r dv y d dy 6. Leonhard Euler was able to find the eact sum of the series in Problem 5. In 736 he proved that n y da e ma, y dy d n In this problem we ask you to prove this fact by evaluating the double integral in Problem 5. Start by making the change of variables 6 8 a b c n y n Copyright 3, Cengage Learning. All rights reserved. This gives a rotation about the origin through the angle 4. You will need to sketch the corresponding region in the uv-plane. [Hint: If, in evaluating the integral, you encounter either of the epressions sin cos or cos sin, you might like to use the identity cos sin and the corresponding identity for sin.] 7. (a) Show that (Nobody has ever been able to find the eact value of the sum of this series.) (b) Show that y y y y y y u v s y u v s yz d dy dz yz d dy dz Use this equation to evaluate the triple integral correct to two decimal places. n n n

2 CHALLENGE PROBLEMS 8. Show that y arctan arctan d ln by first epressing the integral as an iterated integral. 9. If f is continuous, show that y yy yz f t dt dz dy y t f t dt. (a) A lamina has constant density and takes the shape of a disk with center the origin and radius R. Use Newton s Law of Gravitation (see Section.9) to show that the magnitude of the force of attraction that the lamina eerts on a body with mass m located at the point,, d on the positive z-ais is F Gmd d [Hint: Divide the disk as in Figure 4 in Section.3 and first compute the vertical component of the force eerted by the polar subrectangle R ij.] (b) Show that the magnitude of the force of attraction of a lamina with density that occupies an entire plane on an object with mass m located at a distance d from the plane is F Gm Notice that this epression does not depend on d. sr d Copyright 3, Cengage Learning. All rights reserved.

3 CHALLENGE PROBLEMS 3 ANSWERS S Solutions sin 7. (b).9 Copyright 3, Cengage Learning. All rights reserved.

4 4 CHALLENGE PROBLEMS SOLUTIONS E Eercises. Let R 5 i R i,where R i {(, y) + y i +,+ y<i+3, 3, y 5}. R [ + y ] da 5 i R i [ + y]] da 5 i [ + y ] R i da,since [ + y ] constant i +for (, y) R i. Therefore R [ + y ] da 5 i (i +)[A(R i)] 3A(R )+4A(R )+5A(R 3 )+6A(R 4 )+7A(R 5 ) () f ave b a b a f() d cos(t ) dt d cos(t ) dt d t cos(t ) d dt [changing the order of integration] t cos(t ) dt sin t sin 5. Since y <,eceptat(, ), the formula for the sum of a geometric series gives y (y) n,so d dy y (y) n d dy n n n d y n dy n n n (y) n d dy n + (n +) n n n + n 7. (a) Since yz < ecept at (,, ), the formula for the sum of a geometric series gives yz (yz) n, so n Copyright 3, Cengage Learning. All rights reserved. d dy dz yz (yz) n d dy dz n n d y n dy n n n n + n + n + n z n dz (n +) n (yz) n d dy dz

5 CHALLENGE PROBLEMS 5 (b) Since yz <, eceptat(,, ), the formula for the sum of a geometric series gives +yz ( yz) n,so n d dy dz +yz ( yz) n d dy dz n n ( yz) n d dy dz ( ) n n d y n dy n ( ) n n + n n n + n + z n dz ( ) n (n +) n ( ) n To evaluate this sum, we first write out a few terms: s Notice that a 7 <.3. By the Alternating Series Estimation Theorem from Section 8.4, we have 73 s s 6 a 7 <.3. This error of.3 will not affect the second decimal place, so we have s y z f (t) dt dz dy f (t) dv,where E E {(t, z, y) t z, z y, y }. If we let D be the projection of E on the yt-plane then D {(y, t) t, t y }. And we see from the diagram that E {(t, z, y) t z y, t y, t }. So y z f(t) dt dz dy t y t f(t) dz dy dt (y t) f(t) dy dt t y ty f(t) y y t dt t t + t f(t) dt t + t f(t) dt t + t f(t) dt ( t) f(t) dt Copyright 3, Cengage Learning. All rights reserved.