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# h5s - 18.02 HOMEWORK#5 DUE BJORN POONEN 1 Part A(After Oct...

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18.02 HOMEWORK #5, DUE OCTOBER 28, 2010 BJORN POONEN 1. Part A (After Oct. 14) p. 967: 27, 34* (10 pts.) Solution. (to p. 967, 27) The paraboloids intersect on a curve whose projection to the xy -plane is given by x 2 + y 2 = 4 - 3 x 2 - 3 y 2 , which defines the circle x 2 + y 2 = 1. Inside the circle, i.e., when x 2 + y 2 < 1, the value of 4 - 3 x 2 - 3 y 2 is larger than x 2 + y 2 . Thus the answer is ZZ x 2 + y 2 1 (4 - 3 x 2 - 3 y 2 ) - ( x 2 + y 2 ) dA = ZZ x 2 + y 2 1 (4 - 4 x 2 - 4 y 2 ) dA = Z 2 π θ =0 Z 1 r =0 (4 - 4 r 2 ) r dr dθ = Z 2 π θ =0 Z 1 r =0 (4 r - 4 r 3 ) dr dθ = Z 2 π θ =0 (2 r 2 - r 4 ) 1 r =0 = Z 2 π θ =0 1 = 2 π. Solution. (to p. 967, 34) The integral is lim R →∞ I R , where I R is the integral of the same func- tion over the quarter-circle in the first quadrant with center (0 , 0) and radius R . Rewriting 1

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the integral in polar coordinates leads to I R := Z π/ 2 0 Z R 0 1 (1 + r 2 ) 2 r dr dθ = Z π/ 2 0 Z 1+ R 2 1 1 u 2 du 2 (where u := 1 + r 2 ) = 1 2 Z π/ 2 0 - 1 u 1+ R 2 1 = 1 2 - 1 1 + R 2 - ( - 1) Z π/ 2 0 = 1 2 1 - 1 1 + R 2 π 2 = π 4 1 - 1 1 + R 2 . So the answer is lim R →∞ I R = lim R →∞ π 4 1 - 1 1 + R 2 = π 4 . (After Oct. 14) 3B-1b* (5 pts.), 3B-2c, 3B-4* (7 pts.) Solution. (to 3B-1b*) The range for θ is 0 to π , and for each value of θ , dropping a perpen- dicular from the center (1 , 0) to the line at angle θ through (0 , 0) shows that the range for r is [0 , 2 cos( π/ 2 - θ )] = [0 , 2 sin θ ]. Thus the integral takes the shape Z π 0 Z 2 sin θ 0 · · · dr dθ. Solution. (to 3B-4*) The region is a unit square with an inscribed quarter circle removed. The smallest value of r is 1 (along the quarter circle), and the largest value of r is 2 (at the far corner (1 , 1)). For a fixed r [1 , 2], the smallest value of θ is the one for which r cos θ = 1, i.e., θ = cos - 1 (1 /r ). By symmetry, the largest value of θ for that r is π/ 2 - cos - 1 (1 /r ), or equivalently, sin - 1 (1 /r ). Thus the integral takes the shape Z 2 1 Z sin - 1 (1 /r ) cos - 1 (1 /r ) · · · dθ dr. (After Oct. 14) p. 976: 42* (7 pts.) Solution. The area of the region is A = 1 4 ( πr 2 ) = π 4 r 2 . 2
The volume of revolution is the upper solid hemisphere of radius r , whose volume is V = 1 2 4 3 πr 3 = 2 3 πr 3 . The distance travelled by the centroid (¯ x, ¯ y ) is d = 2 π ¯ x . On the other hand, the first theorem of Pappus says that V = Ad , where d is the distance so Substituting all this into the first theorem of Pappus, which says that V = Ad , shows that 2 3 πr 3 = π 4 r 2 (2 π ¯ x ) ¯ x = 4 3 π r. By symmetry, ¯ y = ¯ x , so the centroid is ( 4 3 π r, 4 3 π r ) .

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