This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 xyplane 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 dθ = Z 2 π θ =0 1 dθ = 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 quartercircle in the first quadrant with center (0 , 0) and radius R . Rewriting 1 the integral in polar coordinates leads to I R := Z π/ 2 Z R 1 (1 + r 2 ) 2 r dr dθ = Z π/ 2 Z 1+ R 2 1 1 u 2 du 2 dθ (where u := 1 + r 2 ) = 1 2 Z π/ 2 1 u 1+ R 2 1 dθ = 1 2 1 1 + R 2 ( 1) Z π/ 2 dθ = 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) 3B1b* (5 pts.), 3B2c, 3B4* (7 pts.) Solution. (to 3B1b*) 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 Z 2 sin θ ··· dr dθ. Solution. (to 3B4*) 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....
View
Full
Document
This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.
 Fall '08
 Auroux
 Multivariable Calculus

Click to edit the document details