MAT235PS6-SOLNS

# MAT235PS6-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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Unformatted text preview: UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT235Y–CALCULUS II FALL-WINTER 2008-09 ASSIGNMENT 6. DUE ON MARCH 12. PROBLEMS AND SOLUTIONS Problem 1. Find the limit: lim t →∞ 1 t ZZ ( x- t ) 2 +( y- t ) 2 ≤ 1 p x 2 + y 2 dxdy. Solution. Making the change of variable x = t + tu , and y = t + tv in the double integral ZZ ( x- t ) 2 +( y- t ) 2 ≤ 1 p x 2 + y 2 dxdy, we have ZZ ( x- t ) 2 +( y- t ) 2 ≤ 1 p x 2 + y 2 dxdy = ZZ u 2 + v 2 ≤ 1 t 2 p ( tu + t ) 2 + ( tv + t ) 2 ∂ ( x,y ) u,v dudv = t 3 ZZ u 2 + v 2 ≤ 1 t 2 p ( u + 1) 2 + ( v + 1) 2 dudv. Hence, lim t →∞ 1 t ZZ ( x- t ) 2 +( y- t ) 2 ≤ 1 p x 2 + y 2 dxdy = lim t →∞ t 2 ZZ u 2 + v 2 ≤ 1 t 2 p ( u + 1) 2 + ( v + 1) 2 dudv. By the mean value theorem for double integrals, there is a point ( u, v ) in the disc u 2 + v 2 ≤ 1 t 2 such that ZZ u 2 + v 2 ≤ 1 t 2 p ( u + 1) 2 + ( v + 1) 2 dudv = p ( u + 1) 2 + ( v + 1) 2 ZZ u 2 + v 2 ≤ 1 t 2 dudv = π t 2 p ( u + 1) 2 + ( v + 1) 2 . (1) When t → ∞ , ( u, v ) → (0 , 0). Hence, lim t →∞ p ( u + 1) 2 + ( v + 1) 2 = 1 . Now making t approach infinity in equation (1) we conclude, lim t →∞ t 2 ZZ u 2 + v 2 ≤ 1 t 2 p ( u + 1) 2 + ( v + 1) 2 dudv = π lim t →∞ p ( u + 1) 2 + ( v + 1) 2 = π. Therefore, lim t →∞ 1 t ZZ ( x- t ) 2 +( y- t ) 2 ≤ 1 p x 2 + y 2 dxdy = π. 1 Problem 2. Let us denote by V n the volume of the region in the first octant that lies beneath the surface x a 1 n + y b 1 n + z c 1 n = 1 , where a , b , c , and n are constants. (i) Find V 2 , (ii) Show that lim n →∞ V n = 0. Solution. Let us denote by B n the solid bounded by the surface ( x/a ) 1 n +( y/b ) 1 n +( z/c ) 1 n = 1, and the planes x = 0, y = 0, and z = 0. Then the volume of B n is given by the following triple integral V n = ZZZ B n dxdy dz. (2) Let x = au n , y = bv n and z = cw n . Since ( x/a ) 1 n + ( y/b ) 1 n + ( z/c ) 1 n = 1 we will have that u + v + w = 1. Consider the region of the space D = { ( u,v,w ) ∈ R 3 : u,v,w ≥ ,u + v + w ≤ 1 } , and the transformation T : B n → D given by the equation T ( x,y,z ) = ( au n ,bv n ,cw n ) ....
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## This note was uploaded on 10/16/2009 for the course MATH MAT235 taught by Professor Recio during the Fall '08 term at University of Toronto.

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MAT235PS6-SOLNS - UNIVERSITY OF TORONTO DEPARTMENT OF...

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