Tutorial8_sol_1011

Tutorial8_sol_1011 - 2 e v ± ± ± v =1 v =-1 du = e-e-1 3...

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MATH1211/09-10(1)/Tu8sol s/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2009–2010, First semester) Brief Solution to Tutorial 8 1. The two regions of the two integrals are shown as in the figure at right. By changing the order of integration, we combine the two regions together and rewrite the sum of integrals as one single integral and evaluate: Z 1 0 Z 2 - y y sin xdxdy = Z 1 0 - cos x ± ± ± x =2 - y x = y dy = Z 1 0 ( cos y - cos(2 - y ) ) dy = ( sin y + sin(2 - y ) ) ± ± ± 1 0 = 2 sin 1 - sin 2 . 2. Z 1 0 Z 2 0 Z x 2 0 f ( x,y,z ) dz dxdy = Z 2 0 Z 1 0 Z x 2 0 f ( x,y,z ) dz dy dx = Z 2 0 Z x 2 0 Z 1 0 f ( x,y,z ) dy dz dx = Z 4 0 Z 2 z Z 1 0 f ( x,y,z ) dy dxdz = Z 1 0 Z 4 0 Z 2 z f ( x,y,z ) dxdz dy = Z 4 0 Z 1 0 Z 2 z f ( x,y,z ) dxdy dz. 3. We apply change of variables u = 2 x + y, v = x - y . Then ( u,v ) ( x,y ) = det " ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y # = det ² 2 1 1 - 1 ³ = - 3 , ( x,y ) ( u,v ) = 1 . ( u,v ) ( x,y ) = - 1 3 . As the region D is given by 1 u 4 , - 1 v 1, the integral is computed as Z 4 1 Z 1 - 1 u 2 e v ± ± ± ( x,y ) ( u,v ) ± ± ± dv du = Z 4 1 Z 1 - 1 u 2 e v 1 3 dv du = 1 3 Z 4 1 u
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Unformatted text preview: 2 e v ± ± ± v =1 v =-1 du = e-e-1 3 Z 4 1 u 2 du = 7( e-e-1 ) . 4. First we use cylindrical coordinates to integrate RRR B e z dV : Z 1-1 Z 2 π Z √ 1-z 2 e z r dr dθ dz = Z 1-1 Z 2 π e z 1-z 2 2 dθ dz = π Z 1-1 e z (1-z 2 ) dz = π ² e z ± ± 1-1-( z 2 e z ) ± ± 1-1 + Z 1-1 2 ze z dz ³ = π ² ( e-e-1 )-( e-e-1 ) + (2 ze z ) ± ± 1-1-Z 1-1 2 e z dz ³ = π ´ 2 ( e + e-1 )-2 ( e-e-1 )µ = 4 e-1 π. Volume of the unit ball B is 4 π/ 3 (or it can be computed as R 1-1 R 2 π R √ 1-z 2 r dr dθ dz = R 1-1 π (1-z 2 ) dz = 4 π 3 .) Hence the required answer is 4 e-1 π 4 π 3 = 3 e-1 . 1...
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This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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Tutorial8_sol_1011 - 2 e v ± ± ± v =1 v =-1 du = e-e-1 3...

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