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tutorial8_sol

# tutorial8_sol - e-e-1 3 Z 4 1 u 2 du = 7 e-e-1 3 First we...

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MATH1211/10-11(2)/Tu8sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 8 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. Let W denote the region under consideration. Note that the two surfaces intersect at circle x 2 + z 2 = 9 / 2 , y = 9 / 2. We project W onto the xz -plane, and its projected region D on the xz -plane is x 2 + z 2 9 / 2. Volume of W is then ZZ D Z 9 - x 2 - z 2 x 2 + z 2 dy dx dz = ZZ D (9 - 2 x 2 - 2 z 2 ) dx dz. We use polar coordinates x = r cos θ, z = r sin θ , so that D becomes 0 r 3 2 , 0 θ 2 π . Using change of variables ( x, z ) ( r, θ ), we get the answer ZZ D (9 - 2 x 2 - 2 z 2 ) dx dz = Z 2 π 0 Z 3 / 2 0 (9 - 2 r 2 ) r dr dθ = - π 4 (9 - 2 r 2 ) 2 3 / 2 0 = 81 π 4 2. 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 1 1 - 1 = - 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 ) (
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Unformatted text preview: e-e-1 3 Z 4 1 u 2 du = 7( e-e-1 ) . 3. 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 RRR B e z dV RRR B dV = 4 e-1 π 4 π 3 = 3 e-1 . 4. Z x f ds = Z 2 f ( x ( t )) k x ( t ) k dt = Z 2 6 t 3 p 1 2 + 2 2 + 3 2 dt = 3 √ 14 2 t 4 ± ± ± ± ± 2 = 24 √ 14 . 1...
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