s9 - Solution to Set 9 /2 1 0 0 1x2 /2 1 1. 0 x cos zdydxdz...

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Solution to Set 9 1. Z π/ 2 0 Z 1 0 Z 1 x 2 0 x cos zdydxdz = Z π/ 2 0 cos zdz ! ± Z 1 0 x p 1 x 2 dx ² = 1 3 2. (a) The parabolic sheet cuts the xy-plane at the lines x = ± 1. ZZZ D dv = ZZ R Z 1 x 2 0 dz dA where R is the rectangular region [ 1 , 1] × [ 1 , 2]. The volume is Z 2 1 Z 1 1 (1 x 2 ) dxdy =3 ³ x x 3 3 ´ 1 1 =4. (b) ZZZ D dv = ZZ R Z x 2 + y 2 0 dz dA where R is the triangular region with vertices (0 , 0) , (1 , 0) , (0 , 1). The volume is Z 1 0 Z 1 y 0 ( x 2 + y 2 ) dxdy = Z 1 0 ³ 1 3 (1 y ) 3 + y 2 (1 y ) ´ dy = 1 6 3. (a) ZZZ D ( x 2 + y 2 ) dv = Z 2 π 0 Z 1 0 Z 4 0 r 2 dzrdrdθ =2 π (b) ZZZ D zdv = Z π/ 2 0 Z 1 0 Z 1 r 2 0 zdzrdrdθ = π 16 4. (a) ZZZ D x 2 dv = Z 2 π 0 Z π 0 Z 3 2 ρ 2 sin 2 φ cos 2 θρ 2 sin φdρdφdθ = ± Z 2 π 0 cos 2 θdθ ²± Z π 0 sin 3 φdφ ²± Z 3 2 ρ 4 ² = π × 4 3 × 1 5 (3 5 2 5 ) 56 . 3 π (b) The cone angle (from axis to edge) is tan 1 1 3 = π 6 . ZZZ D 1 x 2 + y 2 + z 2 dv = Z 2 π 0 Z π/ 6 0 Z 9 3 1 ρ 2 ρ 2 sin φdφdθ =2 π × (1 3 2 ) × 6 1 . 6 π 5. (a) x = 1 5 u + 2 5 v, y = 2 5 u + 1
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This note was uploaded on 02/27/2010 for the course MATH MATH101 taught by Professor Chan during the Fall '09 term at HKUST.

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