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S09fnlansl

# S09fnlansl - ρ is used in the spherical coordinate sense...

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Answers for Sample Final Exam BEWARE: The following are just the numerical answers. Since the justification is missing, they would not receive full credit on the real exam. 1. (a) r ( t ) = 1 , 4 , 6 + t 3 , 2 , 1 ; (b) 5 / 14. 2. R ( u ) = j + 2 k + u ( i + j + k ); at ( 2 , 1 , 0). 3. (a)11; (b) 2 x 6 y + 9 z = 19. 4. Maximum value is 3 3 / 2 / 5 5 / 2 , achieved at the four points ( 1 5 , 1 5 , 3 5 ), ( 1 5 , 1 5 , 3 5 ), ( 1 5 , 1 5 , 3 5 ) and ( 1 5 , 1 5 , 3 5 ). 5. 1 / 3 1 / 9 y 1 / 3 f ( x, y ) dx dy + 1 1 / 3 y y f ( x, y ) dx dy . 6. (8 2 2) / 15. 7. 8 π b 4 / 5. 8. (a) f ( x, y ) = x 2 y + xy 3 + 3 y 2 / 2; (b) 105.5. 9. -2/3. 10. div F ( x, y, z ) = xy + 2 y + z 3 ; curl F ( x, y, z ) = xz i + (3 xz 2 yz ) j ; grad(div F )( x, y, z ) = y i + (2 + x ) j + 3 z 2 k . 11. (9 3 8 2 + 1) / 15. 12. 5 π / 2. ******************** Details for question 7. [Beware: The notation is a bit confusing here because the symbol ρ is sometimes used to denote density. In this question
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Unformatted text preview: ρ is used in the spherical coordinate sense, and the density at the point ( ρ, θ, φ ) is given by the function f ( ρ, θ, φ ) = ρ .] The region is described in spherical polars by 0 ≤ ρ ≤ 2 b cos φ , 0 ≤ θ ≤ 2 π , 0 ≤ φ ≤ π/ 2. Then ° ° ° E ρdV = ° π/ 2 ° 2 π ° 2 b cos φ ρ · ρ 2 sin φ dρdθdφ = ° π/ 2 ° 2 π ± ρ 4 4 sin φ ² 2 b cos φ dθdφ = 4 b 4 ° π/ 2 ° 2 π (cos φ ) 4 sin φ d θ d φ = 8 πb 4 ° π/ 2 (cos φ ) 4 sin φ d φ = 8 πb 4 ± − (cos φ ) 5 5 ² π/ 2 = 8 πb 4 ³ − ³ − 1 5 ´´ = 8 πb 4 5 ....
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