Math13S09final_sol

# Z y 2 2z 1 r y 2 1 z 0 22z y 0 1z dx dy dz xy2 3 3 x

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Unformatted text preview: ane. This is got by simply eliminating x from the two equations getting y = 2 − 2z . z y = 2 − 2z 1 R y 2 1 z =0 2−2z y =0 1−z dx dy dz x=y/2 √ 3 √ 3 x dy = 6. C √t −3 2 2 (t − 1)dt = √t −3 √ 4 2 − t dt = 2 2 0 3 t5 t3 − t4 − t2 dt = 2 5 3 √ 0 3 √ 83 = 5 1 1 |r ′ (t)|dt = ds = 7. arclength = C 0 1 0 1 0 0 t4 + 2t2 + 1dt 0 1 t3 +t 3 t2 + 1 dt = (t2 + 1)2 dt = = 1 1 + t2 (t2 + 2)dt = = 0 4 3 8. (a) Use the conversion from spherical to cartesian with ρ = 1 x = sin φ cos θ 0 ≤ θ ≤ 2π π y = sin φ sin θ 2 ≤φ≤π z = cos φ (b) x = x y=y z = − 1 − x2 − y 2 (c) Using (b): dS = R: x2 + y 2 ≤ 1 1+ x2 y 2 + 2 dA = z2 z dA = −(area of R) = −π R −1dA = − R cos φ|rφ × rθ | dA = z dS = S R 1 1...
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## This note was uploaded on 01/20/2014 for the course MATH 13 taught by Professor Weiss during the Spring '07 term at Tufts.

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