Solutions11

# Solutions11 - Math 601 Solutions to Homework 11 1 Consider...

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Math 601 Solutions to Homework 11 1. Consider the curve given by the following parametric equations: x = cos( t ) and y = sin(2 t ) Let D be the region inside this curve and to the right of the y -axis: (a) Use Green’s Theorem to compute ZZ D x 2 dA . (b) Use Green’s Theorem to find the area of the region D . Answer: (a) We need to find a vector field F such that rot( F ) = x 2 . The vector field F = - yx 2 i will work (as would 1 3 x 3 j or lots of other possibilities). Then, by Green’s Theorem: ZZ D x 2 dA = I ∂D - yx 2 dx From the parametrization of the curve, we have: x = cos t y = sin(2 t ) dx = - sin t dt dy = 2 cos(2 t ) dt The bounds for t are - π 2 t π 2 (we can see this by noting that x and y are both 0 when t is a multiple of π/ 2, and then by plotting a few points of the parametric curve to make sure we choose the right-side of the lemniscate). We can now compute the integral (recall that sin(2 t ) = 2 sin t cos t ): ZZ D x 2 dA = I ∂D - yx 2 dx = Z π/ 2 - π/ 2 - sin(2 t ) cos 2 t ( - sin t ) dt = Z π/ 2 - π/ 2 2 sin 2 t cos 3 t dt = Z π/ 2 - π/ 2 2 sin 2 t (1 - sin 2 t ) cos t dt

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We use the substitution u = sin t , du = cos t dt : ZZ D x 2 dA = Z 1 - 1 2 u 2 (1 - u 2 ) du = 2 3 u 3 - 2 5 u 5 1 - 1 = 8 15 Note: If we had used a different vector field, we would have obtained the same answer at the end. (b) To compute the area we need to evaluate the integral Z D dA , so we need to find a vector field F such that rot( F ) = 1. The vector field F = - y i will work (as would x j or lots of other possibilities). Then, by Green’s Theorem: ZZ D dA = I ∂D - y dx As in part (a), we have: x = cos t y = sin(2 t ) dx = - sin t dt dy = 2 cos(2 t ) dt with - π 2 t π 2 . We can now compute the integral (recall that sin(2 t ) = 2 sin t cos t ): ZZ D dA = I ∂D - y dx = Z π/ 2 - π/ 2 - sin(2 t )( - sin t ) dt = Z π/ 2 - π/ 2 2 sin 2 t cos t dt We use the substitution u = sin t , du = cos t dt : ZZ D dA = Z 1 - 1 2 u 2 du = 2 3 u 3 1 - 1 = 4 3
2. Consider the surface integral ZZ S z dA where S is the upper half ( z 0) of the sphere x 2 + y 2 + z 2 = 1. (a) Set up this integral using a parametrization where t = x and u = y . (b) Set up this integral using a parametrization where t = r and u = θ . (c) Set up this integral using a parametrization where t = θ and u = φ . Answer: (a) With t = x and u = y , we have the parametric equations: x = t y = u z = 1 - t 2 - u 2 - 1 - u 2 t 1 - u 2 - 1 u 1 Then, the tangent vectors to the surface are T t = 1 , 0 , - t 1 - t 2 - u 2 T u = 0 , 1 , - u 1 - t 2 - u 2 And, the normal vector to the surface is T t × T u = fl fl fl fl fl fl fl fl i j k 1 0 - t 1 - t 2 - u 2 0 1 - u 1 - t 2 - u 2 fl fl fl fl fl fl fl fl = t 1 - t 2 - u 2 i + u 1 - t 2 - u 2 j + k Thus: dA = k T t × T u k dt du = s t 2 1 - t 2 - u 2 + u 2 1 - t 2 - u 2 + 1 dt du = r 1 1 - t 2 - u 2 dt du Thus: ZZ S z dA = Z 1 - 1 Z 1 - u 2 - 1 - u 2 1 - t 2 - u 2 k T t × T u k dt du = Z 1 - 1 Z 1 - u 2 - 1 - u 2 1 - t 2 - u 2 r 1

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