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Unformatted text preview: 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 yaxis: x y (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 tdt dy = 2cos(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 rightside of the lemniscate). We can now compute the integral (recall that sin(2 t ) = 2sin 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 2sin 2 t cos 3 tdt = Z π/ 2 π/ 2 2sin 2 t (1 sin 2 t )cos tdt We use the substitution u = sin t , du = cos tdt : 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 tdt dy = 2cos(2 t ) dt with π 2 ≤ t ≤ π 2 . We can now compute the integral (recall that sin(2 t ) = 2sin 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 2sin 2 t cos tdt We use the substitution u = sin t , du = cos tdt : 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 , , t √ 1 t 2 u 2 ¶ T u = , 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 +...
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This homework help was uploaded on 04/01/2008 for the course MATH 601 taught by Professor Alndy during the Spring '08 term at A.T. Still University.
 Spring '08
 alndy
 Equations, Parametric Equations

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