362-4

362-4 - MAT 362 Answers to selected exercises, week of...

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MAT 362 Answers to selected exercises, week of Sept. 19 Graded problems 1. Parametrize the surface using polar coordinates: Φ ( r , θ ) = ± r cos θ, r sin θ, - 1 3 1 - r 2 ² . (a) Direct calculation. We need the cross product of the surface tangent vectors to obtain dS : T r × T θ = - r 2 cos θ 3 1 - r 2 , - r 2 sin θ 3 1 - r 2 , r ! is an upward pointing normal to the surface. Also, ∇ × F = (2 x 3 yz , - 3 x 2 y 2 z , - 2) = ± - 2 3 r 4 1 - r 2 cos 3 θ sin θ, r 4 1 - r 2 cos 2 θ sin 2 θ, - 2 ² so that ZZ S ( ∇ × F ) · d S = Z 2 π 0 Z 1 0 ( ∇ × F ) · ( T r × T θ ) dr d θ = Z 2 π 0 Z 1 0 ± 2 9 r 6 cos 4 θ sin θ - 1 3 r 6 cos 2 θ sin 3 θ - 2 r ² dr d θ = - 2 π. You can use an appropriate substitution in the first term of the last integrand and a table of integrals or integration by parts for the second. Alternatively, you can argue by symmetry: the integral is over a full period of each of the first two terms. In the first term, cos 4 θ 0 and - 1 sin θ 1. The portion of the interval over which is sin θ < 0 is the same size as that over which sin θ > 0, and the area under the curve on each portion must be equal, due to the symmetry of the sine function. So the integral of the first term must be zero. A similar argument applies to the second term. Important: we began with a parametrization in terms of polar coordinates. Therefore, the calculation of d S and T r × T θ includes all the factors of r that are needed to evaluate the integral correctly. If we had begun with rectangular coordinates, and evaluated

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This note was uploaded on 10/15/2008 for the course MAT 362 taught by Professor Callahan during the Fall '06 term at ASU.

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362-4 - MAT 362 Answers to selected exercises, week of...

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