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September 17th, 2017, 07:09 PM   #1
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My First Stokes Theorem problem....

Let F = < z, x, y >. The plane z = 2x + 2y - 1 and the paraboloid z = x^2 + y^2 intersect in a closed curve. Stokes's Theorem implies that

$\displaystyle
\int_{}^{} \int_{D1}^{} ( \nabla ~X~ F ) * N ~dS = \int_{C}^{} F *~dr = \int_{}^{} \int_{D2}^{} ( \nabla ~X~ F ) * N ~DS
$


where the line integral is computed over the intersection C of the plane and the paraboloid, and the two surface integrals are computed over the portions of the two surfaces that have boundary C (provided, of course, that the orientations all match). Compute all three integrals

Answer: $\displaystyle -3 \pi $

I did the middle line integral successfully, but I would be much appreciated if someone shows me step-by-step of solving $\displaystyle \int_{}^{} \int_{D1}^{} ( \nabla ~X~ F ) * N ~dS $

Last edited by zollen; September 17th, 2017 at 07:11 PM.
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September 17th, 2017, 08:33 PM   #2
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Let me know if you don't understand this Mathematica sheet

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Last edited by skipjack; September 19th, 2017 at 08:32 AM.
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September 17th, 2017, 08:40 PM   #3
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Doing the surface integrals in cylindrical coordinates in this sheet. It's a bit cleaner.

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September 18th, 2017, 04:11 AM   #4
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Is Wolfram Mathematica home edition free?

Last edited by skipjack; September 19th, 2017 at 08:43 AM.
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September 18th, 2017, 07:44 AM   #5
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I may not have fully understood the syntax. Should you use unit normal vector? Your solutions seem to use (n=Cross(cx,cy)) normal vector.

Last edited by skipjack; September 19th, 2017 at 08:52 AM.
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September 18th, 2017, 09:32 AM   #6
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Quote:
Originally Posted by zollen View Post
I may not have fully understood the syntax. Should you use unit normal vector? Your solutions seem to use (n=Cross(cx,cy)) normal vector.
What happens is that the factor that normalizes the unit vector is exactly the same factor that accounts for the local metric of the surface, so the two cancel each other out.

Last edited by skipjack; September 19th, 2017 at 08:53 AM.
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September 19th, 2017, 08:51 AM   #7
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Quote:
Originally Posted by zollen View Post
Is Wolfram Mathematica home edition free?
In the sense that you mean, no.
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