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Math Help - Surface Integral

  1. #1
    Super Member angel.white's Avatar
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    Surface Integral

    Evaluate the surface integral:
    \int \int_S f(x,y,z) ~dS

    S is the surface with parametric equations x=u^2, y=u~sin~v, z = u~cos~v

    0\leq u\leq1
    0\leq v\leq \frac{\pi}2
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  2. #2
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    Quote Originally Posted by angel.white View Post
    Evaluate the surface integral:
    \int \int_S f(x,y,z) ~dS

    S is the surface with parametric equations x=u^2, y=u~sin~v, z = u~cos~v

    0\leq u\leq1
    0\leq v\leq \frac{\pi}2
    dS = \left|\vec{dS} \right| = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right| du \, dv where \vec{r} = x(u, v) \, i + y(u, v) \, j + z(u, v) \, k.

    For your question, \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = -u \, i + 2u^2 \sin v \, j + 2u^2 \cos v \, k.

    Therefore dS = u \sqrt{1 + 4u^2} \, du \, dv.

    So your surface integral becomes \int_{v=0}^{\pi/2} \int_{u=0}^{1} g(u, v) \, u \sqrt{1 + 4u^2} \, du \, dv where g(u, v) is what you get from f(x, y, z) after making the parametric substitution.
    Last edited by mr fantastic; July 31st 2008 at 02:09 AM. Reason: Forgot about the f(x, y, z) .....
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  3. #3
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    dS = \left|\vec{dS} \right| = \left| \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}\right| du \, dv where \vec{r} = x(u, v) \, i + y(u, v) \, j + z(u, v) \, k.

    For your question, \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v} = -u \, i + 2u^2 \sin v \, j + 2u^2 \cos v \, k.

    Therefore dS = u \sqrt{1 + 4u^2} \, du \, dv.

    So your surface integral becomes \int_{v=0}^{\pi/2} \int_{u=0}^{1} u \sqrt{1 + 4u^2} \, du \, dv.
    Okay, I follow you this far, if f(x,y,z) = yz then I should be finding:

    \int_{v=0}^{\pi/2} \int_{u=0}^{1} u^3\sin v \cos v \sqrt{1 + 4u^2} \, du \, dv

    = \int_{0}^{\pi/2} \sin v \cos v \, dv \int_{0}^{1} u^3 \sqrt{1 + 4u^2} \, du

    Is this correct?
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  4. #4
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    Quote Originally Posted by angel.white View Post
    Okay, I follow you this far, if f(x,y,z) = yz then I should be finding:

    \int_{v=0}^{\pi/2} \int_{u=0}^{1} u^3\sin v \cos v \sqrt{1 + 4u^2} \, du \, dv

    = \int_{0}^{\pi/2} \sin v \cos v \, dv \int_{0}^{1} u^3 \sqrt{1 + 4u^2} \, du

    Is this correct?
    Yes. And you know the substitution to make with the u-integral ....?

    By the way, I edited my earlier reply. I'm glad to see you realised I forgot the f(x, y, z) .......
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Yes. And you know the substitution to make with the u-integral ....?

    By the way, I edited my earlier reply. I'm glad to see you realised I forgot the f(x, y, z) .......
    Right now I'm trying a^2 = 1+4u^2

    I have gotten this far several times, but seem to keep integrating wrong. I needed to verify that I was correct up to this point so I could zero in on my error.
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  6. #6
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    Quote Originally Posted by angel.white View Post
    Right now I'm trying a^2 = 1+4u^2

    I have gotten this far several times, but seem to keep integrating wrong. I needed to verify that I was correct up to this point so I could zero in on my error.
    Try w = 1 + 4u^2.

    You should get \frac{5\sqrt{5}}{24} + \frac{1}{120} = \frac{25\sqrt{5} + 1}{120}.
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