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Thread: Surface integral questions

  1. #1
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    Surface integral questions

    I need to find the surface integrals for these various questions. Some I don't know where to begin, others I don't know how to continue. Help would be appreciated



    Not sure about this one



    I get: $\displaystyle \int \int F \cdot n dS$
    $\displaystyle F = (x,y,2z)$
    $\displaystyle n = \frac {2x}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {2y}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {1}{\sqrt {4x^2 + 4y^2 + 1}}$
    $\displaystyle \int \int F \cdot n dS = \int \int 2 dA $

    But not sure what the limits are to proceed



    I'm guessing similar manner to question above, but do I get z on its own i.e. $\displaystyle z = 3-3x-\frac {3}{2}y$ then proceed. But once again, not sure what the limits would be



    Using divF = 3 then $\displaystyle \int^1_0 \int^1_0 \int^1_0 3 dzdxdy = 3 $ Is that right?



    Need to use divergence again, but not sure how to proceed.



    I could do the same as below, but not sure how to proceed



    I got the curl to be i + j + k and $\displaystyle \sqrt { ( \frac {\partial f}{\partial x})^2 + ( \frac {\partial f}{\partial y})^2 + 1 } = \sqrt {3}$
    Not sure how to proceed
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    I didn't read all your problems - way too many in one post! But I can tell you the first one is easily solved by the divergence theorem. Find $\displaystyle \mbox{div } F$ and integrate $\displaystyle \mbox{div }F\ dV$ inside the cube.
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by godiva View Post
    I need to find the surface integrals for these various questions. Some I don't know where to begin, others I don't know how to continue. Help would be appreciated



    Not sure about this one



    I get: $\displaystyle \int \int F \cdot n dS$
    $\displaystyle F = (x,y,2z)$
    $\displaystyle n = \frac {2x}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {2y}{\sqrt {4x^2 + 4y^2 + 1}} + \frac {1}{\sqrt {4x^2 + 4y^2 + 1}}$
    $\displaystyle \int \int F \cdot n dS = \int \int 2 dA $

    But not sure what the limits are to proceed



    I'm guessing similar manner to question above, but do I get z on its own i.e. $\displaystyle z = 3-3x-\frac {3}{2}y$ then proceed. But once again, not sure what the limits would be



    Using divF = 3 then $\displaystyle \int^1_0 \int^1_0 \int^1_0 3 dzdxdy = 3 $ Is that right?



    Need to use divergence again, but not sure how to proceed.



    I could do the same as below, but not sure how to proceed



    I got the curl to be i + j + k and $\displaystyle \sqrt { ( \frac {\partial f}{\partial x})^2 + ( \frac {\partial f}{\partial y})^2 + 1 } = \sqrt {3}$
    Not sure how to proceed

    for the second problem change it into polar and use the limits

    $\displaystyle \int^{1}_{0}\int^{2\pi}_{0}d\theta dr$

    for the 3rd one the limits are

    $\displaystyle \int^{\frac{3}{2}}_{0} \int^{-2x+3}_{0} dydx$

    Correct on 4

    For question 5 find the divergence and then use

    $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} drd\theta dz$
    Last edited by 11rdc11; May 15th 2010 at 10:26 PM.
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    Quote Originally Posted by 11rdc11 View Post
    For question 5 find the divergence and then use

    $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} drd\theta dz$
    Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

    Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by godiva View Post
    Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

    Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example
    Yes for the 6th one remember what Stokes theorem says. It does not matter what surface you use as long as C is the same. In this case it be alot easier to use the surface of z=1 with the domain being projected to the xy plane and you get the integral

    $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} d\theta dr$

    Another way to do 6 is let

    $\displaystyle r(t)= \cos{t}i +\sin{t}j + k$

    $\displaystyle f(r(t)) = \cos^2{t}i + \sin^2{t}j +k$

    $\displaystyle r'(t) = -\sin{t}i +\cos{t}j + 0k$

    Now just evaluate the line integral
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by godiva View Post
    Thanks. For the above would an r need to be added? I remember something like that when i did coordinates, i.e. should it be $\displaystyle \int^{1}_{0} \int^{2\pi}_{0} \int^{a}_{0} r drd\theta dz$

    Also, any ideas for the final 2? I've gone through my book twice and it doesn't have a proper example

    And for the last one I got a different curl than you did.

    Mine was

    $\displaystyle -i +j -k$

    so

    $\displaystyle \int \int [-Mf_x - Nf_y +P]dxdy$

    f_x = -1

    f_y = -1

    and your integrals would be

    $\displaystyle \int^{1}_{0} \int^{1-x}_{0} dydx$
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