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Math Help - Vector calculus: simple surface integral problem

  1. #1
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    Vector calculus: simple surface integral problem

    At least, it should be simple!

    I am working my way through div, grad, curl and all that by H. M. Schey and have just finished the chapter on surface integrals and the divergence. I don't seem able to get the right answer for one of the problems, which suggests it should be possible to evaluate the surface integral without going through long-winded calculations.

    I need to find
    \iint_{S} \mathbf{F}\cdot \hat{\mathbf{n}} dS,
    the surface integral over surface S.

    The surface S is open and consists of 3 squares with side length b, forming half a cube at the origin. The 3 squares are adjacent and there is one in each of the xy, xz, and yz planes.

    The vector function is:
    \mathbf{F}=\hat{\mathbf{i}}x+\hat{\mathbf{j}}y+\ha  t{\mathbf{k}}z.

    My thinking is as follows: the three squares are identical, so I need only find the surface integral for one square and multiply it by 3.

    I start with S_{1}, the square in the xy plane. The unit normal vector, \hat{\mathbf{n}}, is equal to \hat{\mathbf{k}} and the surface integral becomes:

    \iint_{S_{1}} \mathbf{F} \cdot \hat{\mathbf{k}} dS =\int_{x=0}^{x=b}\int_{y=0}^{y=b} z dx dy = b^{2} z

    I know the surface integral should overall evaluate to zero, but repeating this for the 3 surfaces, I get

    \iint_{S} \mathbf{F} \cdot \hat{\mathbf{n}} dS = b^{2}(x+y+z).

    I'm fairly sure I've missed something here, but not sure what. Any suggestions?

    Thanks for taking the time to read this.
    Last edited by Jnix; May 25th 2011 at 01:03 AM.
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  2. #2
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    Quote Originally Posted by Jnix View Post
    At least, it should be simple!

    I am working my way through div, grad, curl and all that by H. M. Schey and have just finished the chapter on surface integrals and the divergence. I don't seem able to get the right answer for one of the problems, which suggests it should be possible to evaluate the surface integral without going through long-winded calculations.

    I need to find
    \iint_{S} \mathbf{F}\cdot \hat{\mathbf{n}} dS,
    the surface integral over surface S.

    The surface S is open and consists of 3 squares with side length b, forming half a cube at the origin. The 3 squares are adjacent and there is one in each of the xy, xz, and yz planes.

    The vector function is:
    \mathbf{F}=\hat{\mathbf{i}}x+\hat{\mathbf{j}}y+\ha  t{\mathbf{k}}z.
    The "t" in that last is an artifact of the LaTeX. I'm not sure how it got there- I can't find an error in the code!

    My thinking is as follows: the three squares are identical, so I need only find the surface integral for one square and multiply it by 3.

    I start with S_{1}, the square in the xy plane. The unit normal vector, \hat{\mathbf{n}}, is equal to \hat{\mathbf{k}} and the surface integral becomes:

    \iint_{S_{1}} \mathbf{F} \cdot \hat{\mathbf{k}} dS =\int_{x=0}^{x=b}\int_{y=0}^{y=b} z dx dy = b^{2} z
    Have you forgotten that this is in the xy plane? z= 0 in the xy plane!

    I know the surface integral should overall evaluate to zero, but repeating this for the 3 surfaces, I get

    \iint_{S} \mathbf{F} \cdot \hat{\mathbf{n}} dS = b^{2}(x+y+z).

    I'm fairly sure I've missed something here, but not sure what. Any suggestions?

    Thanks for taking the time to read this.
    Follow Math Help Forum on Facebook and Google+

  3. #3
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    Bristol, UK
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    Oh good point!

    I think the mysterious t is something to do with the \hat{} tag not being processed correctly on the k. No idea why, though.

    I supposed this is true for all the other faces, too, as in each case, x,y or z=0 for the entire surface. Thanks
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