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Math Help - Surface Flux Integral

  1. #1
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    Surface Flux Integral

    Hi People

    New here, so please be kind!

    I am doing some work for my maths module, I have got so far but I am not sure if it is correct, and if not, why not. It is important to me to getting it correct because this will also be on my exam so any help you can give me will be very much appreciated.

    Thanks in advance!

    The question:

    Region V of a cylinder is denoted by x^2+y^2\leq 1, 0 \leq z \leq 2 and S is the surface of this cylinder.

    Vector field F(x,y,z) = xi + yi + z(x^2+y^2)k

    Now, I have been instructed earlier in the question to convert F into cylindrical co-ordinates, so...

    F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}

    I am then instructed to find the surface integral. I do this, but am unable to do it in cylindrical co-ordinates. Is this correct?

    I have a feeling it is not, because at the end of it I am left with an answer simply of 8pi

    Once again, thanks for any help and if you need any more workings and such I can provide them.
    Last edited by caliber; April 30th 2010 at 05:27 AM.
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  2. #2
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    Quote Originally Posted by caliber View Post
    Hi People

    New here, so please be kind!

    I am doing some work for my maths module, I have got so far but I am not sure if it is correct, and if not, why not. It is important to me to getting it correct because this will also be on my exam so any help you can give me will be very much appreciated.

    Thanks in advance!

    The question:

    Region V of a cylinder is denoted by x^2+y^2\leq 1, 0 \leq z \leq 2 and S is the surface of this cylinder.

    Vector field F(x,y,z) = xi + yi + z(x^2+y^2)k

    Now, I have been instructed earlier in the question to convert F into cylindrical co-ordinates, so...

    F(\rho,\varphi,z) = \rho \widehat{\rho} + z\rho^{2}\widehat{z}

    I am then instructed to find the surface integral. I do this, but am unable to do it in cylindrical co-ordinates. Is this correct?

    I have a feeling it is not, because at the end of it I am left with an answer simply of 8pi

    Once again, thanks for any help and if you need any more workings and such I can provide them.
    First off, there are 3 surfaces - top, bottom and side. Top and bottom are straight forward leading to \pi (top) and 0 (bottom). The side is a little more complicated. Parametrize the surface

     <br />
{\bf r} = < \cos \theta, \sin \theta, z>,\;\;\; 0 \le \theta \le 2 \pi,\;\;\; 0 \le z \le 2.<br />

    So the flux is given by

     <br />
\iint_D {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z dA = \int_0^2 \int_0^{2 \pi} dz d \theta\ = 4 \pi\;\; \text{since}\; {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z = 1<br />

    Total flux is therefore  4\pi + \pi = 5 \pi.

    To check, use the divergence theorem.
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  3. #3
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    Quote Originally Posted by Danny View Post
    First off, there are 3 surfaces - top, bottom and side. Top and bottom are straight forward leading to \pi (top) and 0 (bottom). The side is a little more complicated. Parametrize the surface

     <br />
{\bf r} = < \cos \theta, \sin \theta, z>,\;\;\; 0 \le \theta \le 2 \pi,\;\;\; 0 \le z \le 2.<br />

    So the flux is given by

     <br />
\iint_D {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z dA = \int_0^2 \int_0^{2 \pi} dz d \theta\ = 4 \pi\;\; \text{since}\; {\bf F} \cdot {\bf r}_{\theta} \times {\bf r}_z = 1<br />

    Total flux is therefore  4\pi + \pi = 5 \pi.

    To check, use the divergence theorem.
    Hi

    Thanks for that. How would I use divergence theorem for this? I am aware of divergence theorem, but not sure how to use it for cylindrical co-ordinates.

    Thanks
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  4. #4
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    Quote Originally Posted by caliber View Post
    Hi

    Thanks for that. How would I use divergence theorem for this? I am aware of divergence theorem, but not sure how to use it for cylindrical co-ordinates.

    Thanks
    I think it's better to use the Div thm. in cartesian and then switch the triple integral. So, if

    {\bf F } = <x,y,(x^2+y^2)z> \; \text{so}\; {\bf \nabla \cdot F} = 1 + 1 + x^2+y^2

    so \iiint\limits_V ( 2 + x^2 + y^2) dV = \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta.
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  5. #5
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    Thanks, However on working through this i get:

    \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta = \int_0^{2 \pi} \int_0^1 \int_0^2 (2r + r^3)  dz\, dr\, d\, \theta

    = \int_0^{2 \pi} \int_0^1 2(2r + r^2) dr\, d\, \theta

    = \int_0^{2 \pi} -(8/3) d\, \theta

    = -16\pi/3

    which is incorrect?
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  6. #6
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    Quote Originally Posted by caliber View Post
    Thanks, However on working through this i get:

    \int_0^{2 \pi} \int_0^1 \int_0^2 (2 + r^2) r dz\, dr\, d\, \theta = \int_0^{2 \pi} \int_0^1 \int_0^2 (2r + r^3) dz\, dr\, d\, \theta

    = \int_0^{2 \pi} \int_0^1 2(2r + r^2) dr\, d\, \theta

    = \int_0^{2 \pi} -(8/3) d\, \theta

    = -16\pi/3

    which is incorrect?
    First let me say that you're integrating 2r + r^3 over a positive range i.e., [0,2 \pi] \times [0,1] \times [0,2] so the answer must be non-negative. Check your integration.
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  7. #7
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    Quote Originally Posted by Danny View Post
    First let me say that you're integrating 2r + r^3 over a positive range i.e., [0,2 \pi] \times [0,1] \times [0,2] so the answer must be non-negative. Check your integration.
    apologies, please disregard the negative sign.
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