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Math Help - flux using divergence theorem

  1. #1
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    flux using divergence theorem

    Let W be the solid bounded by the paraboloid x = y^2 + z^2 and the plane x = 9. Let  \vec{F} = 5x\vec{i}  + y\vec{j}  + z\vec{k} . I know the divF is 7 here.. but after that I will need to multiply this by the volume of the region bounded above to get the flux.. am I right?
    Last edited by mr fantastic; April 28th 2009 at 11:04 PM. Reason: Fixed latex closing tag
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by TheRekz View Post
    Let W be the solid bounded by the paraboloid x = y^2 + z^2 and the plane x = 9. Let  \vec{F} = 5x\vec{i}  + y\vec{j}  + z\vec{k} . I know that div(\vec{F})=7 here.. but after that I will need to multiply this by the volume of the region bounded above to get the flux.. am I right?
    Yes, you want to integrate \int\int\int 7\,dV.

    I would use cylindrical coordinates (note that I'm using x where one would traditionally use z):
    x=x
    r^2=y^2+z^2
    dV=r\,dx\,dr\,d\theta

    So your volume is bounded by x=r^2 and x=9

    \theta=0..2\pi
    r=0..3
    x=r^2..9

    7\int_0^{2\pi}\int_0^3\int_{r^2}^9 r\,dx\,dr\,d\theta

    Spoiler:
    Inner integral: \left[rx\right]_{r^2}^9 = 9r-r^3

    Now you have 7\int_0^{2\pi}\int_0^3 9r-r^3\,dr\,d\theta

    Middle integral: \left[\frac{9r^2}{2}-\frac{r^4}{4}\right]_0^3 = \frac{81}{2}-\frac{81}{4} = \frac{81}{4}

    Now you have \frac{7\cdot 81}{4}\int_0^{2\pi}\,d\theta = \frac{2\pi\cdot 7\cdot 81}{4} = \frac{567\pi}{2}


    PS: The closing tag for LaTeX is "[/tex]", not "[\math]".
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  3. #3
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    hey thanks for that, what if it asks:

    Let S1 be the paraboloid surface oriented in the negative x direction, what is the flux through S1?
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