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Math Help - More Triple Integrals!!!!

  1. #1
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    Exclamation More Triple Integrals!!!!

    i have an idea on how to solve this problem, but it seems like itd be way too much work to even be possible...

    triple integral over R of dV / (x^2 + y^2 + z^2)
    where R is the region between:
    x^2 + y^2 + z^2 = 1
    x^2 + y^2 + z^2 = 4

    am i correct in thinking that first i have put convert to spherical coordinates and then insert pdpdthetadphi?

    ps how do you insert the integral symbol and all that fun stuff?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Dubulus View Post
    i have an idea on how to solve this problem, but it seems like itd be way too much work to even be possible...

    triple integral over R of dV / (x^2 + y^2 + z^2)
    where R is the region between:
    x^2 + y^2 + z^2 = 1
    x^2 + y^2 + z^2 = 4

    am i correct in thinking that first i have put convert to spherical coordinates and then insert pdpdthetadphi?

    ps how do you insert the integral symbol and all that fun stuff?
    I'll answer your last question first. We use LaTeX here at the forums.

    As I answer you're problem, I'll show what code was used.

    So we are to integrate \iiint\limits_G x^2+y^2+z^2 \,dV?

    Code:
    [tex]\iiint\limits_G x^2+y^2+z^2 \,dV[/tex]
    Note that the region is defined as the area between two circles [when looking down at the xy plane]

    So our limits for the radius would then be from 1 to 2.

    The best coordinate system to use would be the spherical coordinate system. We see that \varrho is defined between 1 and 2.

    \varphi would be defined between 0 and \pi

    \vartheta would be defined between 0 and 2\pi

    So our triple integral would be \int_0^{2\pi}\int_0^{\pi}\int_1^2 \varrho^4\sin\varphi \,d\varrho\,d\varphi\,d\vartheta

    Code:
    [tex]\int_0^{2\pi}\int_0^{\pi}\int_1^2 \varrho^4\sin\varphi \,d\varrho\,d\varphi\,d\vartheta[/tex]
    Does this make sense?

    --Chris
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  3. #3
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    answer...

    most of it makes sense yes but where does the
    \varrho^4\sin\varphi\ come from? is that from pluggin in the spherical coordinates into f(x,y,z)?
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    So we are to integrate \iiint\limits_G x^2+y^2+z^2 \,dV?
    ..
    Quote Originally Posted by Dubulus View Post
    triple integral over R of dV / (x^2 + y^2 + z^2)
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  5. #5
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    im afraid i dont know where to begin with taht sum. the point of the problem is using spherical coordinates.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by wingless View Post
    ..
    Shoot...

    \iiint\limits_G\frac{\,dv}{x^2+y^2+z^2}??

    Thus, \iiint\limits_G\frac{\,dv}{x^2+y^2+z^2}\implies\ii  int\limits_{G_{\varrho\varphi\vartheta}}\frac{1}{\  varrho^2}\varrho^2\sin\varphi\,dV_{\varrho\varphi\  vartheta}\implies\iiint\limits_{G_{\varrho\varphi\  vartheta}}\sin\varphi\,dV_{\varrho\varphi\vartheta  }

    To clear things up a little bit, this is the integral setup when we go from rectangular to spherical:

    \int_k^l\int_c^d\int_a^b f(x,y,z)\,dx\,dy\,dz\implies\int_{\vartheta_0}^{\v  artheta}\int_{\varphi_0}^{\varphi}\int_{\varrho_0}  ^{\varrho}f(\varrho,\varphi,\vartheta)\varrho^2\si  n\varphi\,d\varrho\,d\varphi\,d\vartheta

    The \varrho^2\sin\varphi term comes out of the conversion to spherical coordinates.

    I hope this clarifies things for now...for I'm dead tired. I'll post more on this tomorrow...I mean today...

    --Chris
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