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Math Help - Find a volume

  1. #1
    MHF Contributor arbolis's Avatar
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    Find a volume

    I must find the volume within the superficie r=a\sin \varphi (that's spherical coordinates).
    My attempt : If I understand well they ask me to find the volume of a sphere whose radius is a so I should find \frac{4\pi a^3}{3}.

    I set V=4\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^{2a \sin \varphi} rdr d\varphi d\theta.
    I found that the dr integral is worth 2a^2\sin^2 \varphi. The d\varphi is worth \frac{\pi}{4}-\frac{\sin \varphi \cos \varphi}{2}. But I don't think I'm right on this.
    I'd like to know if I set up well the integral for the asked volume and if I made an error calculating the 2 first integrals. Thanks.
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  2. #2
    Senior Member Spec's Avatar
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    That looks a bit strange. You have 0\leq r \leq \alpha \sin \varphi

    I'm not sure which angles the variables are referring to, but my textbooks usually have 0\leq \varphi \leq2\pi and 0\leq \theta \leq \pi. You might need to switch those depending on what the question assumes.

    The area transform for spherical coordinates is r^2\sin\theta (again, assuming 0\leq \theta \leq \pi), not r.
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  3. #3
    MHF Contributor arbolis's Avatar
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    I asked a professor today about this exercise, he told me that r=a\sin \varphi is a torus and not a sphere!
    So I've redone the exercise, I set up V=\int_0^{2\pi} \int_0^{\pi} \int_0^{a\sin \varphi } drd\varphi d\theta =4a\pi. But I don't trust my result, especially because I don't have any \pi ^3.
    Last edited by arbolis; June 16th 2009 at 03:01 PM.
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  4. #4
    Senior Member Spec's Avatar
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    You're still using spherical coordinates, so you have to include the area transform r^2 \sin \theta.

    EDIT: Are you sure it's spherical coordinates and not cylindrical coordinates? I guess you would have some limits for z then ...
    Last edited by Spec; June 16th 2009 at 04:23 PM.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Spec View Post
    You're still using spherical coordinates, so you have to include the area transform r^2 \sin \theta.

    EDIT: Are you sure it's spherical coordinates and not cylindrical coordinates? I guess you would have some limits for z then ...
    Yes I am sure it's spherical coordinates, the problem states so. So I don't have to worry about the z coordinate. It's a strange torus, it looks like a compressed sphere I think.
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    yes, it is a torus, and to imagine things easier, the torus here is the same as the one obtained by rotating the circle x^2+y^2=ax about y axix.

    anyway, to find the volume, just use the spherical coordinates (i'll use the more standard notation \rho instead of r):

    V=\int_0^{2\pi} \int_0^{\pi} \int_0^{a\sin \phi} \rho^2 \sin \phi \ d \rho \ d \phi \ d \theta=\frac{2 \pi a^3}{3} \int_0^{\pi} (\sin \phi)^4 \ d \phi=\frac{\pi^2a^3}{4}.
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  7. #7
    Senior Member Spec's Avatar
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    Ah, different variables used for the angles. Doh! This is the third thread I've seen where there's been confusion because of different standards for which variables describe certain angles. The books I have, have switched those variables around.
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  8. #8
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    yes, it is a torus, and to imagine things easier, the torus here is the same as the one obtained by rotating the circle x^2+y^2=ax about y axix.

    anyway, to find the volume, just use the spherical coordinates (i'll use the more standard notation \rho instead of r):

    V=\int_0^{2\pi} \int_0^{\pi} \int_0^{a\sin \phi} \rho^2 \sin \phi \ d \rho \ d \phi \ d \theta=\frac{2 \pi a^3}{3} \int_0^{\pi} (\sin \phi)^4 \ d \phi=\frac{\pi^2a^3}{4}.
    Thanks a lot for the clarification.
    Are you sure that x^2+y^2=ax and not x^2+y^2=ay?
    I don't know how to check it out. I get x^2+y^2=a^2\sin^2 \varphi.
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    Quote Originally Posted by arbolis View Post
    Thanks a lot for the clarification.
    Are you sure that x^2+y^2=ax and not x^2+y^2=ay?
    I don't know how to check it out. I get x^2+y^2=a^2\sin^2 \varphi.
    we're both right! haha ... x^2+y^2=ax rotated about y axis (the one that i said) and x^2+y^2=ay rotated about x axis give the same torus.
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  10. #10
    MHF Contributor arbolis's Avatar
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    I'm thinking about it and I still can't derive that V=\int_0^{2\pi} \int_0^{\pi} \int_0^{a\sin \phi} \rho^2 \sin \phi \ d \rho \ d \phi \ d \theta. I understand the limits of the integral, but not why I have to integrate \rho^2 \sin \phi.
    \rho^2 \sin \phi=a^2 \sin ^3 \phi.
    I'm also banging my head to the walls to realize that it's a torus. I don't know how to derive that it's a circle rotating. Maybe by fixing a \theta and \phi...
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  11. #11
    Super Member Random Variable's Avatar
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     \rho^2 \sin \phi \ d \rho \ d \phi \ d \theta is the volume element when you're integrating in spherical coordinates
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  12. #12
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    Quote Originally Posted by arbolis View Post
    I'm thinking about it and I still can't derive that V=\int_0^{2\pi} \int_0^{\pi} \int_0^{a\sin \phi} \rho^2 \sin \phi \ d \rho \ d \phi \ d \theta. I understand the limits of the integral, but not why I have to integrate \rho^2 \sin \phi.
    \rho^2 \sin \phi=a^2 \sin ^3 \phi.
    I'm also banging my head to the walls to realize that it's a torus. I don't know how to derive that it's a circle rotating. Maybe by fixing a \theta and \phi...
    ok, in my first post in this thread i tried to simplify things by eliminating z axis but if you want to see things in 3 dimensional here is how it works:

    there's no \theta in the equation \rho=a \sin \phi, which means we only need to find the cross section of the surface with some \theta and then rotate the cross section about z axis because 0 \leq \theta \leq 2\pi.

    so, for example, let \theta=0. then you get x=\rho \sin \phi=a\sin^2 \phi, \ y = 0, \ z=\rho \cos \phi. so the cross section is the circle x^2+z^2=\rho^2=a^2\sin^2 \phi=ax, \ 0 \leq \phi \leq \pi, in the plane y=0, which is

    xz plane. sketch this circle and then rotate it about z axis. what do you get? a torus of course!

    to answer your second question, \rho^2 \sin \phi is the Jacobian of change of variables from cartisian to spherical coordinates, exactly like r in polar. you always have to put it when calculating the

    volume in spherical coordinates. also the tripple integral is over the region bounded by the surface of the torus and not over the surface of the torus. so 0 \leq \rho \leq a \sin \phi and not \rho=a\sin \phi.
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  13. #13
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    ok, in my first post in this thread i tried to simplify things by eliminating z axis but if you want to see things in 3 dimensional here is how it works:

    there's no \theta in the equation \rho=a \sin \phi, which means we only need to find the cross section of the surface with some \theta and then rotate the cross section about z axis because 0 \leq \theta \leq 2\pi.

    so, for example, let \theta=0. then you get x=\rho \sin \phi=a\sin^2 \phi, \ y = 0, \ z=\rho \cos \phi. so the cross section is the circle x^2+z^2=\rho^2=a^2\sin^2 \phi=ax, \ 0 \leq \phi \leq \pi, in the plane y=0, which is

    xz plane. sketch this circle and then rotate it about z axis. what do you get? a torus of course!

    to answer your second question, \rho^2 \sin \phi is the Jacobian of change of variables from cartisian to spherical coordinates, exactly like r in polar. you always have to put it when calculating the

    volume in spherical coordinates. also the tripple integral is over the region bounded by the surface of the torus and not over the surface of the torus. so 0 \leq \rho \leq a \sin \phi and not \rho=a\sin \phi.
    Thank you very much. All is clear now.
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