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Math Help - Triple integral in spherical coordinates

  1. #1
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    Triple integral in spherical coordinates

    Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

    Evaluate \int \int \int_E x^2 dV where E is bounded by the xz-plane and the hemispheres y = \sqrt{9-x^2-z^2} and y = \sqrt{16-x^2-z^2}.

    So my solution is this:

    Using spherical coordinates, we have 3 \le \rho \le 4, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}

    Thus \int \int \int_E x^2 dV = \int_{0}^{\frac{\pi}{2}} \int_0^{\pi} \int_3^4 (\rho \sin(\phi)\cos(\theta))^2\rho^2\sin(\phi) d\rho d\theta d\phi = \frac{781\pi}{15} <br />
    But the answer is \frac{1562\pi}{15} hmmm how did they get this?

    Thanks!
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  2. #2
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    The azimuthal angle \theta will range from 0 to \pi. That you have correct. What's not correct is that the polar angle \phi will also range from 0 to \pi. That may change things a bit for you. I'm not seeing any other obvious problems. See what you get with this.
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  3. #3
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    ohhhhh right, lol omg I sketched the graph wrong, it's y = f(x,z) so y is the axis of symmetry!

    EDIT: thanks Ackbeet!!
    Last edited by usagi_killer; September 23rd 2010 at 08:32 AM. Reason: saying thanks :)
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  4. #4
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    Right. Did you get the correct answer, then?
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Ackbeet View Post
    The azimuthal angle \theta will range from 0 to \pi. That you have correct. What's not correct is that the polar angle \phi will also range from 0 to \pi. That may change things a bit for you. I'm not seeing any other obvious problems. See what you get with this.
    Edit- I am incorrect. I did not see that we were bounded by the xz plane. OOPS!

    The azimuthal will range from 0 to  \frac{ \pi }{2} and theta will range from 0 to  2 \pi ....we want to cover the entire top half the hemisphere so there is no reason why theta would not be 360 degrees. Likewise pi over two corresponds with the top half (measuring from the Z-axis to the XY plane).
    Last edited by AllanCuz; September 23rd 2010 at 11:53 AM.
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    Reply to AllanCuz at Post # 5:

    I think you're a bit confused. The polar angle in this problem is \phi, and inhabits the natural range [0,\pi] in any spherical coordinates problem. The azimuthal angle in this problem is \theta, and inhabits the natural range [0,2\pi] in any spherical coordinates problem. However, in this problem, the shape described, which is that shape bounded by the xz plane, and the two hemispheres \rho=3 and \rho=4, you have the artificial restriction \theta\in[0,\pi]. There is no restriction on the polar angle.

    Hence your limits, while producing the same answer for the integral as my limits, do not describe the shape in the original problem.

    Draw a picture (as I did); that might help you see what I'm talking about.

    [EDIT] Ah, I see you saw your mistake.
    Last edited by Ackbeet; September 23rd 2010 at 11:54 AM. Reason: AllanCuz fixed his post himself.
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Reply to AllanCuz at Post # 5:

    I think you're a bit confused. The polar angle in this problem is \phi, and inhabits the natural range [0,\pi] in any spherical coordinates problem. The azimuthal angle in this problem is \theta, and inhabits the natural range [0,2\pi] in any spherical coordinates problem. However, in this problem, the shape described, which is that shape bounded by the xz plane, and the two hemispheres \rho=3 and \rho=4, you have the artificial restriction \theta\in[0,\pi]. There is no restriction on the polar angle.

    Hence your limits, while producing the same answer for the integral as my limits, do not describe the shape in the original problem.

    Draw a picture (as I did); that might help you see what I'm talking about.
    I am not confused. I just didn't notice we are bounded by the zx plane. You were correct when you changed theta to pi. I thought the problem was to the xy plane which produces a considerably different setup.
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  8. #8
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    Your limits, actually, are a legitimate way of solving the problem if you think of the whole shape as rotated about the x axis through \pi/2 radians such that \hat{j} becomes \hat{k}. So, if you believe that rotations preserve volumes and such, and since the integrand is insensitive to that rotation, your limits produce the same result.
    Last edited by Ackbeet; September 23rd 2010 at 12:00 PM. Reason: Grammar.
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