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Math Help - Triple integral

  1. #1
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    Triple integral

    Hi, I have to compute de volume for the region:

    \{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}

    I've tried to do by two different parametrizations, in spherics and cylindrical coordinates

    For cylindrical coordinates I've made:

    3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}

    For spherics I've considered 4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}

    3 \displaystyle\int_{0}^{Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }

    I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping

    Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Triple integral

    Quote Originally Posted by Ulysses View Post
    For cylindrical coordinates I've made:
    3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}
    You forgot to multiply by r .
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Triple integral

    Quote Originally Posted by Ulysses View Post
    Hi, I have to compute de volume for the region:

    \{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}

    I've tried to do by two different parametrizations, in spherics and cylindrical coordinates

    For cylindrical coordinates I've made:

    3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}

    For spherics I've considered 4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}

    3 \displaystyle\int_{0}^{Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }

    I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping

    Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem

    I will suggesting something, hope it is true.


    Let us look on equation of a circle: x^2+y^2=16.

    We can simplify your problem to the following problem:

    Finding the volume of revolution of x^2+y^2=16 and y=1 around y-axis.
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    Re: Triple integral

    I think you're right Fernando. I haven't used the jacobian in any integral because I thought of it directly in cylindrical and spherical coordinates, I haven't made a change of variables at any step I think, should I involve the jacobian? I think I see it now, but I'm not completely sure, that I actually made a transformation of coordinates at first in both cases, while considering cartesian to polar and cylindrical coordinates. I think that the confusion drives from the fact that in the surface integral the jacobian matrix doesn't appear in the same way that it does on the ordinary integrals. Why is this? What a silly mistake I've made, thank you very much.

    Alsosprach Zaratustra, I don't understand what you purposed
    Last edited by Ulysses; August 8th 2011 at 02:56 PM.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Triple integral

    Quote Originally Posted by Ulysses View Post
    Alsosprach Zaratustra, I don't understand what you purposed

    I will try to explain.

    I say that instead of looking for volume of  \{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}, which is a volume of dome of sphere, so we can to look at the following congruent problem in 2D instead in 3D:

    First, draw yourself a sketch of a circle with radius 4, now draw a line y=1(play the role of z\geq 1).

    Now, compute the volume of revolution with the functions: x^2+y^2=16, and y=1 around y-axis.

    See my sketch...
    Attached Thumbnails Attached Thumbnails Triple integral-vor.png  
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  6. #6
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    Re: Triple integral

    Now I get it. Thanks.
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  7. #7
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    Re: Triple integral

    Well, I added the jacobians, but still giving different, I don't know why. The difference now is more sutil: In cylindrical coord. gives (clic here)

    In spherical coord gives

    I tried to proceed with cylindrical, but the exercise gets more complicated, so I want to be sure that what I'm doing with spherical is ok, so I can proceed that way, but still giving different answers.

    I've also tried using a solid of revolution: http://www.wolframalpha.com/input/?i=Int_-4^4+Pi+%28%2816-x^2%29-1%29

    I don't even know of any of this results is okey, perhaps they are all wrong :P
    Last edited by Ulysses; August 9th 2011 at 07:40 AM.
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