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Math Help - triple integration in spherical

  1. #1
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    triple integration in spherical

    I am having trouble setting up the following integral problem: "Find the volume of the solid that lies inside the cone z^2 = 3x^2 + 3y^2 and between the spheres x^2 + y^2 + z^2 = 1 and x^2 +y^2 + z^2 = 9 "

    I know that it must be done with spherical coordinates and that it will be a triple integral. However, I'm a little unsure how to set up the limits and the integral itself. Thanks for any help
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    \iiint_DdV we have spheres with radii of 1 and 3, and z=\sqrt{3(x^2+y^2)} so converting to spherical we have

    \int_0^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\  int_1^3 p^2\sin(\phi)dp d\phi d\theta

    upon evaluating this (I'll leave that for you to do) you should get \frac{13\pi}{3}.

    if you need more clarification just let me know.
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    Quote Originally Posted by putnam120 View Post
    \iiint_DdV we have spheres with radii of 1 and 3, and z=\sqrt{3(x^2+y^2)} so converting to spherical we have

    \int_0^{2\pi}\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\  int_1^3 p^2\sin(\phi)dp d\phi d\theta

    upon evaluating this (I'll leave that for you to do) you should get \frac{13\pi}{3}.

    if you need more clarification just let me know.
    How you get that?

    For the limits I get,
    0\leq \theta\leq 2\pi
    0\leq \phi \leq \pi/3
    1\leq r\leq 3

    And then you double your result because there are two parts. The solid above and below. And you cannot express that as a single bound on \phi
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    ok i agree i made a mistake. but i think the range for \phi should be 0\to\frac{\pi}{6}

    this is because if we solve the equation of the cone we have

    p^2\cos^2(\phi)=3p^2\sin^2(\phi)\Longrightarrow\ta  n(\phi)=\frac{1}{\sqrt{3}}\Longrightarrow\phi=\fra  c{\pi}{6}
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    Quote Originally Posted by putnam120 View Post
    ok i agree i made a mistake. but i think the range for \phi should be 0\to\frac{\pi}{6}

    this is because if we solve the equation of the cone we have

    p^2\cos^2(\phi)=3p^2\sin^2(\phi)\Longrightarrow\ta  n(\phi)=\frac{1}{\sqrt{3}}\Longrightarrow\phi=\fra  c{\pi}{6}
    If I have done this right this time, hit-and-miss Monte-Carlo gives this
    integral is ~=1.83 +/- ~0.02 (that's +/- 2 sigma error), or about 6-7%
    of the volume of the 3x3x3 circumscribed cube.

    Aparently it's gone wrong - I'll have to check the code when I get the chance.

    OK now its fixed:

    Code:
    >
    >NN=100000;
    >x=(random(NN,3)-0.5)*6;
    >
    >
    >nx=x(:,1)^2+x(:,2)^2+x(:,3)^2;
    >mx=3*x(:,1)^2+3*x(:,2)^2-x(:,3)^2;
    >
    >ll=(nx>1)&&(nx<9)&&(mx<0);ll=sum(ll')/NN*(6^3)
          14.6232
    Integral ~=14.6+/- ~0.34 (that's +/- 2 sigma error)

    RonL
    Last edited by CaptainBlack; November 21st 2006 at 10:07 AM.
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  6. #6
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    I get:

    V=\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{6}}\int_{1  }^{3}{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}=  \frac{26(2-\sqrt{3}){\pi}}{3}\approx{7.295}
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    Quote Originally Posted by galactus View Post
    I get:

    V=\int_{0}^{2{\pi}}\int_{0}^{\frac{\pi}{6}}\int_{1  }^{3}{\rho}^{2}sin({\phi})d{\rho}d{\phi}d{\theta}=  \frac{26(2-\sqrt{3}){\pi}}{3}\approx{7.295}
    I agree with your limits.
    I would just multiply by 2 becaus there are two parts the upper and lower part when you draw these quadradic surfaces.

    How did you get that angle for the cone? I got it from a dot product between the z-axis and the vector on a cone. How did you?
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  8. #8
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    Hey PH.

    I just converted rectangular to spherical coordinates.

    z=\sqrt{3(x^{2}+y^{2})}

    {\rho}cos({\phi})=\sqrt{3[{\rho}^{2}sin^{2}({\phi})cos^{2}({\theta})+{\rho}^  {2}sin^{2}({\phi})sin^{2}({\theta})]}

    {\rho}cos({\phi})=\sqrt{3}{\rho}sin({\phi}), (Actually, this should be \sqrt{3}|psin({\phi})|)

    cot({\phi})=\sqrt{3}

    {\phi}=\frac{\pi}{6}

    May I see your dot product method?.
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    Quote Originally Posted by galactus View Post
    May I see your dot product method?.
    In spherical coordinates the equation,
    \phi = \mbox{konstant} describes a cone (half cone).

    So, \phi=\pi/4 describes a cone above the xy-plane with angle (with z-axis) of 45^o. If \phi=\pi/2 we get the xy-plane. If \phi=3\pi/4 we get the same cone but upside down. Note, these equations only describe a half-cone. A two sided cone appears in this example which is why we need to double the integral.

    The question is the upper side of the cone z^2=3(x^2+y^2) is what equation in spherical coordinates, that is \phi=?. To find the angle we need to find the angle with the lateral side of the cone (which is a straight line) and the z-axis (that is definition of \phi). To do that we can use the dot product. Select a point on the upper-cone, say (0,1,\sqrt{3}). Select a point on the z-axis (0,0,1).
    The vector from the origon to point on cone is \bold{i}+\bold{j}\sqrt{3} and the vector from origon to z-axis is \bold{k}.

    Now you take the dot product to find the angle,
    \bold{u}\cdot \bold{v}=||\bold{u}||\cdot ||\bold{v}|| \cos x
    Which is what you get.
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  10. #10
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    I like it, PH. Good way to go about it.
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