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Math Help - triple integrals

  1. #1
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    Exclamation triple integrals

    the integral from 0 to 2pi, the integral from 2 to 3, the integral from sqrt(4-r^2) to sqrt(9-r^2)

    of (1)/(r^2+z^2)^pi dz dr dtheta


    this is in cylindrical coordinates.
    how do i finish this integral by switching to other coordinates? should i switch to spherical or polar?? thanks!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by holly123 View Post
    the integral from 0 to 2pi, the integral from 2 to 3, the integral from sqrt(4-r^2) to sqrt(9-r^2)

    of (1)/(r^2+z^2)^pi dz dr dtheta


    this is in cylindrical coordinates.
    how do i finish this integral by switching to other coordinates? should i switch to spherical or polar?? thanks!
    Is this what you're asking for help on?

    \int_0^{2\pi}\int_2^3\int_{\sqrt{4-r^2}}^{\sqrt{9-r^2}}\frac{1}{(r^2+z^2)^{\pi}}\,dz\,dr\,d\theta

    If so, I'd go spherical.

    Hint 1:

    r\to\rho\cos\phi
    z\to\rho\sin\phi
    \theta\to\theta

    You have to multiply by the determinant of the Jacobian:

    J(r,z,\theta)=\left|\begin{array}{ccc}\cos\phi&-\rho\sin\phi&0\\\sin\phi&\rho\cos\phi&0\\0&0&1\end  {array}\right|=\rho

    Hint 2:
    Spoiler:
    Since r^2+z^2=\rho^2\cos^2\phi+\rho^2\sin^2\phi=\rho^2, the new integral is:

    \int\int\int\frac{J(r,z,\theta)}{(\rho^2)^{\pi}}\,  d\rho\,d\phi\,d\theta=<br />
\int\int\int\frac{1}{\rho^{2\pi-1}}\,d\rho\,d\phi\,d\theta

    Figure out the new bounds and integrate.
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  3. #3
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    thank you so much! are the bounds
    phi is between 0 and pi
    theta is between 0 and 2 pi
    rho is between 0 and pi/4
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by holly123 View Post
    thank you so much! are the bounds
    phi is between 0 and pi
    theta is between 0 and 2 pi
    rho is between 0 and pi/4
    \rho shouldn't be an angle. It's the radius of the sphere. In the original integral, z=\sqrt{4-r^2}\implies z^2+r^2=4\implies \rho^2=4\implies \rho=2 Similarly, the upper bound is \rho=3.

    I believe the others are right.
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