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Math Help - Triple Integral w/ Spherical Coordinates

  1. #1
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    Triple Integral w/ Spherical Coordinates

    I am having trouble with the following question:

    Evaluate the triple integral over E, where E lies between the spheres rho = 2 and rho = 4, and above the cone phi = pi/3. The density function is given by xyz dV.

    When I change variables I get the following as my density function:
    (rho^5)*(sin^3(phi))*cos(phi)*cos(theta)*sin(theta ). Is this right? This doesn't seem like the easiest integral to work with, since after you integrate with respect to rho, you are left with a bunch of trig. Maybe there is another way to write the density function that I'm not seeing?

    Any ideas would be great. Thanks.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    I am having trouble with the following question:

    Evaluate the triple integral over E, where E lies between the spheres rho = 2 and rho = 4, and above the cone phi = pi/3. The density function is given by xyz dV.

    When I change variables I get the following as my density function:
    (rho^5)*(sin^3(phi))*cos(phi)*cos(theta)*sin(theta ). Is this right? This doesn't seem like the easiest integral to work with, since after you integrate with respect to rho, you are left with a bunch of trig. Maybe there is another way to write the density function that I'm not seeing?

    Any ideas would be great. Thanks.
    You are left with a bunch of very easy trig functions! your integral is
    \int_{r= 2}^4 r^5 dr\int_{\theta= 0}^{2\pi}cos(\theta)sin(\theta)d\theta \int_{\phi= 0}^{\pi/3}cos(\phi) d\phi

    I would make either of the two obvious substitutions in the \theta integral and then not bother with the other integrals!
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    \\
    You are left with a bunch of very easy trig functions! your integral is
    \int_{r= 2}^4 r^5 dr\int_{\theta= 0}^{2\pi}cos(\theta)sin(\theta)d\theta \int_{\phi= 0}^{\pi/3}cos(\phi) d\phi

    I would make either of the two obvious substitutions in the \theta integral and then not bother with the other integrals!
    that middle integral is zero, \sin^2 \theta at 2\pi and 0 is zero.
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