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Evaluate integral by using spherical coordinates

  1. Nov 27, 2012 #1
    030sqrt(9-x2)sqrt(x2+y2)sqrt(18-x2-y2) (x2+y2+z2)dzdxdy

    x=[itex]\rho[/itex]sin[itex]\varphi[/itex]cosθ
    y=[itex]\rho[/itex]sin[itex]\varphi[/itex]sinθ
    z=[itex]\rho[/itex]cos[itex]\varphi[/itex]

    Change the integrand to [itex]\rho[/itex] and integrate wrt d[itex]\rho[/itex]dθd[itex]\varphi[/itex]

    I don't know how to find the limits of integration. Normally I would draw a picture and reason it out, but that doesn't seem plausible here.

    Thanks.
     
  2. jcsd
  3. Nov 27, 2012 #2

    LCKurtz

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    Sure it does. The ##z## limits are ##z=\sqrt{x^2+y^2}## and ##z=\sqrt{18-x^2-y^2}##. What are those two surfaces?
     
  4. Nov 27, 2012 #3
    Z is limited on bottom is a cone and on top a sphere with radius sqrt18.
    Y is limited on the left by the xz-plane and on the right by y=3.
    X... I don't know.
     
  5. Nov 27, 2012 #4

    LCKurtz

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    So where do the sphere and cone intersect? Have you drawn the first octant picture? Once you have that the ##\rho,\phi,\theta## limits are pretty obvious.
     
  6. Nov 27, 2012 #5
    I don't know how to draw the first octant because I'm unsure of how to draw the limiting surface on X.

    Z is a cone and sphere. Y is two planes.

    I feel like the upper limit on X should be sqrt(9-y^2) instead of sqrt(9-x^2). In that case it would be easy to re-parametrize.
     
  7. Nov 27, 2012 #6

    LCKurtz

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    Yes, that is an obvious typo which I didn't notice until you pointed it out. Nevertheless, you have the cone and sphere given.

    Draw the cone and sphere in the first octant. They are going to intersect where their z values are equal. Get that equation. That will tell you the xy domain. But you don't need that anyway. Once you have the picture you can read the spherical coordinate limits from it.
     
  8. Nov 27, 2012 #7
    Ah... yes. Okay.

    rho from 0 to sqrt18
    theta from 0 to pi/2
    phi from 0 to pi/4

    integrating (rho)^3*sin(phi)

    I got 149 but that's too big.

    Edit: Woops. Did theta from 0 to 2pi in calculations. Divided by 4 and got 32.266.
     
  9. Nov 27, 2012 #8

    haruspex

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    I get 37.3. What expression do you get for the indefinite integral?
     
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