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Volume enclosed by two spheres using spherical coordinates

  1. May 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Use spherical coordinates to find the volume of the solid enclosed between the spheres $$x^2+y^2+z^2=4$$ and $$x^2+y^2+z^2=4z$$

    2. Relevant equations
    $$z=\rho cos\phi$$ $$\rho^2=x^2+y^2+z^2$$ $$dxdydz = \rho^2sin\phi d\rho d\phi d\theta$$

    3. The attempt at a solution

    The first sphere is a sphere of radius 2 centered at the origin, and the second is a sphere of radius 2 centered at (0,0,2). So I tried setting the up the triple integral as $$\int_{0}^{2\pi}\int_{0}^{\pi/2}\int_{4cos\phi}^{2} \rho^2sin\phi d\rho d\phi d\theta$$

    Which gives me a negative answer. I'm guessing my bounds for phi or rho are off?

    Additionally, the question suggests I use the iterated order $$d\phi d\rho d\theta$$ but I'm unsure how to change the iterated order around like it suggests. Any help would be appreciated, thanks!
     
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  3. May 11, 2016 #2

    BvU

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  4. May 11, 2016 #3

    LCKurtz

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    First you need to specify exactly what region you are describing as "between" the spheres. In the cross section below, is the the blue or the green region?

    upload_2016-5-11_9-20-11.png

    In either case the limits depend on the ##\phi## value where the curves intersect. If it's the blue region you can go from ##\rho_{inner}## to ##\rho_{outer}##. If it is the green region you have to break it into two integrals depending on whether the ##\rho## radius is above or below the intersection point.
     
    Last edited: May 11, 2016
  5. May 11, 2016 #4
    ##x^2+y^2+z^2=4z##, so ##\rho^2=4\rho cos\phi##

    I've typed exactly what the question asked so the question itself doesn't exactly specify, but I assumed it was the green section. The answer is ##10\pi/3## if that's any help.
     
  6. May 11, 2016 #5

    LCKurtz

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    Well, the answer of ##\frac{10\pi} 3## is consistent with the green region. To help you understand the problem, put your pencil on the origin in the picture. That is where ##\rho = 0##. Now move your pencil in the radial (##\rho##) direction across the green area. Which circle does it hit? Do you see that which circle it hits depends on whether you headed above or below the ##\rho## line I drew? That's why you need two integrals. In each ##\rho## goes from ##0## to the circle it hits. The limits on ##\phi## depend on whether you are above or below the angle I show in the diagram. So set those two integrals up and show us what you get.
     
  7. May 11, 2016 #6

    BvU

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    well, if ##\phi = 0## then ## \rho = 4## seems a strange lower bound to me...
     
  8. May 11, 2016 #7
    Yeah I know something's off with my bounds but I'm not sure how to get the correct bounds in this case; or how to switch around the iterated order like the question suggests.
     
  9. May 11, 2016 #8
    Ah ok I think I understand now, I found the angle of intersection by setting ##4cos\phi=2##, giving ##\phi=\pi/3## , so the integral becomes ##\int_{0}^{2\pi}\int_{0}^{\pi/3}\int_{0}^{2} \rho^2sin\phi d\rho d\phi\ d\theta + \int_{0}^{2\pi} \int_{\pi/3}^{\pi/2} \int_{0}^{4cos\phi} \rho^2sin\phi d\rho d\phi d\theta ##, and evaluating this gives ##10\pi/3##. One last question though, the question suggested that I change the iterated order to ##d\phi d\rho d\theta##, how would I do this and how would this simplify the problem?
     
  10. May 11, 2016 #9

    LCKurtz

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    I fixed your misplaced braces. You should get the correct answer now.

    There is no point in switching the order of integration. And putting the ##\phi## before the ##\rho## integration is a bad idea because as it is, it's easy to solve for ##\rho## in terms of ##\phi##. To reverse the order you will get an inverse cosine in there. Not simpler.
     
  11. May 11, 2016 #10
    Oh ok, strange that the question suggested for me to do so then. Either way, thanks for all the help.
     
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