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Find the volume: once more

  1. Jul 16, 2011 #1
    1. The problem statement, all variables and given/known data
    The textbook I have is horrible in explaining this stuff in the chapter so please bare with me.

    So we have the information: Find the volume V inside both the sphere x^2 + y^2 + z^2 = 1 and the cone z =sqrt(x^2 + y^2)

    2. Relevant equations
    Where the heck do I began? I know this will require triple integration in spherical coordinates and the solution is (2/3)*pi*1^3(1-1/sqrt(2))

    3. The attempt at a solution
    I know the theta integrand is (0,2pi) because there are no bounds, but I have no idea where to find the limits of integration for the phi and zenith integrands. :(

    EDIT (1): Just solved for the intersections of the sphere and cone and got z = 1/sqrt(2))
    EDIT (2): Since phi = 1 and z = (1/sqrt(2)), it's safe to conclude that the zenith angle is pi/4...but what would be it's upper limit?
    EDIT (3): I believe the limits for phi is 0 to pi/4. But I'm not quite sure why though? I think I got the integrands all figured out. I believe it's: for p = (0,1), phi = (0, pi/4) and theta = (0,2pi).

    So overall, could someone just briefly explain to me why phi's limits of integration is from 0 to pi/4. thanks a bunch :D
    Last edited: Jul 16, 2011
  2. jcsd
  3. Jul 17, 2011 #2


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    Looking at the figure from the side (along the negative y-axis), you should see a circle and a "V" (y= 0 so the equations are [itex]x^2+ z^2= 1[/itex], a circle, and [itex]z= \sqrt{x^2}= |x|[/itex]). The "co-latitude", [itex]\phi[/itex], goes from directly up the z-axis, [itex]\phi= 0[/itex], to the line z= x, which bisects the right angle between the x and z axes and so makes angle [itex]\pi/4[/itex] with the z-axis.
  4. Jul 17, 2011 #3


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    I'm sorry. Undressing is forbidden in this forum. :eek:
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