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Triple Integral Problem in Cylindrical Coordinates

  1. Mar 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Use cylindrical coordinates to find the volume of the solid that the cylinder
    r = 3cos/theta cuts out of the sphere of radius 3 centered at the origin.


    2. Relevant equations

    Why do we evaluate theta from 0 to pi instead of from 0 to 2pi? Don't we want to go all the way around the circle?

    3. The attempt at a solution

    x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex] = 9

    so z = [itex]\sqrt{9 - x^2 + y^2}[/itex]

    z = [itex]\sqrt{9 - r^2}[/itex]

    2[itex]\int^{2pi}_{0}\int^{3cos(Θ)}_{0}\int^{sqrt(9 - r^2)}_{0}rdzdrdΘ[/itex]

    Instead of integrating from the bottom on the sphere to the top, I did from the x-y plane to the top and multiplied by two to take advantage of symmetry.

    [itex]\int^{2pi}_{0}\int^{3cos(Θ)}_{0}2rsqrt(9 - r^2)drdΘ[/itex]

    [itex]\int^{2pi}_{0}-(\frac{2}{3})(9 - r^2)^{3/2}|^{3cos(Θ)}_{0}dΘ[/itex]

    [itex]\int^{2pi}_{0}-(\frac{2}{3})(9 - 9cos^{2})^{3/2} - \frac{-2}{3}9^{3/2}dΘ[/itex]

    The integral goes on and eventually reduces to...

    18Θ + 18cosΘ + [itex]\frac{1}{3}[/itex]cos[itex]^{3}[/itex]Θ|[itex]^{2pi}_{0}[/itex]

    This gives a value of 113, which is double the correct answer. I'm assuming this is because I was suppose to integrate from 0 to pi instead of from 0 to 2pi, but I don't know why. Aren't we suppose to go around the whole circle with this thing?
     
  2. jcsd
  3. Mar 29, 2014 #2

    LCKurtz

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    Draw a polar coordinate graph of ##r = 3\cos\theta##. Do ##\theta## from ##-\pi/2## to ##\pi/2## and see what you think.
     
  4. Mar 31, 2014 #3
    Ahh... I see it. Thanks.
     
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