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Homework Help: Double Integral in Polar Coordinates Symmetry Issue

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid lying inside both the sphere [itex]x^2 + y^2 + z^2 = 4a^2[/itex] and the cylinder [itex]x^2 + y^2 = 2ay[/itex] above the xy plane.

    2. Relevant equations
    Polar coordinates:
    [tex] r^2 = x^2 + y^2 [/tex]
    [tex] x = r\cos(\theta)[/tex]
    [tex] y = r\sin(\theta)[/tex]

    3. The attempt at a solution
    So I tried this problem and got it almost perfectly correct. My result was: [itex]V = \frac{8\pi a^3}{3}[/itex], and the right answer is [itex]V = \frac{8\pi a^3}{3} - \frac{32 a^3}{9} [/itex], so I'm just missing one term. The first thing that crossed my mind was that one of my integrals evaluated to 0, so obviously that's where I went wrong. I checked the solution and noticed the only difference from my work was that the book solution used symmetry and integrated [itex]\theta[/itex] from 0 to [itex]\frac{\pi}{2}[/itex], and I just integrated from 0 to [itex]\pi[/itex]. This is what I did:
    [itex]x^2 + y^2 = 2ay[/itex] describes a cylinder with a circle base that is centered at (0,a) and has radius a. Therefore, this cylinder lies entirely in the first and second quadrants, where [itex]\theta[/itex] ranges from [itex](0,\pi)[/itex]. Using cylindrical coordinates, the sphere has the equation [itex] z = \sqrt{4a^2 - r^2} [/itex] and the cylinder has the equation [itex] r^2 = 2ar\sin(\theta) [/itex], or [itex] r = 2a\sin(\theta)[/itex].
    [tex] V = \int_{0}^{\pi}\int_{0}^{2a\sin(\theta)}\sqrt{4a^2 - r^2}rdrd\theta[/tex]
    Making the substitution [itex] u = 4a^2 - r^2, du = -2rdr[/itex]
    [tex] V = \frac{-1}{2} \int_{0}^{\pi}\int_{r = 0}^{r = 2a\sin(\theta)}\sqrt{u}dud\theta[/tex]
    [tex] V = \frac{-1}{3} \int_{0}^{\pi}(4a^2 - r^2)^{3/2}\bigg|_{0}^{2a\sin(\theta)}d\theta[/tex]
    [tex] V = \frac{-1}{3} \int_{0}^{\pi}\left[(4a^2(1 - sin^2(\theta))^{3/2} - (4a^2)^{3/2}\right]d\theta[/tex]
    [tex] V = \frac{-1}{3} \int_{0}^{\pi}\left[(4a^2)^{3/2}(cos^2(\theta))^{3/2} - 8a^3\right]d\theta[/tex]
    [tex] V = \frac{-8a^3}{3} \int_{0}^{\pi}cos^3(\theta)d\theta + \frac{8\pi a^3}{3}[/tex]
    Now, the remaining integral evaluates to 0 because it will be a function of [itex]\sin(\theta)[/itex] where [itex]\theta \in (0,\pi)[/itex].

    I see that if I were to use symmetry like it was done in the book ([itex] \theta \in (0,\frac{\pi}{2})[/itex]) and multiply everything by 2, my integral should return the right answer. I just don't understand why what I did was wrong.
  2. jcsd
  3. Dec 6, 2015 #2


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    Always bear in mind the possibility that the book answer may be wrong.

    The integration over ##\theta## only happens in your last step. So the two methods should be the same up to there. Prior to that integration, your formula is (simplifying a bit):

    V = \frac{8a^3}{3} \int_{0}^{\pi}\left[(1-\cos^3\theta\right]d\theta
    Theirs should be
    $$V = 2\cdot \frac{8a^3}{3} \int_{0}^{\frac{\pi}{2}}\left[(1-\cos^3\theta\right]d\theta$$

    Does their formula match that?

    By the way, note that the integrand is not symmetric around ##\theta=\frac{\pi}{2}##. It is not the case that the integral over ##[0,\pi]## is twice the integral over ##[0,\frac{\pi}{2}]##.
  4. Dec 6, 2015 #3


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    Note that $$(\cos^2\theta)^{\frac 3 2}= \left(\sqrt{\cos^2\theta}\right)^3 = |\cos\theta|^3\ne \cos^3\theta$$
    Last edited: Dec 6, 2015
  5. Dec 6, 2015 #4
    Ahhhh that's why...
    Forgot about those sneaky absolute values. So I would have to split it up into an integral from 0 to pi/2 and a negative one from pi/2 to pi, right?
    That explains why the book was so quick to use symmetry, otherwise the symmetry wouldn't have been all that helpful.
    Thanks a lot for the help.
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