1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Double Integral in Polar Coordinates Symmetry Issue
Loading...