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Triple integral problem help

  1. Sep 25, 2010 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Sep 25, 2010 #2


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    [itex]x^2+ y^2+ z^2= 2az[/itex] when [itex]x^2+ y^2+ z^2- 2az= 0[/itex]. We can complete the square by adding [itex]a^2[/itex] to both sides: [itex]x^2+ y^2+ z^2- 2az+ a^2= x^2+ y^2+ (z- a)^2= a^2[/itex]. That is a sphere with center at (0, 0, a) and radius a. [itex]x^2+ y^2+ z^2= a^2[/itex] is, of course, a sphere with center at (0 0, 0) and radius a. The two spheres intersect when [itex](z- a)^2 - z^2= z^2- 2az+ a^2- z^2= a^2- 2az= 0[/itex] or [itex]z= a/2[/itex], in which case [itex]x^2+ y^2+ z^2= x^2+ y^2+ a^2/4= a^2[/itex] or [itex]x^2+ y^2= 3a^2/4[/itex]. That is, the intersection is a circle in the z= a/2 plane with center at (0, 0, a/2) and radius [itex]a\sqrt{3}/2[/itex].

    In terms of spherical coordinates, each point on that circle of intersection has [itex]\rho= a[/itex] and [itex]\theta= arcsin(\sqrt{3}/2)= \pi/6[/itex]

    The integral, then, should be done in two parts: For the first, [itex]\theta[/itex] goes from 0 to [itex]\pi/6[/itex] and, for each [itex]\theta[/itex], [itex]\rho[/itex] goes to the "upper sphere", [math]x^2+ y^2+ z^2= a^2[/math], from 0 to a.

    For the second part, [itex]\theta[/itex] goes from [itex]\pi/6[/itex] to [itex]\pi/2[/itex] and, for each [itex]\theta[/itex], [itex]\rho[/itex] goes from 0 to the "lower" sphere [itex]x^2+ y^2+ z^2= 2az[/itex] which in polar coordinates is [itex]\rho^2= 2a\rho cos(\theta)[/itex]. That is, for each [itex]\theta[/itex], [itex]\rho[/itex] goes from 0 to [itex]2a cos(\theta)[/itex]. (Note that when [itex]\theta= \pi/6[/itex], [itex]cos(\theta)= 1/2[/itex] so that [itex]\rho= a[/itex].)

    [itex]\phi[/itex] goes from 0 to [itex]2\pi[/itex] for both integrals.

    (In order to be consistent with the original post, I have used "engineering" notation of spherical coordinates in which [itex]\theta[/itex] is the "co-latitude" and [itex]\phi[/itex] is the "longitude" rather than "mathematics" notation in which the two Greek letters are reversed. I have used [itex]\rho[/itex] rather than "r" because that is the convention in either system.)
    Last edited by a moderator: Sep 25, 2010
  4. Sep 25, 2010 #3
    Divide the integral in 2 parts:

    [tex]0< \varphi < \pi/3[/tex] where [tex]\rho = a[/tex]
    [tex]\pi/3< \varphi < \pi[/tex] where [tex]\rho = a\:sen2\varphi[/tex]

    [tex]z = \rho cos\varphi[/tex]

    So you solve

    [tex]\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta
    \int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen2\varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta
    Last edited: Sep 25, 2010
  5. Sep 25, 2010 #4
    Only the firt part, if I didn't make mistakes:

    [tex]=\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{a} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta [/tex]

    [tex]=\int_{0}^{2\pi} \int_{0}^{\pi/3} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_0^a \: d\varphi\:d\theta [/tex]

    [tex]=\frac{ a^5 }{5} \int_{0}^{2\pi} \int_{0}^{\pi/3}(cos\varphi)^2 sen \varphi \: \: d\varphi\:d\theta [/tex]

    [tex]=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ \frac{-(cos\varphi)^3}{3} \: \right]_0^{\pi/3} \: d\theta [/tex]

    [tex]=\frac{ a^5 }{5} \int_{0}^{2\pi} \left[ -\frac{3\sqrt3}{24} +\frac{1}{3} \: \right] \: d\theta [/tex]

    [tex]=\frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right) \int_{0}^{2\pi}\: d\theta [/tex]

    [tex]=2\pi \frac{ a^5 }{5} \left( \frac{8-3\sqrt3}{24} \right) [/tex]
    Last edited: Sep 25, 2010
  6. Sep 25, 2010 #5
    Other part

    [tex]=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \int_{0}^{2asen \varphi} \rho^2 (cos\varphi)^2 \rho^2 sen \varphi \: d\rho \: d\varphi\:d\theta [/tex]

    [tex]=\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left [ \frac{\rho^5}{5} (cos\varphi)^2 sen \varphi \: \right ]_{0}^{2asen \varphi} d\rho \: d\varphi\:d\theta [/tex]

    [tex]=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (cos\varphi)^2 ( sen \varphi)^6 \: d\varphi\:d\theta [/tex]

    [tex]=\frac{32a^5}{5}\int_{0}^{2\pi} \int_{\pi/3}^{\pi} \left (( sen \varphi)^6-( sen \varphi)^8) \: d\varphi\:d\theta [/tex]

    Now have fun solving sine power eight :)
  7. Sep 25, 2010 #6
    Dear HallsofIvy,

    Thank you very much for your help. Yes my solution was wrong. At the end of my book the answer is [tex]\frac{59 \pi a^{5}}{480}[/tex] and when I solved it with your method I got the correct answer. I was actually wrong about the ranges!

    Dear Quinzio,

    I used HallsofIvy's method and I could easily solve it!
  8. Sep 25, 2010 #7

    Attached Files:

  9. Sep 25, 2010 #8
    The two ways are really the same.
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