1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: 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


    User Avatar
    Science Advisor

    [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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook