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Simple Volume Integral

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the volume of the portion of the sphere x^2 + y^2 + z^2 = 4 for which y>1 (or equal to).


    2. Relevant equations

    I would like to do this using a triple integral.

    3. The attempt at a solution

    OK, so I tried integrating the element r^2sin(theta)drd(phi)d(theta) where theta is the angle from the z axis, and theta limits 0 to pi, phi limits 0 to 2pi and r limits 1 to 2 and got 28/3 * pi.

    Am I talking utter rubbish?

    Cheers
    Cpfoxhunt
     
  2. jcsd
  3. Apr 12, 2008 #2

    cepheid

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    In your translation from Cartesian to polar coordinates, your information about the part of the sphere that is cut off (ie that is not included) got lost. The restrictions on y do not translate into the restrictions on the radius that you imposed. If you think about it, by restricting r to be between 1 and 2, you are calculating the volume between two spheres, one of radius 1, and the other of radius 2. That is not the volume you are trying to calculate.
     
  4. Apr 12, 2008 #3

    cepheid

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    You're looking at the portion of the sphere to the "right" (in this picture) of the plane y=1:

    [​IMG]

    I hope that helps visualize it. Now, to accuarately translate y into spherical polar coordinates, consider a point represented by the vector:

    [tex] \vec{r} = (r, \theta, \phi) [/tex]

    In my convention, theta is the polar angle, and phi is the azimuthal angle.

    The magnitude of the projection onto the xy-plane of this vector is given by:

    [tex] r\sin{\theta} [/tex]

    By definition, the angle between this projection and the positive x-axis is phi, so that the x and y components of the projection are given by:

    [tex] x = (r\sin{\theta})\cos \phi [/tex]
    [tex] y = (r\sin{\theta})\sin \phi [/tex]

    Your restriction then becomes:

    [tex] r\sin{\theta}\sin \phi \geq 1 [/tex]

    with the added constraint that, for the point to lie within the sphere,

    [tex] r \leq 2 [/tex]
     
  5. Apr 12, 2008 #4

    cepheid

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    Whether you have enough information to translate those constraints into proper limits of integration on the volume integral, I'm not totally sure.
     
  6. Apr 12, 2008 #5
    Thanks for the help. So I should just apply divergence theorem then? And integrate over the closed surface of the sphere? How would you guys answer the question? I'm currently tying myself in knots over the whole surface and volume integral topic which is annoying because I understood it a few weeks ago.

    When I translate into polars I simply multiply by the jacobian r^2sin(theta) right?
     
  7. Apr 12, 2008 #6

    HallsofIvy

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    Not utter rubbish but using those limits (I had to think for a moment- you're a physicist and have swapped [itex]\theta[/itex] and [itex]\phi[/itex] from what any self respecting mathematician would use) covers the entire sphere- and the volume of a sphere of radius 2 is [itex](32/3)\pi[/itex], not 28/3. But you don't want the entire sphere, only part of it so you don't want to use those limits.

    If it were me, I would alter the problem so we wanted the part with [itex]z\ge 1[/itex]. That way the r and phi limits would be the same with theta altered to give only the part with [itex]z\ge 1[/itex]. And since, in polar coordinates [itex]z= r cos(\theta)[/itex] you want theta between 0 and arccos(1/2).
     
  8. Apr 13, 2008 #7
    Very good point. And yeah, guilty with the swapping the azimuthual with the polar, but thet is apparently the format my exam is going to be in, so im trying to get used to it.

    Cheers for the idea, I like that a lot more.

    Chris
     
  9. Apr 13, 2008 #8
    My brains are clearly falling out of my head, is the answer (8/3)pi? The problem is I don't honestly believe that it is ... using other black arts I get the answer to be 5(pi)/3 but the integration gives 8 ..
     
    Last edited: Apr 13, 2008
  10. Apr 13, 2008 #9

    HallsofIvy

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    [tex]\int_{\rho= 0}^2\int_{\phi= 0}^{2\pi}\int_{\theta= 0}^{arccos(1/2)} \rho^2 sin^2(\theta) d\rho d\phi d\theta[/tex]
    [tex]= \left(\int_{\rho= 0}^2 \rho^2 d\rho\right)\left(\int_{\phi= 0}^{2\pi} d\phi\right)\left(\int_{\theta= 0}^{arccos(1/2)} sin^2(\theta)d\theta\right)[/tex]
    [tex]\int_{\rho= 0}^2 \rho^2 d\rho= \frac{8}{3}[/tex]
    [tex]\int_{\phi= 0}^{2\pi} d\phi= 2\pi[/tex]
    To integrate [itex]sin^2(\theta)d\theta[/itex] use the trig identity [itex]sin^2(\theta)= (1/2)(1+ cos(2\theta))[/itex]
     
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