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Volume enclosed by a spherical coordinate surface

  1. Jul 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the volume enclosed by the spherical coordinate surface ρ = 2sin∅


    2. Relevant equations

    dV = ∫∫∫(ρ^2)sin∅dρd∅dθ

    3. The attempt at a solution


    (Sorry about my notation!)

    Alright, here's what I've done so far...

    Since the region is a torus, centered around the z-axis, I began by finding my limits of integration for ρ, which I think would simply be from 0 to 2sin∅.

    For my limits for ∅, I started at the utmost point on the positive z-axis, which I believe is 0 in regards to ∅, and it covers the entire ∅ "region" down to the negative z axis, so the limits are 0 to pi.

    And for θ, I got simply 0 to 2pi.

    Given those limits, integrating ((ρ^2)sin∅)dρd∅dθ, I wind up with 0, which on the one hand, I've convinced myself makes sense, since it is a torus, and it is symmetrical about the axes (like saying the area under the cosine from 0 to 2pi is 0). On the other hand, a volume of 0 for a physical object simply doesn't make sense.

    I have a feeling I'm simply going about the limits wrong. If I've got those right, I'll post my integration work in case someone can spot the problem.

    Thanks guys!
     
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  3. Jul 14, 2012 #2

    HallsofIvy

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    So what you are saying is that your integral is
    [tex]\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^\pi\int_{\rho= 0}^{2\sin(\phi)} \rho^2sin(\phi) d\rho d\phi d\theta[/tex]
    Clearly, the [itex]\theta[/itex] integral gives a factor or [itex]2\pi[/itex]. Also, the integral of [itex]\rho^2[/itex] is [itex](1/3)\rho^3[/itex] so, evaluating that integral we have
    [tex]\frac{16}{3}\int_0^\pi sin^4(\phi)d\phi[/tex]
    I do not get 0 when I integrate that.
     
  4. Jul 14, 2012 #3
    Ok, so as far as my limits of integration go, those are ok?

    I also get (1/3)[itex]\rho[/itex]^3, and evaluating at 2sin[itex]\phi[/itex] I get 8/3∫∫sin^4([itex]\phi[/itex]) d[itex]\phi[/itex]d[itex]\theta[/itex]. I'm not sure where I ought to be getting 16 from?

    As for ∫∫sin^4([itex]\phi[/itex]) d[itex]\phi[/itex]d[itex]\theta[/itex], I think I'm simply screwing up some basic calculus. I said that integral of sin^4([itex]\phi[/itex]) is -cos^4(([itex]\phi[/itex]). Should I be using some trig substitutions rather than making that leap?

    Then I evaluated that from 0 to pi, which gives me -1 + 1, so I can't even make it to my dθ integral...

    Thinking this is a trig substition I'm forgetting...
     
  5. Jul 14, 2012 #4

    LCKurtz

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    Yes, you should. Start by writing ##\sin^4\phi## as ##(\sin^2 \phi)^2## and using ##\sin^2\phi = \frac {1-\cos(2\phi)} 2##. Then after a little algebra you will need a double angle formula again...
     
  6. Jul 15, 2012 #5
    Ok, so when I expand (1-cos(2[itex]\phi[/itex])/2)^2, I get 1/4 - (cos(2[itex]\phi[/itex]))/2 + (cos^2(2[itex]\phi[/itex]))/4

    Replacing the last term with a half-angle formula, I have

    1/4 - (cos(2[itex]\phi[/itex]))/2 +1/8 + (cos(2[itex]\phi[/itex]))/8

    At that point, I take the integral of that with respect to [itex]\phi[/itex].

    I wind up with 1/4*[itex]\phi[/itex] - 1/2*sin([itex]\phi[/itex])cos([itex]\phi[/itex]) + 1/8*[itex]\phi[/itex] + 1/8*sin([itex]\phi[/itex])cos([itex]\phi[/itex]).

    Evaluating from 0 to pi, I wind up with:

    1/4pi +1/8pi.

    Integrate that with respect to [itex]\theta[/itex] and evaluating from 0 to 2pi, I get

    8/3(1/2*pi^2 + 1/4*pi^2)

    where 8/3 was from the very first integration of [itex]\rho[/itex].

    Hope this is easy enough to follow along in this format. If you could, let me know if the work looks ago.

    Thanks so much by the way, I really appreciate it.
     
  7. Jul 15, 2012 #6

    LCKurtz

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    Hey, it isn't my job to simplify your answer. :smile: You simplify it, then I will tell you whether it agrees with mine.
     
  8. Jul 15, 2012 #7
    Hehe fair enough. I get 2pi^2.

    Looks like a nice and neat enough answer -- hopefully that's what you got!

    I went through the problem a second time, though, and now I'm finding myself confused about one of the trig substitutions I made. Once I've expanded (1-cos(2ϕ)/2)^2, I'm left with a few terms + (cos^2(2ϕ))/4.

    I know that cos^2(x) = (1 + cos(2x))/2. But since mine is essentially (2x), does it become 1+cos(4x)...or am I beginning to overthink this. I think I am...my 2x = x = theta in that trig sub, so I should be alright.

    Oh jeez, this is my problem, haha. Over-thinking.
     
  9. Jul 15, 2012 #8

    LCKurtz

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    Yes, that's correct.
     
  10. Jul 15, 2012 #9
    Thank you so much. I really appreciate all the help.
     
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