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Volume integral with two intersecting shapes

  1. Mar 21, 2008 #1
    The question I am dealing with has to do with the volume contained by two intersecting shapes I have created this integral and can't find a reasonable way of solving it. What is the best approach to solve this:


    [tex]\int_{r=0}^{\pi/2}\int_{u=0}^{2\cos\theta} \sqrt{9-r^2} * r drdu= [/tex]
     
  2. jcsd
  3. Mar 21, 2008 #2

    tiny-tim

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    Your integral doesn't look right. You have u r and theta, but only two ∫s. :confused:

    Do you mean
    [tex]\int_{\theta=0}^{\pi/2}\int_{u=0}^{2\cos\theta} [/tex] ?

    If so, where does r come into it?

    I hope r is right, because
    [tex]\int \sqrt{9-r^2} * r dr[/tex] is very easy … :smile:
     
  4. Mar 24, 2008 #3
    There wasn't supposed to be a 'u' in the question
     
  5. Mar 25, 2008 #4

    tiny-tim

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    … one integral at a time …

    ah! … that makes more sense! :smile:

    right … when we do double integrals, we always do one integral at a time.

    In this case, do the r integral first.

    What is ∫r√(9-r^2)dr?

    It's actually quite easy if t you think about it. :smile:
     
  6. Mar 25, 2008 #5
    [tex]\int_{\theta=0}^{\pi}\int_{r=0}^{2\cos\theta} \sqrt{9-r^2} * r drd\theta= [/tex]

    The first integral didn't cause me much of a problem, however when I insert the 2cos(theta) and the '0' I can't figure out how to work through the second integral.

    I ended up with:

    [tex]\int_{\theta=0}^{\pi}\sqrt{(9-4\cos^2\theta)^3} / (-3) +9 d\theta= [/tex]
     
  7. Mar 25, 2008 #6

    tiny-tim

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    Hi bobsmiters! :smile:

    hmm … don't like the look of that …

    I'll have a think about it later, but before I do …

    You got this ∫∫drdtheta yourself, didn't you?

    I was a little surprised when I first saw it because usually when you convert to polar coordinates, you get an extra sintheta for free (because dxdy = sintheta dr dtheta).
    (and then I got confused by the du, and forgot about it :confused:)

    And that extra sintheta tends to make things much easier.

    Are you sure there shouldn't be a sintheta?
     
  8. Mar 25, 2008 #7

    Vid

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    You're thinking of spherical coordinates.

    dA = dxdy = rdrdtheta
    dV = dxdydz = r^2sin(theta)dr*dtheta*dphi
     
  9. Mar 25, 2008 #8
    The original question read: Find the volume of the solid that lies under x^2 + y^2 + z^2 = 9, above the xy-plane and inside the cylinder x^2 + y^2 = 2x.

    It is possible that I made a mistake somewhere in getting to that integral.
     
  10. Mar 26, 2008 #9

    tiny-tim

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    Indeed I am! :smile:
    Hi bobsmiters! :smile:

    I'm confused … how did you get that … what is your r? :confused:
     
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