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Triple integral volume problem. Please help me!

  1. Apr 8, 2005 #1
    Ok I hve a triple integral problem for find the area of the following:

    cylinder: [tex]x^2 + y^2 = 4[/tex]

    plane: [tex]z = 0[/tex]

    plane: [tex]x + z = 3[/tex]

    It is a cylinder cut at the xy-plane and by the last plane. It looks like a circular wedge standing straight up from the xy-plane.

    I just can't figure out the limits of integration, I know the last variable integrated must have limits that are constants. I just don't know what to do! Help me please! :tongue2:
  2. jcsd
  3. Apr 8, 2005 #2
    Here is what I have so far

    By solving the equation [tex]x + z = 3[/tex] for Z, [tex]z = 3 - x[/tex] I can get the z-limits to be from [tex]0[/tex] to [tex]3 - x[/tex], right?

    I then solve [tex]x^2 + y^2 = 4[/tex] out to be [tex]y = \pm \sqrt{4 - x^2}[/tex], and those are the y-limits?

    After that, the x-limits are from [tex]-2[/tex] to [tex]2[/tex], right?

    Here is that in integral form: [tex]\int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz dy dx = 12\pi[/tex]

    Am I on the right track here?
  4. Apr 8, 2005 #3


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    It looks perfect. :smile:

    If you use cylindrical coordinates: r, phi, z, the integration becomes easier. In these coordinates

    [tex] x=\cos\phi [/tex]
    [tex] y= \sin \phi [/tex]
    [tex] z=z [/tex],

    the volume element is

    [tex] dV= r d \phi dr dz [/tex]

    and your boundaries are

    [tex] 0 \le z \le 3-x = 3 - r \ cos\phi [/tex]

    [tex] 0 \le \phi \le 2\pi [/tex]

    [tex] 0 \le r \le 2 [/tex]

    So you have to calculate the integral

    [tex]\int_{0}^{2} \int_{0}^{2\pi}} \int_{0}^{3 - r \cos(\phi)}r dz d\phi dr [/tex]
  5. Apr 8, 2005 #4
    Here is the triple integral solved step-by-step

    [tex]V = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \int_{0}^{3 - x} dz\; dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[ z \right]_{0}^{3 - x} dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} \left[(3 - x) - (0)\right] dy\; dx = \int_{-2}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} (3 - x) dy\; dx[/tex]

    [tex]V = \int_{-2}^{2} \left[3y - xy\right]_{-\sqrt{4-x^2}}^{\sqrt{4 - x^2}} dx = \int_{-2}^{2} \biggl\{\left[3\left(\sqrt{4 - x^2}\right) - x\left(\sqrt{4 - x^2}\right)\right] - \left[3\left(-\sqrt{4 - x^2}\right) - x\left(-\sqrt{4 - x^2}\right)\right]\biggl\} dx[/tex]

    [tex]V = \int_{-2}^{2} \left(6 \sqrt{4 - x^2} - 2x \sqrt{4 - x^2}\right) dx = 6 \int_{-2}^{2} \left(4 - x^2\right)^{\frac{1}{2}}\; dx + \int_{-2}^{2} -2x \sqrt{4 - x^2}\; dx[/tex]

    [tex]u = 4 - x^2[/tex] and [tex]du = -2x dx[/tex] for that second integral.

    [tex]V = 6\left[\frac{2}{2}\sqrt{4 - x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}\right)\right]_{-2}^{2} +\; 2\left[\frac{2}{3}\left(4 - x^2\right)^{\frac{3}{2}}\right]_{-2}^{2}[/tex]

    [tex]V = 6\biggl\{\left{\sqrt{4 - 4} + 2 \sin^{-1}\left(-1\right)\right] - \left[\sqrt{4 - 4} + 2\sin^{-1}\left(1\right)\right]\biggl\} + 2\biggl\{\left[\frac{2}{3} \left(4 - 2^2\right)\right] - \left[\frac{2}{3} \left(4-(-2)^2\right)\right]\biggl\}[/tex]

    [tex]V = 6\left(2\pi\right) + 2\left(0\right) = 12\pi[/tex]

    Does this check out? Should I post a graph?
    Last edited: Apr 8, 2005
  6. Apr 8, 2005 #5
    Ok, a new triple integral

    Did I do this new Problem correctly? :bugeye: :

    Find volume of the region in the first octant bounded by the plane [tex]y = 1 - x[/tex] and the surface [tex]z = \cos\left(\frac{\pi\;x}{2}\right)[/tex]. And [tex]0\;\le x \;\le1[/tex].


    [tex]\int_{0}^{1} \int_{0}}^{1 - x} \int_{0}^{\cos\bigl\(\frac{\pi\;x}{2}}\bigl\)}\;dz\;dy\;dx = \frac{4}{\pi^2}[/tex]
  7. Apr 8, 2005 #6
    Anyone double check the last calculation yet?
  8. Apr 8, 2005 #7


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    It looks all right.

  9. Apr 8, 2005 #8
    That is indeed correct

  10. Apr 10, 2005 #9
    Many thanks

    Thank you ehild and marlon. I am never sure of some of these calc problems!
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