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Triple Integral - Volume Question

  1. Mar 17, 2007 #1
    1. The problem statement, all variables and given/known data

    So my question is as follows: Find the volume of the solid bounded by z = 3x^2 + 3y^2 and z = 6sqrt(x^2 + y^2)


    3. The attempt at a solution


    I drew the graphs of these out, with the z = 3x^2 + 3y^2 being a circular paraboloid w/ vertex at (0,0,0) and extending in positive z direction.

    The z = 6sqrt(x^2+y^2) is a cone centered on the z axis w/ vertex at (0,0,0).

    I started by solving z = 6sqrt(x^2+y^2) for x^2 + y^2 and then plugging that into my other z = function and got that they intersect at z = 12. After plugging 12 in for z in the equations, i got both of them x^2 + y^2 = 4.

    Circle, radius 2.

    So, when setting up my triple integral, I have the triple integral of dzdydx. I believe my bounds on x are -2 -> 2.

    Now, for my y bounds, I just solved the x^2 + y^2 = 4 for y, found that y = sqrt(4-x^2). I figured since it was a circular figure at this point, the y bounds are -sqrt(4-x^2) -> sqrt(4-x^2).

    For my z bounds, I figured that it went from the lower function to the higher function, so from 3x^2 + 3y^2 -> 6sqrt(x^2+y^2).

    My question comes to this....is any of this right at all? How do I actually integrate this?

    I know I am going to need to probably change it into cylindrical coordinates, but then I run into the problem of how do I find my r integration bounds...

    I know this is probably a jumble, but it's what I've got and hopefully some light can be shed on this.

    thanks
    -twiztidmxcn
     
  2. jcsd
  3. Mar 17, 2007 #2

    HallsofIvy

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    Yes, that's correct.

    Good. In other words, if you project the entire region into the xy-plane, it covers a circle of radius 2 and your integral must cover all of that.

    Good. That's correct.

    Yes, that's also correct.

    It would probably have been easier to set up the integral in cylindrical coordinates from the start. Your figure, projected to the xy-plane forms a circle with center at (0,0) and radius 2. To cover that, take [itex]\theta[/itex] ranging from 0 to [itex]2\pi[/itex], r ranging from 0 to 2. Now express the bounds on z in terms of r and [itex]\theta[/itex].

    Actually, it's very well done and clear.
     
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