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Homework Help: Finding The Volume Of Solid Using Triple Integrals II

  1. Feb 27, 2004 #1
    Hello,

    I am still unsure of my ability to evaluate the volume of a solid using triple integrals. Here is my question:

    Now I know that the intersection of the two paraboloids is

    9 = x^2 + y^2.

    But I am unsure how to set up the triple integral. I was thinking of splitting the volume into two halves.

    But as I am writing this I am not sure that I have to do that. I would say I could set up the triple integral with the following limits of integration as so:

    Let E be the solid in question. Then

    E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, x^2 + y^2 <= z <= 36 - 3x^2 - 3y^2 }

    How does this look? First I would integrate with respect to z, then y then x.

    I was thinking that I could split the solid into two triple integrals as such:

    E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, x^2 + y^2 <= z <= 9 }

    and

    E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, 9 <= z <= 36 - 3x^2 - 3y^2 }

    But I don't see a problem with the first way I set up the triple integral. However I haven't tried to integrate it as of yet.

    Any thoughts on the problem would be appreciated. Thankyou.
     
  2. jcsd
  3. Feb 27, 2004 #2
    I really wouldn't do this volume in cartesian coordinates. This one's just begging for cylindrical coordinates (it's got rotational symmetry about the z-axis), so, without further ado:

    z = x^2 + y^2 = r^2
    z = 36 - 3x^2 - 3y^2 = 36 - 3r^2

    r^2 = 36 - 3r^2
    r = 3
    z = 9

    So the integral intersects when r = 3 and z = 9. It's not easy to figure out that it'd be real difficult do this in only one integral, so let's split the solid up into two smaller solids and add them.

    Volume 1: The solid enclosed by z = r^2 and the plane z = 9.
    Volume 2: The solid enclosed by the plane z = 9 and z = 36 - r^2

    I'll do Volume 1, and you can have your hand at Volume 2.

    Let's start with the obvious and go from there:

    [tex]\int\!\!\!\int\!\!\!\int_{V_1}1\,dV[/tex]

    So we have some kind of triple-integral over some volue. Since we're doing this is cylindrical coordinates,

    [tex]dV = r\,dz\,dr\,d\theta[/tex]
    with [itex]dz\,dr\,d\theta[/itex] arranged in whichever order we want.

    Just because the math works out the easiest, I'll choose the order I wrote. So we have,

    [tex]\int_{\theta_\textrm{low}}^{\theta_\textrm{high}}\!\!\!\int_{r_\textrm{low}(\theta)}^{z_\textrm{high}(\theta)}\!\!\!\int_{z_\textrm{low}(r,\theta)}^{z_\textrm{high}(r,\theta)}r\,dz\,dr\,d\theta[/tex]

    For [itex]V_1[/itex],
    [tex]z_\textrm{low}(r,\theta) = r^2[/tex]
    [tex]z_\textrm{high}(r,\theta) = 9[/tex]
    [tex]r_\textrm{low}(\theta) = 0[/tex]
    [tex]r_\textrm{high}(\theta) = 3[/tex]
    [tex]\theta_\textrm{low} = 0[/tex]
    [tex]\theta_\textrm{low} = 2\pi[/tex]

    So that gives us:

    [tex]\int_0^{2\pi}\!\!\!\int_0^3\!\!\!\int_{r^2}^9r\,dz\,dr\,d\theta = \frac{81\pi}{2} = V_1[/tex]

    Sorry about the 24-hour wait.

    I got [itex]V = V_1 + V_2 = 243\pi[/tex] as the final answer.

    If you want, I can write up a way to do it in Cartesian coordinates but, as I said, it's not exactly pretty.

    cookiemonster
     
  4. Feb 28, 2004 #3
    Yes. I thought so too. However I usually try to set up the integral using cartesian coordinates then I show the conversion to polar coordinates.

    I ended up doing the question in polar coordinates but I don't remember my answer.

    I also ended up doing the question in one volume instead of two. I used the upper limit of

    36 - 3r^2

    and a lower limit of

    r^2.

    However, intuitively, I thought that splitting up the integral into two separate integrals was the way to go. And I think that is what I should have done in the first place.

    Thanks again for the help cookiemonster.

    Chhers.
     
  5. Feb 28, 2004 #4

    HallsofIvy

    User Avatar
    Science Advisor

    I don't see any reason to "break the solid up into two smaller solids and add them". The parabola z= 36- 3x2- 3y2= 36- 3r2 is always above the parabola z= x2+ y2= r2 so that will be the upper and lower limits on the "z" integration. There is complete circular symmetry so the &theta; integration will be from 0 to 2&pi;. Finally, all of the volume lies inside the circle or intersection (which has radius 3) so the r integration will be from 0 to 3.
    In cylindrical coordinates the volume is given by
    [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^3\int_{z=r^2}^{36-3r^2}rdzdrd\theta [/tex].

    That reduces very quickly to [tex]2\pi\int_0^3r(36-3r^2-r^2)dr [/tex] or
    [tex]2\pi\int_0^3(36r- 4r^3)dr [/tex]
     
  6. Feb 28, 2004 #5
    Y'know what, HallsofIvy (and wubie, originally) is absolutely right. No reason to split them up. Silly cookie should pay more attention.

    cookiemonster
     
  7. Feb 28, 2004 #6
    Thanks very much to the both of you for your insights. It's good to be aware of different approaches.

    Thanks for your help.

    Cheers.
     
  8. Nov 20, 2010 #7
    answer is actually 162*pi... not 81/2 pi... you just multiply the 2pi through the equation at the end, not divide.
     
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