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Finding The Volume Of Solid Using Triple Integrals

  1. Feb 26, 2004 #1

    I am having trouble setting up triple integrals to find a volume of a given solid. Here is one of the questions with which I am having trouble.

    Now I can see that the projection of the solid on the xy plane is the circle x^2 + y^2 = 9. And I think I can visualize the plane z = y + 3 with respect to the cylinder - it slices the cylinder in half diagonally. But I am not sure how to set up the triple integral.

    This is the way I would set up the integral

    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, 0 <= z <= y + 3 }

    If I was to integrate using these limits I would then integrate with respect to z first, then y, then x.

    Now I know that it would be easier to eventually convert to polar coordinates but I would like to know if the way I set up the triple integral to find the volume of the solid is correct so far.

    int.[-3,3] int.[-(9-x^2)^1/2,(9-x^2)^1/2] int.[0,y+3] 1*dz*dy*dx

    How does that look so far?
  2. jcsd
  3. Feb 26, 2004 #2

    Tom Mattson

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    I get the same thing.
  4. Feb 26, 2004 #3
    Thanks Tom.
  5. Feb 26, 2004 #4
    I am still having trouble with this question. The easiest part should be integrating the triple integral but I can't seem to get it.

    I first integrate with respect to z.

    int.[0,y+3] 1*dz = y + 3.

    I now integrate with respect to y.

    int.[-(9-x^2)^1/2,(9-x^2)^1/2] y + 3 dy =

    (y^2)/2 - 3y | .[-(9-x^2)^1/2,(9-x^2)^1/2] =

    (((9-x^2)^1/2)^2)/2 + 3((9-x^2)^1/2)) - [((-(9-x^2)^1/2)^2)/2 + 3(-(9-x^2)^1/2)) =

    (9-x^2)/2 + 3(9-x^2)^1/2 - [(9-x^2)/2 - 3(9-x^2)^1/2] = 6(9-x^2)^1/2

    Now I should integrate with respect to x. But I can't seem to do it / remember how to integrate this part:

    int.[-3,3] 6(9-x^2)^1/2 dx

    Any further help is appreciated.

    BTW, would I be able to integrate this integral with trig substitution? If so. I cannot remember how to do that. Could someone guide me / help me recall the process? Thanks again.
    Last edited by a moderator: Feb 26, 2004
  6. Feb 26, 2004 #5
    Right. It's been awhile since I have done trig. substitution, but I believe that that is the way to go here. (Stop me if I am wrong please). 8)

    So I presently have

    int.[-3,3] 6(9-x^2)^1/2 dx

    I now do my trig. sub. by letting x be 3 sin theta. Therefore

    dx = 3 cos theta dtheta

    I also change my limits.

    If x = 3 then theta should be pi/2 and if x = -3 then theta should be -pi/2.

    Now to integrate:

    If I let x = 3 sin theta then

    int.[-3,3] 6(9-x^2)^1/2 dx


    6 * int. [-pi/2, pi/2] (9 - (3sin theta)^2)^1/2 * 3 cos theta dtheta

    which then becomes

    6 * int. [-pi/2, pi/2] (9 - 9 sin^2theta)^1/2 * 3 cos theta dtheta

    which becomes

    6 * int. [-pi/2, pi/2] (9 (1 - sin^2theta))^1/2 * 3 cos theta dtheta

    which becomes

    6 * int. [-pi/2, pi/2] 3 * (1 - sin^2theta)^1/2 * 3 cos theta dtheta

    which then becomes

    54 * int. [-pi/2, pi/2] (cos^2theta)^1/2 * cos theta dtheta


    54 * int. [-pi/2, pi/2] cos theta * cos theta dtheta

    which becomes

    54 * int. [-pi/2, pi/2] cos^2theta dtheta.

    Now using the trig. identity

    cos^2 = (1+cos2theta)/2
    I get

    54 * int. [-pi/2, pi/2] (1+cos2theta)/2 dtheta.

    which becomes

    27 * [ theta + 1/2 * sin2theta] with the limits [-pi/2, pi/2].

    Then sub. in the limits.

    27 * [ pi/2 + 0 - (-pi/2 + 0) ] = 27 * 2pi/2 = 27 pi.

    Anyway, that is what I did and that is what I got. Have I made errors? Or did I do fine? And is this the volume of the solid?
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