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Math Help - Help finding ellipsoid volume with antiderivative?

  1. #1
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    Question Help finding ellipsoid volume with antiderivative?

    Hi, I'm stuck on a homework problem and would appreciate some help, because it's getting late. :-(

    I have to integrate x^2/a^2=y^2/b^2=1 for the volume of this ellipsoid rotating about the x-axis. I have -a<x<a (supposed to be less than or equal to but I don't have that key?) and -b<y<b (also less than or equal to).

    How do I do the Remiens sum to get to the integral problem? We're supposed to do this with the antiderivative method.
    I've got a half ellipse drawn & trying to do the disks, but I'm really stuck.
    I get y=sqare root of b^2(1-x^2/a^2).

    Thanks,
    Jen
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  2. #2
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    Quote Originally Posted by Jenberl View Post
    Hi, I'm stuck on a homework problem and would appreciate some help, because it's getting late. :-(

    I have to integrate x^2/a^2=y^2/b^2=1 for the volume of this ellipsoid rotating about the x-axis. I have -a<x<a (supposed to be less than or equal to but I don't have that key?) and -b<y<b (also less than or equal to).

    How do I do the Remiens sum to get to the integral problem? We're supposed to do this with the antiderivative method.
    I've got a half ellipse drawn & trying to do the disks, but I'm really stuck.
    I get y=sqare root of b^2(1-x^2/a^2).

    Thanks,
    Jen
    Formula: V = \pi \int_{-a}^{a} y^2 \, dx

    From symmetry:

     = 2 \pi \int_0^a y^2 \, dx.

    So you end up with quite a simple integral to do.
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  3. #3
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    ?got a weird answer?

    Thank you sir.
    ok.
    But I tried and I don't get it so simple.
    Is 2*pi*b^2(x-x^3/3a^a) the right answer to plug into?
    ??? it looks funny, but can't see what I'm doing wrong.
    I'm not seeing this right, I don't think.
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  4. #4
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    Quote Originally Posted by Jenberl View Post
    Thank you sir.
    ok.
    But I tried and I don't get it so simple.
    Is 2*pi*b^2(x-x^3/3a^a) the right answer to plug into?
    ??? it looks funny, but can't see what I'm doing wrong.
    I'm not seeing this right, I don't think.
    V = 2 \pi \int_0^a y^2 \, dx = 2 \pi b^2 \int_0^a 1 - \frac{x^2}{a^2} \, dx = 2 \pi b^2 \left[ x - \frac{x^3}{3 a^2}\right]_0^a = 2 \pi b^2 \left( a - \frac{a^3}{3a^2}\right)

    I hope I don't need to continue .....
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