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Math Help - Triple integral

  1. #1
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    Triple integral

    Find the volume of the ellipsoid \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 by solving the triple integral after making the transformation x = au, y = bv and z = cw

    I'm just a bit confused here because this does not seem like a Jacobian transformation. I know what the triple integral would be in cartesian terms, ie, \displaystyle{\int_{-a}^a \int_{-\sqrt{b^2-\frac{b^2x^2}{a^2}}}^{\sqrt{b^2-\frac{b^2x^2}{a^2}}} \int_{-\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}}^  {\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}} 1 dzdydx}

    But clearly, we don't want to integrate that, so what does it mean by "after making the transformation x = au, y = bv and z = cw"? Do I just simply replace all the x with au and y with bv etc, then what happens to the dz dy and dx?

    Thanks heaps!
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  2. #2
    A Plied Mathematician
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    Do I just simply replace all the x with au and y with bv etc
    Yes.

    then what happens to the dz dy and dx?
    You get the same thing you'd normally get in a single integration: dx = a du, dy = b dv, and dz = c dw. Your limits change, your integrand changes, and your differentials change. So what's your new integral, and how do you think you should go about computing it?
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  3. #3
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    Ahhh okay, thanks heaps, yeah because I was just evaluating triple integrals with transformations involving the Jacobian so I got confused when I saw this question (I am studying ahead of class and this question was on one of my classes exercise sheets).

    I understand it now, thanks Ackbeet with all your help! I really appreciate it!!
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  4. #4
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    Perhaps the simplest thing to do is NOT to make the transformation in the integral or, if fact, to do the integral at all. If you let x= au, y= bv, and z= cw, then the equation of the ellipse becomes
    \frac{a^2u^2}{a^2}+ \frac[b^2v^2}{b^2}+ \frac{c^2w^2}{c^2}= u^2+ v^2+ w^2= 1 which is a sphere, of radius 1, in the uvw-coordinate system. The Jacobian (I'm not sure what you mean by a "Jacobian transformation" but every transformation has a Jacobian) is
    \left|\begin{array}{ccc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w}  \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{array}\right|= \left|\begin{array}{ccc}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right|= abc so that volume of the ellipse is the volume of the sphere, \frac{4}{3}\pi 1^3= \frac{4}{3}, times abc.

    (Since you have been "evaluating triple integrals with transformations involving the Jacobian" that's exactly what happens here: dxdydz= abc dudvdw.)
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  5. #5
    A Plied Mathematician
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    You're welcome. Have a good one!
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