# Thread:

1. ## 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!

2. 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?

3. 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!!

4. 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.)

5. You're welcome. Have a good one!
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