# Triple integral

Printable View

• Sep 17th 2010, 06:10 AM
usagi_killer
Triple integral
Find the volume of the ellipsoid $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ by solving the triple integral after making the transformation $\displaystyle x = au, y = bv$ and $\displaystyle z = cw$

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, $\displaystyle \displaystyle{\int_{-a}^a \int_{-\sqrt{b^2-\frac{b^2x^2}{a^2}}}^{\sqrt{b^2-\frac{b^2x^2}{a^2}}} \int_{-\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}}^ {\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}} 1 dzdydx}$

But clearly, we don't want to integrate that, so what does it mean by "after making the transformation $\displaystyle x = au, y = bv$ and $\displaystyle z = cw$"? 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!
• Sep 17th 2010, 06:22 AM
Ackbeet
Quote:

Do I just simply replace all the x with au and y with bv etc
Yes.

Quote:

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?
• Sep 17th 2010, 11:58 PM
usagi_killer
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!!
• Sep 18th 2010, 04:10 AM
HallsofIvy
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
$\displaystyle \frac{a^2u^2}{a^2}+ \frac[b^2v^2}{b^2}+ \frac{c^2w^2}{c^2}= u^2+ v^2+ w^2= 1$ 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
$\displaystyle \left|\begin{array}{ccc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{array}\right|= \left|\begin{array}{ccc}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right|= abc$ so that volume of the ellipse is the volume of the sphere, $\displaystyle \frac{4}{3}\pi 1^3= \frac{4}{3}$, times abc.

(Since you have been "evaluating triple integrals with transformations involving the Jacobian" that's exactly what happens here: dxdydz= abc dudvdw.)
• Sep 18th 2010, 06:02 AM
Ackbeet
You're welcome. Have a good one!