Let $E$ be the solid ellipsoid $E = ${$(x,y,z)$ | $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1$} where $a > 0,\: b > 0,\: c > 0$
Evaluate $\int\int \int xyz\: dxdydz$ over:
a. the whole ellipsoid
b. that part of it in the first quadrant, $x \ge 0,\: y \ge 0,\: z \ge 0$
So for the first part, I made the change of variables $u = \frac{x}{a}$, $v = \frac{y}{b}$, $w = \frac{z}{c}$. The ellipse became a sphere. I then made another change of variables into spherical coordinates. I found that the integral evaluated to $$\frac{-2\pi*a^2b^2c^2}{9}$$
Now I am not sure about part b.) Since I change the ellipsoid into a unit sphere, do I just change the bounds to what they are in the first quadrant (ie, theta ranges from 0 to $\frac{\pi}{2}$ instead of 0 to $2\pi$), then integrate the same way I did in a.)?
(Also, if someone could check my answer for a.) that would be great. I have an exam coming and I'm weak in this area. Thanks!)
Best Answer
You should have gotten $0$ as the answer for the first part. Since $xyz$ is odd w.r.t. $x$ and the ellipsoid is symmetric about the plane $x = 0$, the integral over the whole ellipsoid is $0$. Note this argument can also be used if the integrand is odd w.r.t. $y$ or $z$ and the region is symmetric about the planes $y = 0$ to $z = 0$ respectively.
For the second part, note that $xyz > 0$ over the part of the ellipsoid in the first octant, so the integral should evaluate to something positive.
Be sure to double check your bounds. The first octant has bounds of $0 \le \theta \le \dfrac{\pi}{2}$ and $0 \le \phi \le \dfrac{\pi}{2}$.