Integral of Ellipsoid by Integral over Disk

Question

I have a problem where we need to determine the Volume of a Ellipsoid given by

$$\frac{x^2}2+\frac{y^2}2+\frac{z^2}9 \le 1$$ and the triple Integral $$\iiint_D z^2 dV$$

my approach was to solve for $z$ to get the surface $z(x,y)$ and integrate over the area $D$ given by the disk which is given by the intersection of the ellipsoid and the $xy$-plane, which means:

i determined the radius of the disk by setting $z = 0$ which gives me (with shifting to polar coordinates):

$$r = \sqrt2$$

if we now substitute $$x = r\cos(\phi)$$ and $$ y = r\sin(\phi)$$ and $$z = z $$ we get that $$z = \pm\sqrt{9-\frac92r^2}$$

if we now take the Integral $$\iint z^2(r,\phi) dA $$ we should get the Volume of the ellipsoid which then would be $$\int_0^{2\pi} \int_0^{\sqrt2} \left(9-\frac92r^2\right) *r \,dr\,d\phi $$ $$ \frac12V = 9\pi$$ as we only calculated for the positive value of $z$ which gives us the volume of the upper half of the ellipsoid, as the ellipsoid is symmetric the overall Volume should then be $$2*9\pi = 18\pi$$ however this solution is not correct, so i wonder if i am even allowed to use the formula i used when integrating over another function.

Thanks for ur help

(for some reason some $$ dont work i am new to this and cant figure it out i hope the problem is still understandable)

Edit: I believe i might have understood my mistake, the double Integral over a surface $D$ with $z(x,y)$ can be Interpreted as the Volume but is not the same as $dV$

Answer 1

Not an answer, but volume integrals over ellipsoids of the form

$ \displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 $

might be easier to handle by introducing the modified spherical coordinates

$ \begin{align} x&=ar\cos\phi\sin\theta\\ y&=br\sin\phi\sin\theta\\ z&=cr\cos\theta \end{align} $

where $r$ is the radial variable, and $\theta,\phi$ are the aziumthal and polar angles, respectively. In the present case we have

$ \displaystyle\int dV z^2 =a^2c\int_0^1 r^2 dr \int_0^{\pi/2}\sin\theta d\theta\int_0^{2\pi}d\phi (cr\cos\theta)^2 $

where $a=\sqrt{2},c=3$, and $a^2cr^2\sin\theta$ is the Jacobian of the transformation.

Answer 2

With your choice of cylindrical coordinates, the ellipsoid $D$ is the set

$$D = \left\{(r,\theta,z) \mid r\in[0,\sqrt2]\land\theta\in[0,2\pi]\land z\in\left[-3\sqrt{1-\frac{r^2}2},3\sqrt{1-\frac{r^2}2}\right]\right\}$$

Then the triple integral is

$$\iiint_D z^2 \, dV = \int_0^{\sqrt2} \int_0^{2\pi} \color{red}{\int_{-3\sqrt{1-\tfrac{r^2}2}}^{3\sqrt{1-\tfrac{r^2}2}}} r \color{red}{z^2 \, dz} \, d\theta \, dr$$

You've overlooked the innermost integral with respect to $z$; the triple integral does not reduce to the double integral as you've described. You should instead be getting

$$\int_a^b z^2 \, dz = \frac{b^3-a^3}3$$

or if you take advantage of the symmetry of $a=-b$,

$$2\int_0^b z^2 \, dz = \frac{2b^3}3$$