Finding $\iiint x^2\,{\rm d}x{\rm d}y{\rm d}z$ over the volume bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2 }= 1$ 
Finding the value of $$\iiint x^2\,\mathrm dx\mathrm dy\mathrm dz$$ over the volume bounded by the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1.$$

I am taking limits as follow:
$$-a \leq x \leq a$$
$$-\frac{b}{a} (a^2-x^2)^{1/2} \leq y \leq \frac{b}{a} (a^2-x^2)^{1/2}$$
$$-\frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} \leq z \leq \frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $\mathrm dz$ first then $\mathrm dy$ and then $\mathrm dx$. But it is just becoming harder and harder with each step.
I do not have a tutor. So, I highly need help of you masters. Please help me.

First, rewrite the integral as
$$\int_{-a}^a x^2 \int\int_{E_x}1\ dy\ dz\ dx $$
where $E_x$ is the ellipse in the $yz$-plane satisfying $\frac{y^2}{b^2} + \frac{z^2}{c^2} = (1-\frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $\frac{y^2}{b^2\left(1-\frac{x^2}{a^2}\right)} + \frac{z^2}{c^2\left(1-\frac{x^2}{a^2}\right)}=1$, we see that the area of said ellipse is $\pi bc(1-\frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).
Thus, the triple integral simplifies to the single-variable integral
$$\int_{-a}^a x^2 \pi bc(1-\frac{x^2}{a^2})\  dx= \frac{4}{15}\pi a^3 bc$$
