So, I asked this question and got downvotes and someone saying I should present my attempt. Here lies my thought process:
I know the volume is $\iiint dxdydz$ on any W$\subset R^3$ and I "just" had to find the proper values for the W, namely x, y, z, their range.
I said $-1≤x≤1$, $-10≤y≤10$ and $-2≤z≤2$.
Expressing them on the integral, for every variable we set the next one to 0 and the previous as constant so it's:
$-1≤x≤1$
$-10 \sqrt{1-x^2} ≤y≤10 \sqrt{1-x^2}$
$-2 \sqrt{1-x^2-\frac{y^2}{100}} ≤ z ≤ 2 \sqrt{1-x^2-\frac{y^2}{100}}$
and it actually went from $dxdydz$ to $dzdydz$
Any thoughts? Thanks in advance.
What you did is correct. Translating it into an integral in Cartesian coordinates, it becomes$$\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}\int_{-2\sqrt{1-x^2-\frac{y^2}{100}}}^{2\sqrt{1-x^2-\frac{y^2}{100}}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$You can compute it as follows:\begin{align}\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}\int_{-2\sqrt{1-x^2-\frac{y^2}{100}}}^{2\sqrt{1-x^2-\frac{y^2}{100}}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx&=\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}4\sqrt{1-x^2-\frac{y^2}{100}}\,\mathrm dy\,\mathrm dx\\&=\int_{-1}^120\pi(1-x^2)\,\mathrm dx\\&=\frac{80}3\pi.\end{align}However, the computation becomes much easier if you do the change of coordinates$$\left\{\begin{array}{l}x=X\\y=10Y\\z=2Z\end{array}\right.$$Then your integral becomes$$20\int_{-1}^1\int_{-\sqrt{1-^2}}^{\sqrt{1-X^2}}\int_{-\sqrt{1-X^2-Y^2}}^{\sqrt{1-X^2-Y^2}}1\,\mathrm dZ\,\mathrm dY\,\mathrm dX,\tag1$$since, other than the factor $20$, $(1)$ is the volume of the unit sphere, which is $\frac43\pi$. So you get, again, that the answer is $\frac{80}3\pi$.