here:
$z$ varies from $1$ to $0$,
$y$ varies from $\sqrt{1−x^2}$ to $0$,
$x$ varies from $1$ to $0$
$$\begin{align}\iiint x^2y\ \mathrm{d}x\ \mathrm{d}y\ \mathrm{d}z=&\iint x^2yz\ \mathrm{d}x\ \mathrm{d}y \\ =&\iint x^2y\ \mathrm{d}x\ \mathrm{d}y\\ =&\int \frac{x^2y^2}{2}\mathrm{d}x\\ =&\frac{1}{2} \int x^2(1−x^2)\ \mathrm{d}x\\ =&\frac{1}{2}\int(x^2−x^4) \ \mathrm{d}x\\ =&\frac{1}{2}\left(\frac{x^3}{3}−\frac{x^5}{5}\right)\\ =&\frac{1}{6}−\frac{1}{10}\\ =&\frac{1}{15} \end{align}$$
But the answer supposed to be is $0$.
I don't know where I did wrong.
You can simply compute the integral using cylindrical coordinates: $$\iiint_Ef(x,y,z)d^3\textbf x=\int_0^1dz\iint_{\mathbb B_1(0)}f(x,y,z)dxdy=\int_0^1dz\iint_{(0,1)\times(0,2\pi)}\rho^4\sin\theta\cos^2\theta d\rho d\theta\\ =\int_0^1dz\int_0^1\rho^4d\rho\int_0^{2\pi}\sin\theta\cos^2\theta d\theta.$$