Different results for the same triple integral: where is the mistake? Let $\Omega$ be the portion of the unit sphere lying in the first octant, I want to compute
$$
\int_{\Omega}xyz^2\,dx\,dy\,dz
$$
First method: using cylindrical coordinates, $\Omega$ becomes
$$
\Omega=\{(\rho,\theta,z):0\leq\rho\leq1,\,0\leq z\leq1,\,0\leq\theta\leq\pi/2\}
$$
In this coordinates system we have $dx\,dy\,dz=\rho\,d\rho\,d\theta\,dz$, so the integral is
$$\int_0^1\left(\int_0^1\left(\int_0^{\pi/2}z^2\rho^3\sin\theta\cos\theta\,d\theta\right)\,dz\right)\,d\rho$$
$$=\left(\int_0^1 \rho^3\,d\rho\right)\left(\int_0^1 z^2\,dz\right)\left(\int_0^{\pi/2} \frac{1}{2}\sin(2\theta)\,d\theta\right)\\
    =\frac{1}{24}$$
Second method: using cartesian coordinates $\Omega$ is
$$
\Omega=\{(x,y,z):0\leq z\leq 1,\,0\leq y\leq\sqrt{1-z^2},\,0\leq x\leq\sqrt{1-z^2-y^2}\}
$$
If I compute the integral the result is
\begin{align*}
\int_0^1\left(\int_0^{\sqrt{1-z^2}}\left(\int_0^{\sqrt{1-z^2-y^2}}xyz^2\,dx\right)dy\right)dz &=\frac{1}{2}\int_0^1z^2\left(\int_0^{\sqrt{1-z^2}}y(1-z^2-y^2)dy\right)dz\\
&=\frac{1}{8}\int_0^1 z^2(1-z^2)^2\,dz\\
&=\frac{1}{105}
\end{align*}
Why the two results are different? Where is the mistake? I think the second result is the correct one, so why using cylindrical coordinates I don't get the same?
 A: In the first method, the parametrization of $\Omega$ is not correct.
Indeed, using cylindrical coordinates, $\Omega$ is
$$
\left\{(\rho,\theta,z):0\leq\rho\leq1, 0\leq\theta\leq\pi/2,\,0\leq z\leq\sqrt{1-\rho^2}\right\}.
$$
Therefore, after setting $t=1-\rho^2$, we get
\begin{align*}
\int_{\Omega}xyz^2\,dx\,dy\,dz&=\int_0^1\left(\int_0^{\sqrt{1-\rho^2}}\left(\int_0^{\pi/2}z^2\rho^3\sin\theta\cos\theta\,d\theta\right)\,dz\right)\,d\rho\\&=
\left[\frac{\sin^2(\theta)}{2}\right]_0^{\pi/2}
\frac{1}{3}\int_0^1\rho^3(1-\rho^2)^{3/2}\,d\rho\\
&=\frac{1}{12}\int_0^{1} (1-t)t^{3/2}\, dt\\
&=\frac{1}{12}\left[\frac{t^{5/2}}{5/2}-\frac{t^{7/2}}{7/2}\right]_0^{1}=\frac{1}{105}
\end{align*}
that coincides with the result obtained by the second method (which is correct).
A: You should use spherical coordinates with
$$x=\rho  \sin (\theta ) \cos (\phi )$$
$$y=\rho  \sin (\theta ) \sin (\phi )$$
$$z = \rho cos(\theta)$$
Then the integral is
$$\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _0^1 x y z^2 \rho ^2\sin (\theta )d\rho d\theta d\phi $$
Evaluate the volume integral
$$\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _0^1\rho ^6 \sin ^3(\theta ) \cos ^2(\theta ) \sin (\phi ) \cos (\phi )d\rho d\theta d\phi =\frac{1}{105}$$
