Triple integral of portion of cone (cylindrical polar coordinates)? $V$ is the portion of the cone $z=\sqrt{x^2+y^2}\;$ for $\; x\geq 0.$
Find $$\iiint\limits_{V} xe^{-z} dV.$$
I am trying to solve this question.  The answer is supposed to be just $4.$ 
I have worked out the limits as 
$0\leq z \leq \infty\;$ and 
$\;-\pi/2\leq  \theta \leq \pi/2\;$ and $\;0\leq R \leq z.$ What am I doing wrong? 
 A: With the bounds you have stated, the integral is in fact finite:
\begin{align*}
\iiint_Vxe^{-z}dV&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^\infty\int_0^ze^{-z}r^2\cos\theta\,drdzd\theta\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_0^\infty\frac{z^3}{3}e^{-z}\cos\theta\, rdrdzd\theta\\
&=\frac{1}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(-e^{-z}(z^3+3z^2+3z+6)\right)\big\rvert_0^\infty\cos\theta\, rdrdzd\theta\\
&=\frac{1}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}6\cos\theta\, rdrdzd\theta\\
&=2(\sin\theta)\big\rvert_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\
&=4
\end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.
A: According to the limit you have indicate, which seems to be correct, we should have
$$\int\int\limits_{V}\int xe^{-z} dV=\int_0^\infty dz \int_{-\pi/2}^{\pi/2} d\theta \int_0^z R^2\cos \theta e^{-z}dR$$
Maybe you forgot the $R \,dR \,dz \,d\theta$ term or simply you have made a wrong evaluation.
Note that the one presented here works fine: Integral evaluation.
