Triple integral of portion of cone (cylindrical polar coordinates)?

Question

$V$ 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?

Answer 1

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.

Answer 2

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.