Btukfyl

$V$ is the portion of the cone $z=sqrt{x^2+y^2};$ for $; xgeq 0.$
Find $$intintlimits_{V}int xe^{-z} dV.$$
I am trying to solve this question. The answer is supposed to be just $4.$

I have worked out the limits as

$0leq z leq infty;$ and
$;-pi/2leq theta leq pi/2;$ and $;0leq R leq z.$ What am I doing wrong?

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^inftyint_0^ze^{-z}r^2costheta,drdzdtheta\
&=int_{-frac{pi}{2}}^{frac{pi}{2}}int_0^inftyfrac{z^3}{3}e^{-z}costheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}left(-e^{-z}(z^3+3z^2+3z+6)right)bigrvert_0^inftycostheta, rdrdzdtheta\
&=frac{1}{3}int_{-frac{pi}{2}}^{frac{pi}{2}}6costheta, rdrdzdtheta\
&=2(sintheta)bigrvert_{-frac{pi}{2}}^{frac{pi}{2}}\
&=4
end{align*}
The $z$ integration requires integrating by parts a few times, and then taking a limit.

According to the limit you have indicate, which seems to be correct, we should have

$$intintlimits_{V}int xe^{-z} dV=int_0^infty dz int_{-pi/2}^{pi/2} dtheta int_0^z R^2cos theta e^{-z}dR$$

Maybe you forgot the $R ,dR ,dz ,dtheta$ term or simply you have made a wrong evaluation.

Note that the one presented here works fine: Integral evaluation.