<s>[INST] Trouble with a triple integral on a region bounded by a sphere and two planes 
I would like to compute the integral $\int_A zdzdydx,$ where $A$ is the region bounded by the sphere $x^2+y^2+z^2=R^2,$ plane $\frac{x}a+\frac{y}b=1$ and coordinate planes (which doesn't contain the origin on its boundary and is in the first quadrant).

I considered switching to either spherical or cylindrical coordinates, but I don't see any symmetry or pattern.
In spherical coordinates, I tried expressing the lower bound for radius $r$ in terms of $\theta\in[0,\pi/2]$ $$r(\theta, \varphi)=\frac{ab}{a\sin\varphi\sin\theta+b\cos\varphi\sin\theta}$$  and the integral becomes
\begin{aligned}&\color{white}=\int_0^{\pi/2}\int_0^{\pi/2}\int_{\frac{ab}{a\sin\theta\sin\varphi+b\sin\theta\cos\varphi}}^Rr^2\sin\theta r\cos\theta drd\theta d\varphi\\&=\int_0^{\pi/2}\int_0^{\pi/2}\sin\theta\int_{\frac{ab}{a\sin\theta\sin\varphi+b\sin\theta\cos\varphi}}^Rr^3drd\theta d\varphi\end{aligned}
In cylindrical coordinates
$$r(\varphi)=\frac{ab}{a\sin\varphi+b\cos\varphi}$$ and the integral is
$$\int_0^{\pi/2}\int_{\frac{ab}{a\sin\varphi+b\cos\varphi}}^R\int_0^{\sqrt{R^2-r^2}}zrdzdrd\varphi$$
However, I have to deal with a powers of $a\sin\theta\sin\varphi+b\sin\theta\cos\varphi$ and $a\sin\varphi+b\cos\varphi$ too early, and if I swap the order of integration, I need to express the angles in terms of $r$ which seems worse.
I saw the substitution in this answer, but it isn't so smooth here and I've seen the reduction formula.
How should one attack this task?
 [/INST] Proceed in cylindrical coordinates as follows
\begin{align}
&\int_0^{\pi/2}\int_{r(\varphi)}^R\int_0^{\sqrt{R^2-r^2}}z \>rdzdrd\varphi\\
= & \> \frac12 \int_0^{\pi/2}\int_{\frac{ab}{a\sin\varphi+b\cos\varphi}}^R (R^2-r^2)r dr d\varphi\\ 
= & \> \frac12 \int_0^{\pi/2}\left( \frac14R^4–\frac12\frac{a^2b^2R^2}{(a\sin\varphi+b\cos\varphi)^2}
 + \frac14\frac{a^4b^4}{(a\sin\varphi+b\cos\varphi)^4 } \right)d\varphi\\ 
 = & \> \frac\pi{16}R^4 -\frac 14abR^2 +\frac1{24}ab(a^2+b^2)\\ 
\end{align}
where
$$\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^2} =\frac1{ab}, \>\>\>
\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^4} =\frac{a^2+b^2}{3a^3b^3}$$
</s>