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?

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
$$\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^2} =\frac1{ab}, \>\>\>
\int_{0}^{\frac{\pi}{2}} \frac{d\varphi}{(a\sin\varphi+b\cos\varphi)^4} =\frac{a^2+b^2}{3a^3b^3}$$
