Surface integral over part of cylinder which is between two intersecting planes

Find the surface integral $\int\int_\Gamma(x-y^2+z^2)d\sigma$,
where $\Gamma$ is part of cylinder $x^2+y^2=a^2$, which is between $x-z=0$ and $x+z=0$, $x \ge0$.

I tried to parametrize the surface using $y=r\cos\phi$, $z=r\sin\phi$, $x=\sqrt{a^2-y^2}$, where $0<\phi<2\pi$ and $0<r<a$, then turned double integral into single integral, but it looks too hard to be correct.

Did I do something wrong or this is correct? And what is the easiest way to solve this surface integral?

Edit:
As I understood, I need the part of green cylinder between red and blue planes. The projection onto Oyz would be circle

Please note that the surface area element of a cylinder is $dS = a ~d\theta ~dz$, where $a$ is the radius of the cylinder.

As $x \geq 0, - \pi/2 \leq \theta \leq \pi/2$.

Also, the surface is bound between $~z = - x = - a \cos\theta$ and $z = x = a \cos\theta$

So the integral becomes,

$ \displaystyle \int_{-\pi/2}^{\pi/2} \int_{-a\cos\theta}^{a\cos\theta} a \cdot (a \cos\theta - a^2 \sin^2\theta + z^2) ~ dz ~ d\theta$