MHB Implicit surface area, sphere cut by two plane

[skate_nerd]

Got a little problem here, just want to make sure what I'm doing is right because it's a little different from anything I've done that is using the same formula initially.
So there's a sphere denoted as \(\Omega\):
$$x^2+y^2+z^2=R^2$$
and its cut by two planes \(z=a\) & \(z=b\) where \(0\leq{a}\leq{b}\leq{R}\). I have to find the area of the portion of \(\Omega\) cut by those two planes.

So I got to setting up this integral. Not so bad...
$$\int\int_{\Omega}^{ }\frac{|\nabla{F}|}{|\nabla{F}\bullet{p}|}\,dA$$
where $$\nabla{F}=2x\hat{i}+2y\hat{j}+2z\hat{k}$$
so now $$|\nabla{F}|=\sqrt{4x^2+4y^2+4z^2}=2\sqrt{x^2+y^2+z^2}=2R$$
and then $$\nabla{F}\bullet{p}$$ where \(p=\hat{k}\) (normal to the surface)
$$=(2x\hat{i}+2y\hat{j}+2z\hat{k})\bullet{\hat{k}}=2z$$
so now the integral looks like $$\int\int_{\Omega}^{ }\frac{2R}{2z}\,dA=R\int\int_{\Omega}^{ }\frac{1}{z}\,dA$$
This is where my problem comes up...I don't really have anything I can do with z...I have the two planes \(z=a\) & \(z=b\) but then how would I find my bound of integration? I feel like polar coordinates would be optimal but even so, what would the inner \(dr\) bounds be? The two connecting circles are
$$x^2+y^2=R^2-a^2$$ and $$x^2+y^2=R^2-b^2$$
Would integrating this integrand I found over this two circles make sense for finding the area?

 

[skate_nerd]

A little update, I made some progress but am still not sure if what I did would work correctly so if somebody could verify that would be nice.
So I used the given sphere \(x^2+y^2+z^2=R^2\) and solved for \(z\) getting \(z=\sqrt{R^2-(x^2+y^2)}\). Plugged this into the integrand I got above, and put it in polar coordinates, giving
$$R\int_{0}^{2\pi}\int_{a}^{b}\frac{r}{\sqrt{R^2-r^2}}\,drd\theta$$
I solved the inner integral using a u-substitution, and got
$$-\frac{R}{2}\int_{0}^{2\pi}2(\sqrt{R^2-b^2}-\sqrt{R^2-a^2})\,d\theta$$
and finally ended up with the formula for this area:
$$-2\pi{R}(\sqrt{R^2-b^2}-\sqrt{R^2-a^2})$$ which is equivalent to
$$2\pi{R}(\sqrt{R^2-a^2}-\sqrt{R^2-b^2})$$
And there is my best guess. Anybody up for some proof reading? (Rock)

 

[skate_nerd]

Ugh. Consider this post solved (Wasntme). I made the bounds of integration of the dr integral \(\sqrt{R^2-a^2}\) to \(\sqrt{R^2-b^2}\) and got the answer \(2\pi{R}(b-a)\) which is what it should be.
Cool problem though. Ultimately is meant to show that this arbitrary portion of a sphere's area is always going to be equal to the portion cut by the same two planes out of the infinit cylinder \(x^2+y^2=R^2\). Worked out.