Triple Integral over a disk How do I integrate $$z = \frac{1}{x^2+y^2+1}$$ over the region above the disk $x^2+y^2 \leq R^2$?

Use polar coordinates: $x^2+y^2 = r^2$, etc.  An area element is $dx\, dy = r \, dr \, d\theta$.  The integral over the disk is
$$\int_0^R dr \, r \: \int_0^{2 \pi} d\theta \frac{1}{1+r^2} = 2 \pi \int_0^R dr \frac{r}{1+r^2}$$
You can substitute $u=r^2$ to get for theintegral
$$\pi \int_0^{R^2} \frac{du}{1+u}$$
I trust that you can evaluate this.
