# Evaluate this Integral Using Change of Variables

I have this integral here $$\iiint_R\sqrt{x^2+y^2}dV$$, $$R$$ is the region bounded by $$z=x^2+y^2$$ and $$z=8-x^2-y^2$$.

I immediately noticed the $$\sqrt{x^2+y^2}$$, which is a sign that it would be easier to use cylindrical coordinates. Converting everything, I got $$z=r$$, $$z=8-r$$, and $$\iiint r$$. This means $$r \leq z \leq 8-r$$, $$0 \leq r \leq 8$$, $$0 \leq \theta \leq 2 \pi$$.

So the final form of the integral that I need to evaluate should be $$\int_{0}^{2\pi}\int_{0}^{8}\int_{r}^{8-r}r^2dzdrd\theta$$ right? Though I am not so sure about $$r$$.

You are right that it is easier to evaluate it in cylindrical coordinates but your integral bounds are not correct. You have two paraboloids, $$z = x^2 + y^2$$ opening up and $$z = 8 - x^2 - y^2$$ opening down.
Note that in polar coordinates, $$x^2 + y^2 = r^2$$
So at the intersection of both paraboloids, $$z = 8 - r^2 = r^2 \implies r = 2$$
That means the projection of the region in xy-plane is a circle of radius $$2$$.
$$\displaystyle \int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r^2 ~ dz ~ dr ~ d\theta = \frac{256 \pi}{15}$$