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 $r$.


1 Answer 1


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 $2$.

So the integral should be,

$ \displaystyle \int_0^{2\pi} \int_0^2 \int_{r^2}^{8-r^2} r^2 ~ dz ~ dr ~ d\theta = \frac{256 \pi}{15}$


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