Finding the Centroid of the Paraboloid $z=x^2 +y^2$ Find the centroid of the solid paraboloid inside $z = x^2 + y^2$, $ \ 0<z<c$
I am not really sure where to start. I know the the centroid is 1 over the volume  times $\int z\, dz$. But I am un sure how to find the volume. I was thinking of integrating $\pi r^2z$ and that would be the volume. Is that correct? 
 A: We need to say something about the size of the paraboloid, like $z$ ranges from $0$ to (say) $b$. Imagine the solid paraboloid is made of material of density $1$. So mass is equal to volume. 
The volume is obtained by slicing.  Consider a thin slice, parallel to the $x$-$y$ plane, "at" height $z$, and of thickness "$dz$." This thin slice has approximate volume $\pi(x^2+y^2)\,dz$. For the full volume, "add up" (integrate) from $z=0$ to $z=b$. The volume is therefore $\displaystyle\int_{z=0}^{b} \pi(x^2+y^2)\,dz$, where $x^2+y^2=z$. The integration is easy. We get $\dfrac{\pi b^2}{2}$. 
For the height of the centroid, you need to find the moment of the paraboloid about the $x$-$y$ plane.
The  cross-sectional area at height $z$ is $\pi(x^2+y^2)=\pi z$. A thin slice of thickness $dz$ at that height has approximate volume $\pi z\,dz$, and therefore moment about the $x$-$y$ plane approximately $z(\pi z \,dz)$. 
So the full moment about the $x$-$y$ plane is $\displaystyle\int_{z=0}^b \pi z^2\,dz$. 
Finally, divide the moment by the volume. 
