parabolic volume integration mistake

Question

While dealing with volume integral of paraboloid bowl, say the volume within the surface $x^2 + y^2 = r^2 $ and $z = r^2$, where $ 0\le r\le \sqrt{3} $. One classic way seems to firstly have infinitesimal volume via slicing area and infinitesimal height: $$dV =Adz= \pi r^2dz=\pi z\,dz,$$ and then integrate for total volume $$V= \int {dV}= \int_{z=0}^{3} \pi z \,dz =\frac{1}{2} \pi z^2 |_{0}^{3}=\frac{9}{2} \pi .$$ When I practiced integration via cylinder coordinate $(R,\theta,z)$, I got wrong result but unable to spot where I did wrong. The infinitesimal volume I derived: $$dV=dA*dz=(R\,d\theta\, dR)\,dz,$$ and then I got the total volume$$V= \int {dV}= \int_{z=0}^{3}\int_{\theta=0}^{2\pi}\int_{R=0}^{\sqrt{3}} R\,d\theta \,dR\,dz =\frac{1}{2} \pi z^2 |_{0}^{3}=9 \pi.$$ Sincerely hope my mistake can be pointed out.

Answer 1

The correct integral should be

$\displaystyle \int_{z=0}^{3}\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt{z}} r \ dr \ d\theta \ dz$

Please note that the radius is not $\sqrt3$ throughout. It increases with $z$ and for any given $z$, the upper bound of radius is $\sqrt z \ $. If you take upper bound of $r$ independent of $z$, you will get volume of the cylinder of radius $\sqrt3$ and of height $3$.