If $C$ is the region obtained by intersecting, the surfaces $z=x^2+y^2$ and $z=2-x^2-y^2$.
Express the volume of $C$ as the triple integral in cylindrical coordinates.
My try:
$x^2+y^2=2-x^2-y^2$
$x^2+y^2=1$ which a unit circle of radius $1$.
So $0\le r\le1$ and $0\le\theta\le\ 2\pi$ and $r^2\le z\le2-r^2$
So, the integral is $\int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{2-r^2}r(2-r^2-r^2)dz\ dr\ d\theta$
Is my above integral correct?
You're correct with those bounds.
Also the integral should be $$\int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{2-r^2}r dz\ dr\ d\theta$$ as you haven't integrated yet with respect to $z$, or $$\int_{0}^{2\pi}\int_{0}^{1}r(2-r^2-r^2) dr\ d\theta$$ once you do integrate with respect to $z$