Calculate the volume enclosed by the paraboloid $z=x^2+y^2$ and the plane $z=2y$.
When $x = 0$ it seems that the region of interest lies between $y=2$ and $z=4$. If I'm not mistaken I would need to calculate $$\int_D(x^2+y^2-2y) \ dA.$$
Also $D$ seems to be given by $D=\{(x,y) \in \mathbf{R}^2 \mid x^2+y^2 \le 2y \}$
My question is how can I found the bounds for the integrals? Any hints would be appreciated.
Best Answer
You found the intersection correctly. Here is a 2D cross section of how it would look when seen from the side on $YZ$ plane. It is bound below by the paraboloid $z = x^2+y^2$ and above by the plane $z = 2y$,
From $z = x^2 + y^2, z = 2y,$ you get $x^2 = 2y-y^2 \implies x = \pm \sqrt{2y-y^2}$
So your integral becomes,
$\displaystyle \int_{0}^{2}\int_{-\sqrt{2y-y^2}}^{\sqrt{2y-y^2}}(2y - x^2 - y^2) dx \, dy$
In cylindrical coordinates,
$x = r \cos \theta, y = r \sin \theta$
So $z = x^2 + y^2 =r ^2 = 2y = 2r \sin \theta \implies r = 2 \sin \theta$
So your integral becomes,
$\displaystyle \int_{0}^{\pi}\int_{0}^{2\sin\theta}(2r\sin \theta - r^2) \, r \, dr \, d\theta$