What would be the volume of the region bounded by paraboloid $z=x^2 + y^2$ and the plane $z=4y$ ?
I am unable to imagine this region in 3D. Though we can say $z$ goes from $4y$ to $x^2 + y^2$, but what about $x$ and $y$?
I have tried with putting $z=4y$ in equation of paraboloid and then getting equation of circle, and hence the limits for $x$ and $y$. The answer I am getting is $76\pi/3$. Is this correct approach?
The intersection curve of the paraboloid and the plane projected on the $xy$-plane is given by $$4y=z=x^2+y^2\leftrightarrow x^2+(y-2)^2=2^2$$ that is the circle of centre $(0,2)$ and of radius $2$. Note that over the disc $x^2+(y-2)^2\leq 2^2$, the plane stays above the paraboloid. Hence the volume of the region is equal to \begin{align*} V&=\iint_{x^2+(y-2)^2\leq 2^2}\left(\int_{z=x^2+y^2}^{4y}1dz\right)\,dxdy\\ &= \iint_{x^2+(y-2)^2\leq 2^2} (4-x^2-(y-2)^2)\,dxdy\\ &=\int_{\theta=0}^{2\pi}\int_{\rho=0}^2 (4-\rho^2)\rho d\rho d\theta\\ &=2\pi\left[4\frac{\rho^2}{2}-\frac{\rho^4}{4}\right]_0^2=8\pi \end{align*} where in the last step we used the polar coordinates $x=\rho\cos(\theta)$, $y=2+\rho\sin(\theta)$.