Find the volume in the first octant bounded by the parabolic cylinder, bounded by $x-1 = y^2$, the circular cylinder $4z + 4y -z^2 - y^2 = 7$ and the plane $x = 0$..
My attempt: I first took the circular cylinder and completed the square to get $(y-2)^2 + (z-2)^2 =1$. However now I am not sure where to really go... Any help would be appreciated.
Call the bounded region $D$, which is given by $$D=\{(x,y,z)\,|\,0\le x\le y^2+1,\,2-\sqrt{1-(z-2)^2}\le y\le2+\sqrt{1-(z-2)^2},\,1\le z\le3\}$$. Here's a plot rendered in Mathematica.
The volume is given by the integral $$\iiint_D \mathrm{d}V=\int_1^3\int_0^{y^2+1}\int_{2-\sqrt{1-(z-2)^2}}^{2+\sqrt{1-(z-2)^2}}\mathrm{d}y\,\mathrm{d}x\,\mathrm{d}z$$(Note that this is not the only integral setup you can use.)
Convert to cylindrical coordinates, using $$\begin{cases}x=x\\[1ex]y=r\sin t\\[1ex]z=r\cos t\end{cases}$$which yields the Jacobian $J$ $$J=\begin{vmatrix}\dfrac{\partial x}{\partial r}&\dfrac{\partial x}{\partial t}&\dfrac{\partial x}{\partial x}\\[1ex] \dfrac{\partial y}{\partial r}&\dfrac{\partial y}{\partial t}&\dfrac{\partial y}{\partial x}\\[1ex] \dfrac{\partial z}{\partial r}&\dfrac{\partial z}{\partial t}&\dfrac{\partial z}{\partial x}\end{vmatrix}=\begin{vmatrix}0&0&1\\[1ex] \sin t&r\cos t&0\\[1ex] \cos t&-r\sin t&0\end{vmatrix}=-r$$so that the integral becomes $$\iiint_D\mathrm{d}V=\iiint_{D^*} |J|\,\mathrm{d}r\,\mathrm{d}t\,\mathrm{d}x=\iiint_{D^*} r\,\mathrm{d}r\,\mathrm{d}t\,\mathrm{d}x$$ where $D^*$ is used to denote the same region as $D$, but in cylindrical coordinates; that is, $$D^*=\{(r,t,z)\,|\,0\le r\le1,\,0\le t\le2\pi,\,0\le x\le r^2\sin^2t+1\}$$ This leaves you with $$\int_0^{2\pi}\int_0^1\int_0^{r^2\sin^2t+1} r\,\mathrm{d}x\,\mathrm{d}r\,\mathrm{d}t$$