Find the volume of the wedge cut from the first octant by the cylinder $z = 12 - 3y^2$ and the plane $x+y=2$.
What I did- sketched parabola and repeated in all of x axis, drew the plane and found the domain in which its to be integrated which is $0<y<2$ and $2-y<x<2$. The problem is it's wrong. It would be great if anyone could show the right direction to solve the problem
In Cartesian coordinates, one integral for the volume could be
$$\int_0^2 \int_0^{2-x} \int_0^{12-3y^2} dz \, dy \, dx = \int_0^2 \int_0^{2-x} (12-3y^2)\, dy \, dx$$
or by swapping $x$ and $y$ (permissible by symmetry),
$$\int_0^2 \int_0^{2-y} (12-3y^2) \, dx \, dy$$