A little bit rusty on the topic of surface integrals and perhaps some basic calculus after a while. A little help would be appreciated
Find the volume of the solid under the surface $$z=3x^{2}+y^{2} $$ and above the region bounded by $$y=x$$ and $$y^{2}-y=x.$$
The main problem I'm facing is with the parabola. I've found the parabola to cut the y-intercept at $y=0$ and $y=1$. However, this does not square with the solutions I am looking at.
Any help is appreciated.
It might be easier here to view the region in the $x-y$ plane, which comprises the domain for the integral, having $y$ playing the role of the independent variable and having $x$ as a function of $y$.
The region in the $x-y$ plane is bounded from $x=y^2-y$ to $x=y$ as $y$ goes from $0$ to $2$.
Therefore, the volume integral is
$$V=\int_0^2\int_{y^2-y}^y(3x^2+y^2)dx\,dy$$
Can you finish?