Find the volume between $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral

Question

So the volume of $z=4-x^2-y^2$ and $z=4-2x$ as a triple integral shall look similar to $$\int^2_0\int^{y=?}_{y=?}\int^{4-x^2-y^2}_{4-2x} dz dy dx$$ but how do I find the limits on $y$?

Answer 1

The point is that you are integrating over a domain in which $4-2x<4-x^2-y^2$. In that domain, $$ -2x < -x^2-y^2\\ 2x>x^2+y^2\\ y^2 < 2x -x^2 \\ -\sqrt{2x-x^2} < y < +\sqrt{2x-x^2} $$ So the integral is $$ \int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (4- x^2 -y^2 -(4-2x) )dy\,dx = \int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (2x- x^2 -y^2 )dy\,dx $$