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

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
$$
\int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (4- x^2 -y^2 -(4-2x) )dy\,dx = \int_{x=0}^2 \int_{y=-\sqrt{2x-x^2}}^{-\sqrt{2x-x^2}} (2x- x^2 -y^2 )dy\,dx 
$$
