So I have just started studying double integrals and the book I have bought hasn’t many solved examples. An exercise asks to show that the volume of the solid created by the intersection of the surfaces $z=x^2$ and $z=4-x^2-y^2$ is $4\pi\sqrt{2}$. I have solved $z_1=z_2$ but don’t know how to find the limits of integration. Sometimes they are just numbers and other times they are functions so I’m confused. If anyone can solve it in Cartesian or in parametric coordinates please help and explain.
Thanks in advance.
If $z_1=z_2$ you have $x^2=4-x^2-y^2$, then $$ 2x^2+y^2=4$$ $$\dfrac{x^2}{2}+\dfrac{y^2}{4}=1$$ This is an ellipse with vertxs $(-\sqrt{2},0)$ and $(\sqrt{2},0)$ and also $$y= \pm \sqrt{4-2x^2}$$ $$V= \int_{-\sqrt{2}}^{\sqrt{2}} \int_{- \sqrt{4-2x^2}}^{\sqrt{4-2x^2}} (4-x^2-y^2-x^2 ) \ dy \ dx$$