Q:

Double integral in Cartesian coordinates

Using double integrals
The volume of solid limited by the cylinder $x^2\ + y^2 = 9$ , and plans $z = 0$ and $z = 3 - x$.

A:

$$\int_{-3}^{3}\int_{-\sqrt{3-x^2}}^{\sqrt{3-x^2}} (z_2 - z_1) dy dx $$
Where $z_2=3-x$ is upper surface and $z_1=0$ is lower surface.

