# Find the volume of the solid in the firsr quadrant bounded by the coordinate planes, the cylinder $x^{2} + y^{2}=4$, and the plane $z+y=3$

Find the volume of the solid in the first quadrant bounded by the coordinate planes, the cylinder $x^{2} + y^{2}=4$, and the plane $z+y=3$.

If we draw the graph, then the integral will be calculated should be

$$\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} (3-y) \: dy dx$$

with $3-y = z = f(x,y)$.

The boundary $\sqrt{4-x^{2}}$ is from the cylinder. Is this correct? Thanks.