Answer
a. $\frac{32}{3}$
b. $\frac{32}{3}$
Work Step by Step
a. See figure. Find the intersections between the two functions. Letting $y=3-x^2=-1$, we have $x=\pm2$ and the intersections are $(-2,-1),(2,-1)$. Use the symmetry and integrate with respect to $x$. We have the enclosed area as:
$A=2\int_{0}^2 (3-x^2-(-1))dx =8x|_{0}^2-\frac{2}{3}x^3|_{0}^2=8(2)-\frac{2}{3}(2)^3=\frac{32}{3}$
b. Notice that the maximum of $y=3-x^2$ is at $(0,3)$. Integrate with respect to $y$ for $x=\sqrt {3-y}$ in the first quadrant. Use the function symmetry and we have the enclosed area as:
$A=2\int_{-1}^3 (\sqrt {3-y})dy=-2\int_{-1}^3 (3-y)^{1/2}d(3-y)=-\frac{4}{3}(3-y)^{3/2}|_{-1}^3=-\frac{4}{3}(3-3)^{3/2}+\frac{4}{3}(3-(-1))^{3/2}=\frac{32}{3}$
