7. Find the area of the region in the $x y$-plane satisfying $x^{6}-x^{2}+y^{2} \leq 0$.
Solution: $\pi / 2$
Rewrite the condition as $|y| \leq \sqrt{x^{2}-x^{6}}$. The right side is zero when $x$ is $-1,0$, or 1 , and it bounds an area symmetric about the $x$ - and $y$-axes. Therefore, we can calculate the area by the integral
$$
2 \int_{-1}^{1} \sqrt{x^{2}-x^{6}} d x=4 \int_{0}^{1} x \sqrt{1-x^{4}} d x=2 \int_{0}^{1} \sqrt{1-u^{2}} d u=\pi / 2 .
$$