Solution
= Solution
First perform the inner integral, which gives the vertical width of the region:
$$
I=\int_{-a}^{a}\big(\sqrt{1-x^2}-x^2\big)\,dx
=2\int_0^a\big(\sqrt{1-x^2}-x^2\big)\,dx.
$$
Using $\int\sqrt{1-x^2}\,dx=\tfrac12[x\sqrt{1-x^2}+\arcsin x]$ gives
$$
I=a\sqrt{1-a^2}+\arcsin a-\frac23a^3.
$$
At the intersection, $\sqrt{1-a^2}=a^2$, so
$$
\boxed{I=\arcsin a+\frac13a^3,\qquad A=1,\quad B=\frac13}.
$$