$\displaystyle V = \int_{a}^{b} A(x) dx$The solid lies between planes perpendicular to the x axis at $\displaystyle x = -1$ and $\displaystyle x = 1$. The cross-sections perpendicular to thexaxis between these planes are squares whose bases run from the semi-circle $\displaystyle y = -\sqrt{1 - x^{2}}$ to the semi-circle $\displaystyle y = \sqrt{1 - x^{2}}$

$\displaystyle x^{2}$ is the area of a square

$\displaystyle A = x^{2}$

$\displaystyle V = \int_{-1}^{1} [2 \sqrt{1 - x^{2}}]^{2}$ On the right track here?

$\displaystyle V = \int_{-1}^{1} [4 (1 - x^{2})]$

$\displaystyle V = \int_{-1}^{1} 4 - 4x^{2}$

The book says the answer will come out to $\displaystyle \dfrac{16}{3}$