Answer
=$\frac{8}{3}$
Work Step by Step
A(x) = $\frac{(diagonal)^2}{2}$
=$\frac{[\sqrt{1-x^2}-(-\sqrt{1-x^2})]^2}{2}$
=$2[\frac {2\sqrt {1-x^2}}{2}]$
=$2(1-x^2)$ a=-1, b=1;
V=$\int^b_a A(x) dx =2\int^1_{-1}(1-x^2)dx$
=$2[x-\frac{x^3}{3}]^1_{-1}$
=$4(1-\frac{1}{3})$
=$\frac{8}{3}$
