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