Answer
$\frac{16}{3}$
Work Step by Step
Step 1: We find edge: $[\sqrt{ 1-x^2} - (-\sqrt {1-x^2})] $
Step 2: We find area: $A(x)=(edge)^2 = [\sqrt{ 1-x^2} - (-\sqrt {1-x^2})]^2 = (2\sqrt {1-x^2})^2 = 4(1-x^2)$;
Step 3: We find volume: V=$\int^b_a A(x) dx $ in
here $ a=-1, b=1$;
V = $\int^1_{-1} 4 (1-x^2)dx$
$4[x-\frac{x^3}{3}]^1_{-1}$ = $8(1-\frac{1}{3})$ = $\frac{16}{3}$
