Answer
= $\frac{16\pi}{15}$
Work Step by Step
A(x) = $\frac{\pi(diameter)^2}{4}=\frac{\pi[2(1-x^2)]^2}{4}$
=$\pi(1-2x^2+x^4)$
a=-1 , b=1
V=$\int^b_a A(x)dx$
=$\int^{1}_{-1} \pi(1-2x^2+x^4)dx$
=$\pi[x-\frac{2}{3}x^3+\frac{x^5}{5}]^1_{-1}$
=$2\pi (1-\frac{2}{3}+\frac{1}{5})$
=$\frac{16 \pi}{15}$
