Answer
$$
\left( f(x)=4-x^{2}, \quad y =0, \quad x=-1, \quad x=1 \right)
$$
The volume of the solid formed by revolving the function, $f(x)=4-x^{2} $ about the x-axis from $x=-1$ to $x=1$, is
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\
&=\pi \int_{-1}^{1}\left(4-x^{2}\right)^{2} d x\\
&=\frac{406 \pi}{15} \\
&\approx 85.03 \end{aligned}
$$
Work Step by Step
$$
\left( f(x)=4-x^{2}, \quad y =0, \quad x=-1, \quad x=1 \right)
$$
If $f(x) $ is nonnegative and $R$ is the region between $f(x) $ and the x-axis from $x=a$ to $x=b$, the volume of the solid formed by rotating $R$ about the x-axis is given by
$$
\int_{a}^{b} \pi [f(x)]^{2}dx
$$
So,the volume of the solid formed by revolving the function, $f(x)=4-x^{2} $ about the x-axis from $x=-1$ to $x=1$, is
$$
\begin{aligned} V &=\int_{a}^{b} \pi [f(x)]^{2}dx\\
&=\pi \int_{-1}^{1}\left(4-x^{2}\right)^{2} d x \\ &=\pi \int_{-1}^{1}\left(16-8 x^{2}+x^{4}\right) d x \\ &=\left.\pi\left(16 x-\frac{8 x^{3}}{3}+\frac{x^{5}}{5}\right)\right|_{-1} ^{1} \\
&=\pi\left(16-\frac{8}{3}+\frac{1}{5}+16-\frac{8}{3}+\frac{1}{5}\right) \\
&=\pi\left(32-\frac{16}{3}+\frac{2}{5}\right) \\
&=\frac{406 \pi}{15} \\
&\approx 85.03 \end{aligned}
$$