Answer
$$
y=f(x)=\sqrt {1-x^{2}}
$$
The function has as its graph a semicircle of radius $r=1$ with center at$(0,0)$.
The volume that results when the semicircle is rotated about the x-axis is given by :
$$
\begin{aligned} V &=\pi \int_{-1}^{1}(f(x))^{2} d x\\
&=\pi \int_{-1}^{1}(\sqrt{1-x^{2}})^{2} d x \\
&=\frac{4 \pi}{3}. \end{aligned}
$$
Work Step by Step
$$
y=f(x)=\sqrt {1-x^{2}}
$$
The function has as its graph a semicircle of radius $r=1$ with center at$(0,0)$.
The volume that results when the semicircle is rotated about the x-axis is given by :
$$
\begin{aligned} V &=\pi \int_{-1}^{1}(f(x))^{2} d x\\
&=\pi \int_{-1}^{1}(\sqrt{1-x^{2}})^{2} d x \\ &=\pi \int_{-1}^{1}\left(1-x^{2}\right) d x \\ &=\left.\pi\left(x-\frac{x^{3}}{3}\right)\right|_{-1} ^{1} \\ &=\pi\left(1-\frac{1}{3}\right)-\pi\left(-1+\frac{1}{3}\right) \\
&=2 \pi-\frac{2}{3} \pi\\
&=\frac{4 \pi}{3}. \end{aligned}
$$
