Answer
$$
y=f(x)=\sqrt {36-x^{2}}
$$
The function has as its graph a semicircle of radius $r=6$ with center at$(0,0)$.
The volume that results when the semicircle is rotated about the $x$-axis is given by :
$$
\begin{aligned} V &=\pi \int_{-6}^{6}(f(x))^{2} d x\\
&=\pi \int_{-6}^{6}(\sqrt{36-x^{2}})^{2} d x \\
&=288 \pi.
\end{aligned}
$$
Work Step by Step
$$
y=f(x)=\sqrt {36-x^{2}}
$$
The function has as its graph a semicircle of radius $r=6$ with center at$(0,0)$.
The volume that results when the semicircle is rotated about the $x$-axis is given by :
$$
\begin{aligned} V &=\pi \int_{-6}^{6}(f(x))^{2} d x\\
&=\pi \int_{-6}^{6}(\sqrt{36-x^{2}})^{2} d x \\
&=\pi \int_{-6}^{6}\left(36-x^{2}\right) d x \\
&=\left.\pi\left(36x-\frac{x^{3}}{3}\right)\right|_{-6} ^{6} \\ &=\pi\left(216-\frac{216}{3}\right)-\pi\left(-216+\frac{216}{3}\right) \\
&=288 \pi.
\end{aligned}
$$
