# How do you find the volume of the solid y=sqrt(9-x^2) revolved about the x-axis?

Feb 12, 2017

Volume $= 36 \pi \setminus {\text{unit}}^{3}$

#### Explanation:

graph{(y-sqrt(9-x^2))=0 [-6, 6, -2, 4]}

The Volume of Revolution about $O x$ is given by:

$V = {\int}_{x = a}^{x = b} \setminus \pi {y}^{2} \setminus \mathrm{dx}$

So for for this problem, Noting that $9 - {x}^{2} = 0 \implies x = \pm 3$, and that by symmetry we can double the volume for the region $x \in \left[0 , 3\right]$

$V = 2 {\int}_{0}^{3} \setminus \pi {\left(\sqrt{9 - {x}^{2}}\right)}^{2} \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \pi \setminus {\int}_{0}^{3} \setminus \left(9 - {x}^{2}\right) \setminus \mathrm{dx}$
$\setminus \setminus \setminus = 2 \pi \setminus {\left[9 x - {x}^{3} / 3\right]}_{0}^{3}$
$\setminus \setminus \setminus = 2 \pi \setminus \left\{\left(27 - \frac{27}{3}\right) - \left(0 - 0\right)\right\}$
$\setminus \setminus \setminus = 36 \pi$