# How do you find the volume of the region bounded above by the line y = 16, below by the curve y = 16-x^2, and on the right by the line x = 4 about the line y = 16?

Sep 21, 2015

$\frac{1024 \pi}{5}$

#### Explanation:

Intersection of the curve $y = 16 - {x}^{2}$ and x-axis is:
$0 = 16 - {x}^{2} \iff {x}^{2} = 16 \iff x = 4 \vee x = - 4$
It's the same bounding line as the one given in the task.

$V = \pi {\int}_{0}^{4} \left({r}_{1}^{2} - {r}_{2}^{2}\right) \mathrm{dx}$

where ${r}_{1}$ and ${r}_{2}$ are outer and inner radius of the solid, respectively.

${r}_{1} = 16 - \left(16 - {x}^{2}\right) = 16 - 16 + {x}^{2} = {x}^{2}$
${r}_{2} = 16 - 16 = 0$

$V = \pi {\int}_{0}^{4} \left({\left({x}^{2}\right)}^{2} - {0}^{2}\right) \mathrm{dx} = \pi {\int}_{0}^{4} {x}^{4} \mathrm{dx} = \pi {x}^{5} / 5 {|}_{0}^{4}$

$V = \frac{1024 \pi}{5}$