# How do you find the volume bounded by x^2y^2+16y^2=6 and the x & y axes, the line x=4 revolved about the x-axis?

Nov 16, 2016

$\frac{3}{8} {\pi}^{2}$

= 3.7011 cubic units, nearly.

#### Explanation:

The graph of ${x}^{2} {y}^{2} + 16 {y}^{2} = 6$ is inserted.

In ${Q}_{1}$, The curve meets x = 0 ( y-axis ) at $\left(0 , \frac{1}{2} \sqrt{\frac{3}{2}}\right)$ and

the parallel x = 4 at $\left(4 , \frac{\sqrt{3}}{4}\right) .$

The volume is

$V = \pi \int {y}^{2} \mathrm{dx}$, from x = 0 to x = 4

$= \pi \int \frac{6}{{x}^{2} + {4}^{2}} \mathrm{dx}$, between the limits

$= 6 \pi \left(\frac{1}{4}\right) \left[{\tan}^{- 1} \left(\frac{x}{4}\right)\right]$, between the limits

$= \frac{3}{2} \pi \left[{\tan}^{- 1} 1 - {\tan}^{- 1} 0\right]$

$\frac{3}{2} \pi \left(\frac{\pi}{4}\right)$

$= \frac{3}{8} {\pi}^{2}$

= 3.7011 cubic units, nearly.-

graph{x^2y^2+16y^2-6=0 [-10, 10, -5, 5]}