Question

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?

Answer 1

`3/8pi^2`

= 3.7011 cubic units, nearly.

Explanation:

The graph of `x^2y^2+16y^2=6` is inserted.

In `Q_1`, The curve meets x = 0 ( y-axis ) at `(0, 1/2sqrt(3/2))` and

the parallel x = 4 at `( 4, sqrt3/4 ).`

The volume is

`V = pi int y^2 dx`, from x = 0 to x = 4

`=pi int 6/(x^2+4^2) dx`, between the limits

`=6pi(1/4)[tan^(-1)(x/4)]`, between the limits

`=3/2pi[tan^(-1)1-tan^(-1)0]`

`3/2pi(pi/4)`

`=3/8pi^2`

= 3.7011 cubic units, nearly.-

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