Question

How do you find the volume of a solid that is enclosed by `y=sqrt(4+x)`, x=0 and y=0 revolved about the x axis?

Answer 1

The volume `=8pi`

Explanation:

The volume of a disc of thickness dx is `dV=piy^2dx`
as `y=sqrt(4+x)` `=>` `dV=pi(4+x)dx`
The limits of integration are `x=0` and `x=-4`
As `4+x>=0`
so the volume `V=int_-4^0pi(4+x)dx=pi(4x+x^2/2)_-4^0`
`=-pi(-16+8)=8pi`
graph{sqrt(x+4) [-10, 10, -5, 5]}