Question

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

Answer 1

Volume `= 36pi \ "unit"^3`

Explanation:

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

The Volume of Revolution about `Ox` is given by:

`V= int_(x=a)^(x=b) \ pi y^2 \ dx`

So for for this problem, Noting that `9-x^2=0 => x=+-3`, and that by symmetry we can double the volume for the region `x in [0,3]`

`V= 2int_0^3 \ pi (sqrt(9-x^2))^2 \ dx`
`\ \ \= 2pi \ int_0^3 \ (9-x^2) \ dx`
`\ \ \= 2pi \ [9x-x^3/3]_0^3`
`\ \ \= 2pi \ {(27-27/3) - (0-0)}`
`\ \ \= 36pi`