Question

How do you find the volume of the solid obtained by rotating the region bounded by y=5x^2 ,x=1 , and y=0, about the x-axis?

Answer 1

Volume `= 5pi " unit"^3`

Explanation:

graph{(y-5x^2)(x-1)(y)=0 [-3, 3, -2, 6]}

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

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

So for for this problem:

`V= int_0^1 \ pi (5x^2)^2 \ dx`
`\ \ \= pi \ int_0^1 \ 25x^4 \ dx`
`\ \ \= pi \ [(25x^5)/5]_0^1`
`\ \ \= 5pi \ [x^5]_0^1`
`\ \ \= 5pi (1-0)`
`\ \ \= 5pi`