Question

How do you find the volume of the solid `y=x^2, y=x^3` revolved about the x-axis?

Answer 1

Volume `(2pi)/35 " unit^3`

Explanation:

graph{(y-x^2)(y-x^3)=0 [-0.1, 1.5, -0.1, 1.5]}

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

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

So, the volume of revolution bounded by two curves `f(x)` and `g(x)` with `f(x) gt g(x)` is given by the difference between the individual solids of revolution, thus:

`V = int_(x=a)^(x=b) \ pi f^2(x) \ dx - int_(x=a)^(x=b) \ pi g^2(x) \ dx`
`\ \ = int_(x=a)^(x=b) \ pi {f^2(x) - g^2(x)} \ dx`

We can see by observation that the points of intersection of the curves `y=x^2` and `y=x^3` are `(0,0)` and `(1,1)`.

So for for this problem:

`V= int_0^1 \ pi { (x^2)^2 -(x^3)^2} \ dx`
`\ \ \= pi \ int_0^1 \ x^4-x^6 \ dx`
`\ \ \= pi \ [x^5/5-x^7/7]_0^1`
`\ \ \= pi \ { (1/5-1/7) - (0)}`
`\ \ \= pi * 2/35`

`\ \ \= (2pi)/35`