Question

How do you find the volume of a solid that is enclosed by `y=3x^2` and y=2x+1 revolved about the x axis?

Answer 1

`V=pi(4/3+2+1-9/5-4/3(-1/27)-2(1/9)-(-1/3)+1/5(-1/27))=pi1088/405`

Explanation:

Let's find the intersections between the line and the parabola so that we have to solve `3x^2-2x-1=0` this gives two solutions `x_1=-1/3` and `x_2=1`.

From the first Guldino theorem, the sought volume is given by the integral `2piint_(-1/3)^1dxint_(3x^2)^(2x+1)ydy`

that can be rewritten as `2piint_(-1/3)^1(2x+1)^2/2-(3x^2)^2/2dx`

Rewrite after developing the binomial square
so that it gets `piint_(-1/3)^1(4x^2+4x+1-9x^4)dx`
This solved gets
`pi(4/3x^3+2x^2+x-9/5x^5)|_(-1/3)^1`