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?

1 Answer
Nov 15, 2016

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