Question

How do you find the volume of the solid obtained by rotating the region bounded by the curves `f(x) = 3x^2` and `f(x) = 5x+2` about the x axis?

Answer 1

`317617/810 pi`

Explanation:

The region bounded by the two functions, a vertical parabola and a straight line is shown in the picture. On solving the two equations the points of intersection can be easily found to be`(-1/3, 1/3)` and (2,12)

If `y_1= 3x^2` and `y_2`= 5x+2, consider an element of length `y_2 -y_1` and width dx, of the region bounded by the two functions. If this element is rotated about x axis, the volume of the elementary disc so formed would be `pi(y_2-y_1)^2` dx.

The volume of the solid formed by rotation of the whole region, about x axis would be

`int_(-1/3)^ 2 pi(y_2-y_1)^2 dx`

`int_(-1/3) ^2 pi(5x+2-3x^2)^2 dx`

`int_(-1/3)^ 2 pi(9x^4 -30x^3 + 13x^2 +20x +4)dx`

On solving, this integral would work out to be=`317617/810 pi`