# 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?

Aug 20, 2015

$\frac{317617}{810} \pi$

#### 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$\left(- \frac{1}{3} , \frac{1}{3}\right)$ and (2,12)

If ${y}_{1} = 3 {x}^{2}$ 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 $\pi {\left({y}_{2} - {y}_{1}\right)}^{2}$ dx.

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

${\int}_{- \frac{1}{3}}^{2} \pi {\left({y}_{2} - {y}_{1}\right)}^{2} \mathrm{dx}$

${\int}_{- \frac{1}{3}}^{2} \pi {\left(5 x + 2 - 3 {x}^{2}\right)}^{2} \mathrm{dx}$

${\int}_{- \frac{1}{3}}^{2} \pi \left(9 {x}^{4} - 30 {x}^{3} + 13 {x}^{2} + 20 x + 4\right) \mathrm{dx}$

On solving, this integral would work out to be=$\frac{317617}{810} \pi$