Surface Area by Integration

Question

$$2\pi\int_{3}^6\left(\frac{1}{3}x^\frac{3}{2}-x^\frac{1}{2}\right)\left(1+\left(\frac{1}{2}x^\frac{1}{2}-\frac{1}{2}x^\frac{-1}{2}\right)^2\right)^\frac{1}{2}dx$$

I've managed to simplify this down to the equation below (not sure if it'll help), but I still can't integrate it.

Please help.

$$\frac{\pi}{3}\int_{3}^6(x^4-4x^3-2x^2+12x+9)^\frac{1}{2}dx$$

PS. The answer you should get is 9$\pi$.

Answer 1

HINT

I would say

$x^4-4x^3-2x^2+12x+9=(x^2-2x-3)^2$

Answer 2

You did a good bit of the hard work yourself, but these other computations may help: \begin{align} \frac{\pi}{3}\int_3^6\sqrt{x^4-4x^3-2x^2+12x+9}\;dx &= \frac{\pi}{3}\int_3^6 (x^2-2x-3)\;dx\tag{simplify}\\[1em] &= \frac{\pi}{3}\int_3^6 x^2\;dx - \frac{2\pi}{3}\int_3^6x\;dx -\pi\int_3^61\;dx\\[1em] &= \left.\frac{\pi x^3}{9}\right|_3^6-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\tag{FTC}\\[1em] &= 21\pi-\frac{2\pi}{3}\int_3^6x\;dx-\pi\int_3^61\;dx\\[1em] &= 21\pi+\left.\left(-\frac{\pi x^2}{3}\right)\right|_3^6-\pi\int_3^61\;dx\tag{FTC}\\[1em] &= 12\pi-\pi\int_3^61\;dx\\[1em] &= 12\pi+\left.(-\pi x)\right|_3^6\\[1em] &= 9\pi. \end{align} Hence, you can see the integral evaluates to $9\pi$, as desired.