Find the area under the curve I apologize, I have asked this question previously and I expect I will be down voted but thats okay because I really need help on this problem not just the answer.
Find the area of the figure rotated about the x-axis $0 \leq t \leq 1$
$x=9t-3t^3$ and $y=9t^2$
Setting up the integral:
$\int_0^1 2 \pi 9t^2 \sqrt{(9-9t^2)^2+(18t^2)}dt$ 
$18 \pi \int_0^1 t^2 \sqrt{(9(1-t^2)+9(2t)^2)}dt$
$18 \pi \int_0^1 t^2 \sqrt{9((1-t^2)+(2t)^2)}dt$
$18(3) \pi \int_0^1 t^2 \sqrt{((1-t^2)+(2t)^2)}dt$ $\to$ $54 \pi \int_0^1 t^2 \sqrt{((1-t^2)+(2t)^2)}dt$
Algebra inside the integral square root:
$(1-t^2)^2= t^4-2t^2+1$ and $(2t)^2=4t^2$ Combining we get:
$t^4+2t^2+1 \to (t^2+1)^2$ So:
$54 \pi \int_0^1 t^2(t^2+1)^2 dt \to$ $54\pi \int_0^1 t^6+2t^4+t^2 dt$
$54 \pi ( \frac{1}{7}t^7+\frac{2}{5}t^5+\frac{1}{3}t^3 \vert_0^1)$
$\frac{92}{105} \times \frac{54\pi}{1} = \frac{4968\pi}{105}$ 
This of course was not the accepted final answer

I would say:
$\displaystyle S=18\pi\int_0^1 t^2 \sqrt{(9-9t^2)^2+(18t)^2}\,dt=18\pi\int_0^1 t^2 \sqrt{81(1-t^2)^2+81\cdot 4t^2}\,dt=$
$\displaystyle=162\pi\int_0^1 t^2 \sqrt{(1-t^2)^2+4t^2}\,dt=162\pi\int_0^1 t^2 \sqrt{1-2t^2+t^4+4t^2}\,dt=$
$\displaystyle =162\pi\int_0^1 t^2 \sqrt{(1+t^2)^2}\,dt=162\pi\int_0^1 t^2 
(1+t^2)\,dt=162\pi\int_0^1 (t^2+t^4)\,dt=$
$\displaystyle =162\pi \left[\frac{t^3}{3}+\frac{t^5}{5}\right]_0^1=\cdots$
