$0 \leq t \leq 1$ and $y=9t^2$ and $x=9t-3t^3$
setting up the integral:
$\int_0^1 2\pi 9t^2 \sqrt{[9(1-t^2)^2]+[9(2t)^2]}$
Factor out the 9 to give: $\int_0^1 54 \pi t^2 \sqrt{(1-t^2)^2 + 4t^2}$ This should get simplified to $162 \pi \int_0^1 ((1-t^2)+2t)t^2$
should give:
$162 \pi \int_0^1 t^2-t^4+2t^3 \to \frac{1}{3}t^3-\frac{1}{5}t^5+\frac{1}{2}t^4 \vert_0^1 \to (\frac{1}{3}-\frac{1}{5}+\frac{1}{2}) \times 162 \pi$
but this this gives a negative answer...if someone could explain in detail what happens inside the square root that would be great
I actually get for the integral
$$2 \pi \int_0^1 dt \sqrt{(9-9 t^2)^2 + (18 t)^2} (9 t^2) $$
which can be simplified down to
$$162 \pi \int_0^1 dt \sqrt{(1-t^2)^2+4 t^2} (t^2) $$
which can be very simplified down to$:
$$162 \pi \int_0^1 dt \, (1+t^2) t^2 $$
which at this point I assume you can do.