Calculate the area of the rotational surface that arise when the curve $\displaystyle y=^\frac{x^2}{8}, \, 0\leqslant y\leqslant 6$ rotates around the y-axis.

My approach is something along the lines of:

Use an integral to calculate the area.

Since the curve is rotating around the y-axis, we need to (not sure about the english term), "extract" x from the formula (we need to get the height between the curve and the y-axis).

$\displaystyle y=\frac{x^2}{8}\Leftrightarrow \frac{y}{8}=x^2 \Rightarrow x=\pm\sqrt{8y}$

Since the curve is symmetrical around the y-axis we don't have to care the negative x; we simply multiply by 2 to get the total area.

So, we now have the following

$\displaystyle \int_0^6 \sqrt{8y} \, dy$

which simplified should be

$\displaystyle 2\sqrt{2}\int_0^6 \sqrt{y} \, dy = 2\sqrt{2}[\frac{2\sqrt[3]{6}}{3}]$

Something isn't right though, and I can't figure out what.