<s>[INST] Rotation area of a function Im faced with the following problem. 
Calculate the area of the "cone" that is created when the function 
$$y = \frac{x^2}{32}$$
$$0≤y≤24$$ rotates around the y-axis. 
If we draw the cone that is created we can see that the hight will be 24. However we have to change our function. 
$$y = \frac{x^2}{32} \implies \sqrt{32y} = x = f(y)$$
The formula for the rotation area is the following:
$$Area = 2\pi\cdot\int f(x)\cdot \sqrt{1+f'(x)}dx $$
I've tried to go continue from this however, i cant manage the integral part. Can any1 help me?
/John  
 [/INST] Your formula is missing a square in the derivative.
The rotational area is given by
$$ 2\pi \int x ds = 2\pi \int x \sqrt{dx^2 + dy^2} $$
where $ds$ is the arc length element of the curve $y = \dfrac{x^2}{32}$ 
If you choose to integrate w.r.t $y$, the integral looks like this
\begin{align} 
2\pi \int x \sqrt{dx^2 + dy^2} &= 2\pi \int_0^{24} x(y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}dy \\
&= 2\pi \int_0^{24}4\sqrt{2y} \sqrt{1 + \left(\frac{2\sqrt{2}}{\sqrt{y}}\right)^2} dy \\
&= 8\pi\sqrt{2} \int_0^{24} \sqrt{y} \sqrt{1+\frac{8}{y}} dy \\
&= 8\pi\sqrt{2} \int_0^{24} \sqrt{y+8} \ dy \\
&= 8\pi\sqrt{2}\cdot \frac{2}{3} (y+8)^{3/2}\Bigg|_0^{24}
\end{align}
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