Would like to prove the following:
$\lim\limits_{n\to \infty; n\in \mathbb R^+}\text{surface area} \{(x,y): y= x^n(1-x)^n,x\in[0,1]\}=\pi.$
The surface of revolution of the function is being revolved about the line of revolution, $x=1/2.$
What I know is that as $n\to\infty$ this results in an arc length formula:
$\lim\limits_{n\to \infty; n\in \mathbb R^+}\text{length} \{(x,y): y= x^n(1-x)^n,x\in[0,1]\}=1,$ by using the inequality $x(1-x) \leq \frac 1 4,$ and seeing that the arc length $s$, is $s \leq \sqrt {1+\frac {n^{2}} {4^{n-1}}}.$
So that means one is essentially calculating the area of a disc with radius $1$. And the area of such a disc is equal to $\pi.$
I think I have the right parts for the proof. Am I on the right track?