This meant to be a relatively easy problem but I cannot get my head around it. It is from Burkill's "First course in Analysis", book $4$(f), $10$.
An open bowl is in the form of a segment of a sphere of metal of negligible thickness. Find the shape of the bowl if its volume is the greatest for a given area of metal. (Solution: Hemisphere)
Could anyone help me with the solution of the problem?
Here is one of my attempts. I assumed that the problem is circumferentially symmetric so I considered the planar problem instead. I took the area of the segment of a disk with radius R, central angle $\theta$, and area A which I calculated as follows:
$A = \text{sector area} - \text{area of triangle} = \frac{R^2\pi}{2\pi} \theta - 2 \frac{1}{2} R \cos\left(\frac{\theta}{2}\right) R\sin\left(\frac{\theta}{2}\right) = \frac{R^2\theta}{2}-\frac{R^2}{2}\sin\theta.$
This is constrained by the area that is the length of material we have say $L$:
$L = R\theta.$
Substituting in for $R$:
$A = \frac{L^2}{2\theta} - \frac{L^2}{2\theta^2} \sin\theta.$
Differentiate to find turning value:
$\frac{dA}{d\theta} = \frac{L^2}{2}\left(-\frac{1}{\theta^2} + \frac{2}{\theta^3} \sin\theta + \frac{1}{\theta^2} \cos\theta\right) = 0.$
I am bit stuck now how to get $\theta$ out of this and I am questioning whether my method is really correct. Could anyone help me out? Thank you!