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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!

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    The sine and cosine functions are denoted `\sin` and `\cos` in $\TeX$, respectively.2012-05-01
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    Do you mean a segment as in http://mathworld.wolfram.com/SphericalSegment.html? Also I don't understand your jump to a 'planar' problem. There is spherical symmetry, but that just eliminates a variable, you can't just flatten the object into a disc annulus and triangle.2012-05-01

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