Question
My attempt
The region between the dotted lines shows the hollow region.
The distance MP = $ \sqrt{R^2 - x^2} $
Hollow area created = 2πx×2MP
= $4πx\sqrt{R^2- x^2}$
The area removed due to hole = 2×(area of the cap )
What I did is I calculated solid angle $\Omega$ = ${ πx^2\over {R^2 -x^2}}$
And I took the curved surface area of each cap to be ${ \Omega × 4πR^2\over 4π}$ = $\Omega R^2$ = ${ πx^2R^2\over {R^2 -x^2}}$
So, New Total Surface Area
(TSA) = 4π$R^2$ - 2×area of each cap + hollow area
= 4π$R^2$ - 2×${ πx^2R^2\over {R^2 -x^2}}$ + 4πx$\sqrt{R^2- x^2}$
For this to be maximum, I differentiated this and equated to 0 and after removing denomiator I got
$R^2x^2$ + 2${(R^2 - x^2)}^{5/2}$ = 2$R^2x(R^2 - x^2)$ + 2$x^2(R^2 - x^2)^{3/2}$
Which I am unable to solve further.
Any help , hint or solution would be appreciated. Also I want to know whether my expression of TSA is right.
Please use the given hint. The surface area of a segment of height $h$ on a sphere of radius $R$ is $2 \pi R h$. Here $h = R - \sqrt{R^2 - x^2}$. There are two such segments removed by drilling the hole.
So surface area removed from the sphere: $4 \pi R^2 - 4 \pi R \sqrt{R^2 - x^2}$
Surface area of the sphere remaining: $4 \pi R \sqrt{R^2 -x^2}$
Surface area of the cylindrical surface through the sphere: $2 \pi x \cdot 2 \sqrt{R^2 - x^2}$
Total surface $S = 4 \pi R \sqrt{R^2 -x^2} + 4 \pi x \sqrt{R^2 - x^2}$
$S = 4 \pi (R + x) \sqrt{R^2 - x^2}$
$ \displaystyle \frac{dS}{dx} = \frac{4 \pi (R^2 - R x - 2x^2)}{\sqrt{R^2- x^2}}$
Solving $R^2 - R x - 2x^2 = 0$, we find that $S$ is maximized when $x = R/2$.