Lara Browning

2023-03-13

a) Show that the formula for the surface area of a sphere with radius r is $4\pi {r}^{2}$.
b) If a portion of the sphere is removed to form a spherical cap of height h then then show the curved surface area is $2\pi h{r}^{2}$?

Emencisydeessyo2

A sphere's area element has a constant radius r and two angles. One is longitude $\varphi$, which varies from 0 to $2\pi$. The other one is the angle with the vertical. To avoid counting twice, that angle only varies between 0 and $\pi$.
Thus, the area element is $dA=rd\theta r\mathrm{sin}\theta d\varphi ={r}^{2}\mathrm{sin}\theta d\theta d\varphi$
Integrated over the whole sphere gives
$A={\int }_{0}^{\pi }\mathrm{sin}\theta d\theta {\int }_{0}^{2\pi }d\varphi {r}^{2}=-\mathrm{cos}\left(\theta \right){\mid }_{0}^{\pi }2\pi {r}^{2}=4\pi {r}^{2}$
In part b, $\mathrm{cos}\left(\theta \right)$ varies between $\frac{a}{r}$ and $\frac{b}{r}$ which is such that $b-a=h$
Thus $A=\mathrm{cos}\theta {\mid }_{b}^{a}2\pi {r}^{2}=\left(\frac{b}{r}-\frac{a}{r}\right)2\pi {r}^{2}=\left(\frac{h}{r}\right)2\pi {r}^{2}=2\pi rh$
Every other derivation of this result that I found uses cylindrical coordinates and is far more involved than this one.

smakkie8iz

Spherical coordinates are more convenient to use than cylindrical or rectangular coordinates. This solution appears lengthy because I have broken down each step, but it can be computed in a few lines of code.
With spherical coordinates, we can define a sphere of radius r by all coordinate points where $0\le \varphi \le \pi$
(Where $\varphi$ is the angle measured down from the positive z-axis), and $0\le \theta \le 2\pi$ (just the same as it would be polar coordinates), and $\rho =r$).
The Jacobian for Spherical Coordinates is given by $J={\rho }^{2}\mathrm{sin}\varphi$
And so we can calculate the surface area of a sphere of radius r using a double integral:

where $R=\left\{\left(x,y,z\right)\in {ℝ}^{3}\mid {x}^{2}+{y}^{2}+{z}^{2}={r}^{2}\right\}$

When we look at the inner integral, we get:

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left({r}^{2}\mathrm{sin}\varphi \right)\left(2\pi -0\right)$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=2\pi {r}^{2}\mathrm{sin}\varphi$
Thus, our integral becomes:

QED
By trigonometry $\mathrm{cos}\varphi =\frac{h}{r}⇒\varphi =\mathrm{arccos}\left(\frac{h}{r}\right)$, and so we must restrict $\varphi$ to $\mathrm{arccos}\left(\frac{h}{r}\right)\le \varphi \le \frac{\pi }{2}$, which gives us:

QED

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