# a) Show that the formula for the surface area of a sphere with radius r is 4pir^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 2pihr^2?

## 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 $2 \pi h {r}^{2}$

Apr 4, 2017

A = int dA

#### Explanation:

An area element on a sphere has constant radius r, and two angles. One is longitude $\phi$, 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 $\pi$.

So the area element is $\mathrm{dA} = r d \theta r \sin \theta d \phi = {r}^{2} \sin \theta d \theta d \phi$
Integrated over the whole sphere gives
$A = {\int}_{0}^{\pi} \sin \theta d \theta {\int}_{0}^{2 \pi} \mathrm{dp} h i {r}^{2} = - \cos \left(\theta\right) {|}_{0}^{\pi} 2 \pi {r}^{2} = 4 \pi {r}^{2}$

In part b, $\cos \left(\theta\right)$ varies between $\frac{a}{r}$ and $\frac{b}{r}$ which is such that $b - a = h$
Then $A = \cos \theta {|}_{b}^{a} 2 \pi {r}^{2} = \left(\frac{b}{r} - \frac{a}{r}\right) 2 \pi {r}^{2} = \left(\frac{h}{r}\right) 2 \pi {r}^{2} = 2 \pi r h$

Note: Every derivation I found of this result uses cylindrical coordinates and is far more involved than this one. Can someone else check?

Apr 4, 2017

a) $A = 4 \pi {r}^{2}$
b) $A = 2 \pi h {r}^{2}$

#### Explanation:

It is easier to use Spherical Coordinates, rather than Cylindrical or rectangular coordinates. This solution looks long because I have broken down every step, but it can be computer in just a few lines of calculation

With spherical coordinates, we can define a sphere of radius $r$ by all coordinate points where $0 \le \phi \le \pi$ (Where $\phi$ 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 $J = {\rho}^{2} \sin \phi$

And so we can calculate the surface area of a sphere of radius $r$ using a double integral:

$A = \int {\int}_{R} \setminus \setminus \mathrm{dS} \setminus \setminus \setminus$

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

$\therefore A = {\int}_{0}^{\pi} \setminus {\int}_{0}^{2 \pi} \setminus {r}^{2} \sin \phi \setminus d \theta \setminus d \phi$

If we look at the inner integral we have:

${\int}_{0}^{2 \pi} \setminus {r}^{2} \sin \phi \setminus d \theta = {r}^{2} \sin \phi \setminus {\int}_{0}^{2 \pi} \setminus d \theta$
$\text{ } = {r}^{2} \sin \phi {\left[\setminus \theta \setminus\right]}_{0}^{2 \pi}$
$\text{ } = \left({r}^{2} \sin \phi\right) \left(2 \pi - 0\right)$
$\text{ } = 2 \pi {r}^{2} \sin \phi$

So our integral becomes:

$A = {\int}_{0}^{\pi} \setminus 2 \pi {r}^{2} \sin \phi \setminus d \phi$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} {\left\{\cos \phi\right]}_{0}^{\pi}$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(\cos \pi - \cos 0\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(- 1 - 1\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(- 2\right)$
$\setminus \setminus \setminus = 4 \pi {r}^{2} \setminus \setminus \setminus$ QED

For the area of the portion of a sphere we have a similar set-up: By trigonometry $\cos \phi = \frac{h}{r} \implies \phi = \arccos \left(\frac{h}{r}\right)$, and so we must restrict $\phi$ to $\arccos \left(\frac{h}{r}\right) \le \phi \le \frac{\pi}{2}$, which gives us:

$A = {\int}_{\arccos} {\left(\frac{h}{r}\right)}^{\frac{\pi}{2}} \setminus {\int}_{0}^{2 \pi} \setminus {r}^{2} \sin \phi \setminus d \theta \setminus d \phi$
$\setminus \setminus \setminus = {\int}_{\arccos} {\left(\frac{h}{r}\right)}^{\frac{\pi}{2}} \setminus \left({r}^{2} \sin \phi\right) \left(2 \pi - 0\right) \setminus d \phi$
$\setminus \setminus \setminus = {\int}_{\arccos} {\left(\frac{h}{r}\right)}^{\frac{\pi}{2}} \setminus 2 \pi {r}^{2} \sin \phi \setminus d \phi$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} {\left[\cos \phi\right]}_{\arccos} {\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(\cos \left(\frac{\pi}{2}\right) - \cos \left(\arccos \left(\frac{h}{r}\right)\right)\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(0 - \frac{h}{r}\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(- \frac{h}{r}\right)$
$\setminus \setminus \setminus = 2 \pi h {r}^{2} \setminus \setminus \setminus$ QED