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

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

Beginner2023-03-14Added 2 answers

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 $A=\mathrm{cos}\theta {\mid}_{b}^{a}2\pi {r}^{2}=(\frac{b}{r}-\frac{a}{r})2\pi {r}^{2}=\left(\frac{h}{r}\right)2\pi {r}^{2}=2\pi rh$

Every other derivation of this result that I found uses cylindrical coordinates and is far more involved than this one.

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 $A=\mathrm{cos}\theta {\mid}_{b}^{a}2\pi {r}^{2}=(\frac{b}{r}-\frac{a}{r})2\pi {r}^{2}=\left(\frac{h}{r}\right)2\pi {r}^{2}=2\pi rh$

Every other derivation of this result that I found uses cylindrical coordinates and is far more involved than this one.

smakkie8iz

Beginner2023-03-15Added 5 answers

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:

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

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

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

When we look at the inner integral, we get:

${\int}_{0}^{2\pi}{r}^{2}\mathrm{sin}\varphi d\theta ={r}^{2}\mathrm{sin}\varphi {\int}_{0}^{2\pi}d\theta$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={r}^{2}\mathrm{sin}\varphi {[\theta ]}_{0}^{2\pi}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left({r}^{2}\mathrm{sin}\varphi \right)(2\pi -0)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=2\pi {r}^{2}\mathrm{sin}\varphi$

Thus, our integral becomes:

$A={\int}_{0}^{\pi}2\pi {r}^{2}\mathrm{sin}\varphi d\varphi$

$=-2\pi {r}^{2}{\left\{\mathrm{cos}\varphi \right]}_{0}^{\pi}$

$=-2\pi {r}^{2}(\mathrm{cos}\pi -\mathrm{cos}0)$

$=-2\pi {r}^{2}(-1-1)$

$=-2\pi {r}^{2}(-2)$

$=4\pi {r}^{2}$ QED

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

$A={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}{\int}_{0}^{2\pi}{r}^{2}\mathrm{sin}\varphi d\theta d\varphi$

$={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}\left({r}^{2}\mathrm{sin}\varphi \right)(2\pi -0)d\varphi$

$={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}2\pi {r}^{2}\mathrm{sin}\varphi d\varphi$

$=-2\pi {r}^{2}{\left[\mathrm{cos}\varphi \right]}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}$

$=-2\pi {r}^{2}(\mathrm{cos}\left(\frac{\pi}{2}\right)-\mathrm{cos}\left(\mathrm{arccos}\left(\frac{h}{r}\right)\right))$

$=-2\pi {r}^{2}(0-\frac{h}{r})$

$=-2\pi {r}^{2}(0-\frac{h}{r})$

$=2\pi h{r}^{2}$ QED

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:

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

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

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

When we look at the inner integral, we get:

${\int}_{0}^{2\pi}{r}^{2}\mathrm{sin}\varphi d\theta ={r}^{2}\mathrm{sin}\varphi {\int}_{0}^{2\pi}d\theta$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}={r}^{2}\mathrm{sin}\varphi {[\theta ]}_{0}^{2\pi}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\left({r}^{2}\mathrm{sin}\varphi \right)(2\pi -0)$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=2\pi {r}^{2}\mathrm{sin}\varphi$

Thus, our integral becomes:

$A={\int}_{0}^{\pi}2\pi {r}^{2}\mathrm{sin}\varphi d\varphi$

$=-2\pi {r}^{2}{\left\{\mathrm{cos}\varphi \right]}_{0}^{\pi}$

$=-2\pi {r}^{2}(\mathrm{cos}\pi -\mathrm{cos}0)$

$=-2\pi {r}^{2}(-1-1)$

$=-2\pi {r}^{2}(-2)$

$=4\pi {r}^{2}$ QED

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

$A={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}{\int}_{0}^{2\pi}{r}^{2}\mathrm{sin}\varphi d\theta d\varphi$

$={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}\left({r}^{2}\mathrm{sin}\varphi \right)(2\pi -0)d\varphi$

$={\int}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}2\pi {r}^{2}\mathrm{sin}\varphi d\varphi$

$=-2\pi {r}^{2}{\left[\mathrm{cos}\varphi \right]}_{\mathrm{arccos}\left(\frac{h}{r}\right)}^{\frac{\pi}{2}}$

$=-2\pi {r}^{2}(\mathrm{cos}\left(\frac{\pi}{2}\right)-\mathrm{cos}\left(\mathrm{arccos}\left(\frac{h}{r}\right)\right))$

$=-2\pi {r}^{2}(0-\frac{h}{r})$

$=-2\pi {r}^{2}(0-\frac{h}{r})$

$=2\pi h{r}^{2}$ QED

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