Chapter 16.P, Problem 1P

Let S be a smooth parametric surface and P be a point such that each line that starts at P intersects S at most once. The Solid angle $\Omega \left(S\right)$ subtended by S at P is the set of lines starting at P and passing through S. Let S(a) be the intersection of $\Omega \left(S\right)$ with the surface of the sphere with center P and radius a. Then the measure of the solid angle (in steradians) is defined to be

$|\Omega \left(S\right)|=\frac{\text{area of }S\left(a\right)}{a^2}$

Apply the Divergence Theorem to the part of $\Omega \left(S\right)$ between S(a) and S to show that

$|\Omega \left(S\right)|={\displaystyle \underset{s}{\iint }\frac{\text{r}\cdot \text{n}}{r^3}ds}$

Where r is the radius vector from P to any point $S,r=\left|\text{r}\right|$, and the unit normal vector n is directed away from P.

This shows that the definition of the measure of a solid angle is independent of the radius a of the sphere. (Note the analogy with the definition of radian measure.) The total solid angle subtended by a sphere at its center is thus $4\pi$ steradians.

To determine

To show:

$\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\left|\mathrm{\Omega }\left(S\right)\right|$

Answer to Problem 1P

Solution:

$\left|\mathrm{\Omega }\left(S\right)\right|=\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS$

Explanation of Solution

1) Concept:

i) The Divergence Theorem:

Let $E$ be a simple solid region and let $S$ be the boundary surface of $E$, given with the positive (outward) orientation. Let $F$ be a vector field whose component functions have continuous partial derivatives on an open region that contains $E$. Then

${\iint }_{S}F·\mathrm{n}dS={\iiint }_E\mathrm{d}\mathrm{i}\mathrm{v} FdV$

ii) $\mathrm{d}\mathrm{i}\mathrm{v} F=\nabla ·F$

2) Given:

The measure of solid angle is

$\left|\mathrm{\Omega }\left(S\right)\right|=\frac{1}{a^2}\left(\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f} S\left(a\right)\right)$

3) Calculation:

We are given that, $S$ is the smooth parametric surface, and $P$ is any point such that each line starting from $P$ intersects $S$ at most once. The solid angle $\mathrm{\Omega }\left(S\right)$ subtended by $S$ at $P$ is the set of lines starting at $P$ and passing through $S$. And

$S\left(a\right)$ be the intersection of $\mathrm{\Omega }\left(S\right)$ with the surface of the sphere with centre $P$ and radius $a.$

The measure of solid angle is

$\left|\mathrm{\Omega }\left(S\right)\right|=\frac{1}{a^2}\left(\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f} S\left(a\right)\right)\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(*\right)$

Let $S_1$ be the portion of $\mathrm{\Omega }\left(S\right)$ between $S\left(a\right)and S$.

Let $\partial S_1$ be the boundary of $S_1.$

Let $S_L$ be the surface of $S_1$ except $S and S\left(a\right)$ that is the lateral surface of $S_1$

Let $F=\frac{\mathrm{r}}{r^3}$

By using divergence theorem,

$\underset{\partial S_1}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\underset{S_1}{\overset{}{\iiint }}\nabla · \frac{\mathrm{r}}{r^3} dV$

$=\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S_L}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS$

Note that on $S_L, \mathrm{r}·\mathrm{n}=0$ hence,

$\underset{S_L}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=0$

Therefore,

$\underset{\partial S_1}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS\dots \dots \dots \dots \dots \dots ..\left(1\right)$

Consider,

$\nabla ·\frac{\mathrm{r}}{r^3}=〈\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z}〉·〈\frac{x}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}},\frac{y}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}},\frac{z}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}〉$

By definition of dot product,

$=\frac{\partial }{\partial x}\left(\frac{x}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\right)+\frac{\partial }{\partial y}\left(\frac{y}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\right)+\frac{\partial }{\partial z}\left(\frac{z}{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}}\right)$

By using quotient rule of differentiation,

$=\frac{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}\left(1\right)-\frac{3}{2}x{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(2x\right)}{{\left(x^2+y^2+z^2\right)}^3}+\frac{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}\left(1\right)-\frac{3}{2}y{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(2y\right)}{{\left(x^2+y^2+z^2\right)}^3}+\frac{{\left(x^2+y^2+z^2\right)}^{\frac{3}{2}}\left(1\right)-\frac{3}{2}z{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(2z\right)}{{\left(x^2+y^2+z^2\right)}^3}$

Simplifying,

$=\frac{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(x^2+y^2+z^2\right)-3x^2}{{\left(x^2+y^2+z^2\right)}^3}+\frac{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(x^2+y^2+z^2\right)-3y^2}{{\left(x^2+y^2+z^2\right)}^3}+\frac{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}\left(x^2+y^2+z^2\right)-3z^2}{{\left(x^2+y^2+z^2\right)}^3}$

Factor out common terms and then cancel out common factor from the numerator and denominator,

$=\frac{\left(x^2+y^2+z^2-3x^2\right)+\left(x^2+y^2+z^2-3y^2\right)+\left(x^2+y^2+z^2-3z^2\right)}{{\left(x^2+y^2+z^2\right)}^{\frac{5}{2}}}$

$=\frac{3x^2+{3y}^2+3z^2-3x^2-3y^2-3z^2}{{\left(x^2+y^2+z^2\right)}^{\frac{5}{2}}}$

$=0$

Therefore,

$\underset{ \partial S_1}{\overset{}{\iint }}\frac{r·n}{r^3}dS=\underset{S_1}{\overset{}{\iiint }}\nabla · \frac{r}{r^3} dV$ gives,

$=\underset{S_1}{\overset{}{\iiint }}\left(0\right) dV=0$

Therefore,

$\underset{\partial S_1}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=0$

From $\left(1\right)$

$\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS+\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=0$

$2\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=-2\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS$

$\underset{ S}{\overset{}{\iint }}\frac{r·n}{r^3}dS=-\underset{ S\left(a\right)}{\overset{}{\iint }}\frac{r·n}{r^3}dS\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right)$

On $S\left(a\right),r=a$

$n=-\frac{r}{r^{}}=-\frac{r}{a} and r·r=r^2=a^2$

Therefore,

$\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\left(-\frac{\mathrm{r}}{a}\right)}{r^3}dS$

$=\underset{S\left(a\right)}{\overset{}{\iint }}\frac{-a^2}{a^4}dS$

$-\underset{S\left(a\right)}{\overset{}{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\frac{1}{a^2}\underset{S\left(a\right)}{\overset{}{\iint }}dS$

$=\frac{1}{a^2}\left(\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f} S\left(a\right)\right)$

From $\left(*\right)$

$=\left|\mathrm{\Omega }\left(S\right)\right|$

Therefore, from $\left(2\right)$

$\underset{S}{\overset{\square }{\iint }}\frac{\mathrm{r}·\mathrm{n}}{r^3}dS=\left|\mathrm{\Omega }\left(S\right)\right|$

Therefore,

$\underset{ S}{\overset{}{\iint }}\frac{r·n}{r^3}dS=\left|\Omega \left(S\right)\right|$

Conclusion:

$\left|\Omega \left(S\right)\right|=\underset{ S}{\overset{}{\iint }}\frac{r·n}{r^3}dS$