calculating the solid angle of disc

We determine the solid angle formed by a disc when one is looking at it on the normal line of its plane set to the center of it.

Let us look the disc from the origin and let the disc with radius $R$ situate such that its plane is parallel to the $xy$-plane and the center is on the $z$-axis at  $(0,\mathrm{\hspace{0.17em}0},h)$  with  $h>0$.  Into the

$\mathrm{\Omega }=-{\displaystyle {\int }_a}\overrightarrow{da}\cdot \nabla {\displaystyle \frac{1}{r}}={\displaystyle {\int }_a}\overrightarrow{da}\cdot {\displaystyle \frac{\overrightarrow{r}}{{|\overrightarrow{r}|}^3}}$ (1)

of the parent entry (http://planetmath.org/SolidAngle), we may substitute the position vector  $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+h\overrightarrow{k}$  of the directed surface element  $d\overrightarrow{a}=\overrightarrow{k}da,$ getting

$$\mathrm{\Omega }={\int }_a\frac{hda}{{(x^2+y^2+h^2)}^{3/2}}.$$

Now we can use a annulus (http://planetmath.org/Annulus2)-formed surface element  $da=2\pi \varrho d\varrho$  where  ${\varrho }^2=x^2+y^2$, whence the surface integral may be calculated as

$$\mathrm{\Omega }=\pi h{\int }_0^R\frac{2\varrho d\varrho }{{({\varrho }^2+h^2)}^{3/2}}=\frac{\pi h}{-2}\underset{\varrho =0}{\overset{R}{/}}\frac{1}{\sqrt{{\varrho }^2+h^2}}.$$

Thus we have the result

$$\mathrm{\Omega }=2\pi h\left(\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right).$$

Title	calculating the solid angle of disc
Canonical name	CalculatingTheSolidAngleOfDisc
Date of creation	2013-03-22 18:19:36
Last modified on	2013-03-22 18:19:36
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	6
Author	pahio (2872)
Entry type	Example
Classification	msc 15A72
Classification	msc 51M25
Related topic	SubstitutionNotation
Related topic	AngleOfViewOfALineSegment