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Version 1 |
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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.\\
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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. |
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| 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\, $(0,\,0,\,h)$\, with\, $h > 0$.\, Into the \PMlinkescapetext{formula} |
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\, $(0,\,0,\,h)$\, with\, $h > 0$.\, Into the \PMlinkescapetext{formula} |
| \begin{align} |
\begin{align} |
| \Omega = -\int_a \vec{da}\cdot\nabla\frac{1}{r} = \int_a \vec{da}\cdot\frac{\vec{r}}{|\vec{r}|^3} |
\Omega = -\int_a \vec{da}\cdot\nabla\frac{1}{r} = \int_a \vec{da}\cdot\frac{\vec{r}}{|\vec{r}|^3} |
| \end{align} |
\end{align} |
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of the \PMlinkname{parent entry}{SolidAngle}, we may substitute the position vector \,$\vec{r} = x\vec{i}+y\vec{j}+h\vec{k}$\, of the directed surface element \,$d\vec{a} = \vec{k}\,da,$
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of the \PMlinkname{parent entry}{SolidAngle}, we may substitute the position vector \,$\vec{r} = x\vec{i}+y\vec{j}+h\vec{k}$\, of the directed surface element \,$d\vec{a} = \vec{k}da,$
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| getting |
getting |
| $$\Omega = \int_a \frac{h\,da}{(x^2+y^2+h^2)^{3/2}}.$$ |
$$\Omega = \int_a \frac{h\,da}{(x^2+y^2+h^2)^{3/2}}.$$ |
| Now we can use a ring-formed surface element \,$da = 2\pi\varrho\;d\varrho$\, where\, $\varrho^2 = x^2+y^2$, whence the surface integral may be calculated as |
Now we can use a ring-formed surface element \,$da = 2\pi\varrho\;d\varrho$\, where\, $\varrho^2 = x^2+y^2$, whence the surface integral may be calculated as |
| $$\Omega = \pi h\int_0^R \frac{2\varrho\;d\varrho}{(\varrho^2+h^2)^{3/2}} = |
$$\Omega = \pi h\int_0^R \frac{2\varrho\;d\varrho}{(\varrho^2+h^2)^{3/2}} = |
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\frac{\pi h}{-2}\sijoitus{\varrho=0}{\quad R}\frac{1}{\sqrt{\varrho^2+h^2}}.\\$$
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\frac{\pi h}{-2}\sijoitus{\varrho=0}{\quad R}\frac{1}{\sqrt{\varrho^2+h^2}}.$$
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| Thus we have the result |
Thus we have the result |
| $$\Omega = 2\pi h\left(\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right)\!.$$ |
$$\Omega = 2\pi h\left(\frac{1}{h}-\frac{1}{\sqrt{R^2+h^2}}\right)\!.$$ |
