Gauss integral (solid angle)

Let $\varphi (x,y)$ be a function with continuous second order partial derivatives such that

Compute the Gauss integral $u(x,y)=\displaystyle \oint_{C} \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r}\, ds$, where $r=\sqrt{(\xi -x)^{2}+(\eta -y)^{2}}$ is the length of the vector $\overrightarrow {r}$, this vector being obtained by joining the point $A(x,y)$ and a moving point $M(\xi ,\eta )$ on a simple closed smooth contour $C$, and $(\overrightarrow {r},\overrightarrow {n})$ is the angle between the vector $\overrightarrow {r}$ and the outward normal $\overrightarrow {n}$ at the point $M$ on the curve $C$.

Using the polar coordinates $\rho$ and $\varphi$, compute the logarithmic double-layer potentials $K_{1}=\displaystyle \int_{0}^{2\pi } \cos m\psi \, \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r}\, d\psi$ and $K_{2}=\displaystyle \int_{0}^{2\pi } \sin m\psi \, \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r}\, d\psi$, where $r$ is the distance between the point $A(\rho ,\varphi )$ and the moving point $M(1,\psi )$, $(\overrightarrow {r},\overrightarrow {n})$ is the angle between the direction $\overrightarrow {AM}=\overrightarrow {r}$ and the radius $\overrightarrow {OM}=\overrightarrow {n}$ drawn from the point $O(0,0)$, and $m$ is a natural number.

Given the function

where $S$ is the area bounded by the contour $C$, $\overrightarrow {n}$ is the normal to the surface $S$, and $\overrightarrow {r}$ is the radius vector joining the point $M(x,y,z)$ in space to the moving point $A(\xi ,\eta ,\zeta )$ on the surface $S$, prove that this function is the potential of the magnetic field $\overrightarrow {H}$ produced by the current $i$ flowing through the contour $C$.

Setup of problem 4340+: by the Biot--Savart law, the current $i$ through the line element $ds$ produces, at the point $M(x,y,z)$ in space, a magnetic field whose intensity is $d\overrightarrow {H}=ki\, \dfrac {\overrightarrow {r}\times d\overrightarrow {s}}{r^{3}}$, where $\overrightarrow {r}$ is the vector joining the element $d\overrightarrow {s}$ to the point $M$, and $k$ is a proportionality constant; for a closed conductor $C$, the intensity of the magnetic field $\overrightarrow {H}$ at the point $M$ is $\overrightarrow {H}=ki\displaystyle \oint_{C} \dfrac {\overrightarrow {r}\times d\overrightarrow {s}}{r^{3}}$ (see problem 4340+).

Prove the formula

where $S$ is the closed surface enclosing the volume $V$, $\overrightarrow {n}$ is the outward normal at the moving point $(\xi ,\eta ,\zeta )$ on the closed surface $S$, and

with $\overrightarrow {r}$ the radius vector from the point $(x,y,z)$ to the point $(\xi ,\eta ,\zeta )$.

Compute the Gauss integral

where $S$ is a simple closed smooth surface containing the volume $V$, $\overrightarrow {n}$ is the outward normal to the surface $S$ at the point $(\xi ,\eta ,\zeta )$, $\overrightarrow {r}$ is the radius vector connecting the point $(x,y,z)$ and the point $(\xi ,\eta ,\zeta )$,

Consider two cases: (a) when the surface $S$ does not enclose the point $(x,y,z)$; (b) when the surface $S$ encloses the point $(x,y,z)$.

Find the flux of the vector $\overrightarrow {a}=m\dfrac {\overrightarrow {r}}{r^{3}}$ ($m$ a constant) through a closed surface $S$ surrounding the origin.

Given the vector

where $e_{i}$ are constants and $r_{i}$ is the distance from the point $M_{i}$ (the starting point) to the moving point $M(\overrightarrow {r})$. Find the flux of this vector through a closed surface $S$ surrounding the points $M_{i}$ $(i=1,2,\ldots ,n)$.

Show that the integral $\displaystyle w=\iint \limits_{S} \dfrac {\cos (\overrightarrow {r},\overrightarrow {n})}{r^{2}} \, ds=\iint \limits_{S} \dfrac {\partial \frac{1}{r}}{\partial n} \, ds$, taken over the surface $S$, equals the solid angle at which the surface $S$ is seen from the origin. Here $\overrightarrow {r}$ is the radius vector from the origin to the surface element $ds$; $\overrightarrow {n}$ is the normal to the surface, forming an acute angle with the vector $\overrightarrow {r}$ drawn from the origin. The quantity $\dfrac {\partial u}{\partial n}$ is defined by the equality

Prove that

according to whether the point from which the radius vectors $\overrightarrow {r}$ are drawn to the points of the surface lies inside the surface or outside it. The surface is assumed to be simply connected, closed, and without singular points.

Prove the equality

where $\varphi$ is the angle at which the line $AB$ is seen from the point $(\xi ,\eta )$, $r=\sqrt{(x-\xi )^{2}+(y-\eta )^{2}}$, and $\overrightarrow {n}$ is the normal to the line $AB$.

Prove the identity

where $\sigma$ is the surface of the body of volume $\omega$; $\overrightarrow {n}$ is the outward normal to $\sigma$; $\overrightarrow {r}$ is the vector from the point $(x,y,z)$ to the point $(\xi ,\eta ,\zeta )$; $r$ is its length.

Under the same conditions as in the previous problem ($\sigma$ is the surface of the body of volume $\omega$; $\overrightarrow {n}$ is the outward normal to $\sigma$; $\overrightarrow {r}$ is the vector from the point $(x,y,z)$ to the point $(\xi ,\eta ,\zeta )$; $r$ is its length), prove the identity

Prove the equality $\displaystyle \int_{S} \frac{\cos (\overrightarrow {r},\overrightarrow {n})}{r} \, ds = 0$, if the point $(x,y)$ lies outside the region bounded by the contour $S$, and $=2\pi$, if the contour $S$ makes one complete circuit around the point $(x,y)$.

Compute the Gauss integral

where $\partial G$ is the piecewise smooth boundary of the region $G, \mathbf{r} = \overline{M N}, M(x ; y) \in R^{2}$, $N(\xi ; \eta ) \in \partial G, r = \sqrt{(\xi -x)^{2}+(\eta -y)^{2}}, \mathbf{n}$ is the outward normal to $\partial G$, $(\widehat{\mathbf{r}, \mathbf{n}})$ is the angle between $\mathbf{r}$ and $\mathbf{n}$, assuming that:

Let $G \in R^{3}$ be a bounded domain with smooth boundary $S$, $\mathbf{n}$ the outward normal to $S, \mathbf{r} = (\xi -x) \mathbf{i}+(\eta -y) \mathbf{j}+(\zeta -z) \mathbf{k}$ :

Let $S$ be a smooth surface, oriented by the normal $\mathbf{n}$, and suppose the closure of $S$ does not contain the origin. Show that the integral

is the flux of some field through $S$.

Let $G$ be a bounded domain with piecewise smooth boundary, oriented by the outward normal $\mathbf{n}, O \notin \bar{G}, \mathbf{r} = (x ; y ; z), r = \left|\mathbf{r}\right|$.

Prove that:

Let a bounded region $G$ have a piecewise smooth boundary $\partial G$, oriented by the outward normal, $M_{0}$ a fixed point of $G$,

$S_{\varepsilon }\left(M_{0}\right)$ is the sphere with center $M_{0}$ and radius $\varepsilon$, lying in $G$, oriented by the outward normal. Prove that the flux of a through $\partial G$ equals the flux of $a$ through $S_{\varepsilon }\left(M_{0}\right)$.

Let a bounded region $G$ have a piecewise smooth boundary $\partial G$, oriented by the outward normal, $M_{0}$ a fixed point,

Find the flux of the field a through $\partial G$, if:

Under the conditions of problem 83, let $M_{0} \in \partial G$ and, in a neighborhood of $M_{0}$, the boundary $\partial G$ be twice continuously differentiable. Let $\partial G_{\varepsilon }$ be the part of the boundary $\partial G$ lying inside the ball $\left|\overline{M_{0} M}\right| \leqslant \varepsilon$, and let $\Pi_{\varepsilon }$ be the flux of the field a through $\partial G \backslash \partial G_{\varepsilon }$. Find $\lim_{\varepsilon \rightarrow 0} \Pi_{\varepsilon }$.

Let $\gamma$ be a smooth plane simple (p. 295) curve whose closure does not contain the origin, $\mathbf{n}$ being a continuous unit normal to $\gamma$. Show that the Gauss integral

is the flux of some field through $\gamma$.

Under the conditions of problem $87$, let $\gamma$ be the boundary of a bounded region $G$. Calculate the Gauss integral, if:

Show that the value of the Gauss integral from problem 87 equals the polar angle at which the curve $\gamma$ is seen from the origin.