I did not quite understand the latest lecture I've been to and would like a thorough explanation if possible. A field vector is given by $F=(\cos(xyz), \sin(xyz), xyz)$. Calculate the flux through a square that's parallel to the $XY$ plane, its center is in $(0,0,a)$, and its side's length is $a$. Thanks and regards.
[Math] Calculating flux through a square
integrationVector Fieldsvectors
Related Solutions
The solid angle can be expressed in terms of complete elliptic integrals of first and third kind.
I believe this is first derived by F. Paxton in 1959.
THE REVIEW OF SCIENTIFIC INSTRUMENTS. VOLUME 30, NUMBER 4 APRIL, 1959.
Solid Angle Calculation for a Circular Disk. F. PAXTON
A copy of the article can be found here.
Below is my attempt to derive the same result in an alternate manner.
The integral is indeed trickier than I thought.
To avoid confusion with the usage of $x,y,z$ as coordinates, let us change the problem and assume the circular disk $C$ we are dealing with lies in the plane $z = b$, centered at $(a,0,b)$ with radius $s$. We will assume $a, b > 0$ and $a \ne s$.
In terms of ordinary spherical polar coordinates $(r,\theta,\phi)$, points on the plane $z = b$ can be parametrized as:
$$(x,y,z) = (b\tan\theta \cos\phi,\,b\tan\theta\sin\phi,\, b)$$ Let $\rho = \sqrt{x^2+y^2} = b\tan\theta$, the "element" for integrating the solid angle is given by:
$$d\Omega = \sin\theta d\theta \wedge d\phi =\frac{\tan\theta d\tan\theta}{\sqrt{1 + \tan\theta^2}^3} \wedge d\phi = \frac{b \rho d\rho \wedge d\phi}{\sqrt{\rho^2+b^2}^3} = \frac{b\;dx \wedge dy}{\sqrt{\rho^2+b^2}^3} $$ Introduce complex coorindates $\eta = x + iy$ and $\bar{\eta} = x - iy$, the solid angle extended by $C$ can be rewritten as:
$$\begin{align} \Omega_{C} = & \int_{C} d\Omega = \frac{b}{2i}\int_{C} \frac{d\bar{\eta} \wedge d\eta}{\sqrt{\eta\bar{\eta}+b^2}^3} =\color{blue}{^{[1]}} \frac{b}{i} \int_C d\left(\frac{1}{\eta}\left( \frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta}+b^2}}\right)\right) \wedge d\eta\\ = & \frac{b}{i} \int_{\partial C} \left(\frac{1}{b} - \frac{1}{\sqrt{\eta\bar{\eta} + b^2}}\right) \frac{d\eta}{\eta} = 2\pi \delta_{C} + ib \int_{\partial C} \frac{d\eta}{\eta \sqrt{\eta\bar{\eta} + b^2}} \end{align}$$ where $\delta_{C} = 1 \text{ or } 0$ depends on whether $s > a$ or $< a$. In other words, whether $\partial C$ contains $0$ or not.
On $\partial C$, we can parametrize $\eta$ as $s e^{it} + a$ and $\bar{\eta}$ as $s e^{-it} + a$. Substitute this into above expression, we obtain:
$$\begin{align} &\Omega_C - 2\pi \delta_{C}\\ = &ib \int_{-\pi}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}}\frac{ is e^{it}}{s e^{it} + a }\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \left( \frac{s e^{it}}{s e^{it} + a } + \frac{s e^{-it}}{s e^{-it} + a } \right)\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \frac{2s(s + a\cos t)}{s^2 + a^2 + 2sa\cos t}\\ = & -b \int_{0}^{\pi} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos t}} \left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos t} \right)\\ = & -2b \int_{0}^{\pi/2} \frac{ dt }{\sqrt{ s^2 + a^2 + b^2 + 2sa\cos(2t)}} \left( 1 + \frac{s^2 - a^2}{s^2 + a^2 + 2sa\cos(2t)} \right)\\ = & -2b \int_{0}^{\pi/2} \frac{ dt }{\sqrt{ (s+a)^2 + b^2 - 4sa\sin^2(t)}} \left( 1 + \frac{s^2 - a^2}{(s+a)^2 - 4sa\sin^2(t)} \right) \end{align}$$ The last integral can be expressed in terms of the complete elliptic integral of first and third kind: $$ K(k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{\sqrt{1-k^2\sin^{2}t}} \quad\text{ and }\quad \Pi(n,k) = \int_{0}^{\frac{\pi}{2}} \frac{dt}{(1-n\sin^{2}t)\sqrt{1-k^2\sin^{2}t}} $$
The final results are: $$\Omega_{C} = 2\pi\delta_C -\frac{2b}{\sqrt{(s+a)^2+b^2}} \left(\;K(k) + \left(\frac{s-a}{s+a}\right) \Pi(n,k)\;\right) $$ where $\displaystyle \;\;k = \sqrt{\frac{4sa}{(s+a)^2+b^2}} \quad\text{ and }\quad n = \frac{4sa}{(s+a)^2} $.
Notes
$\color{blue}{[1]}$ When $C$ contains $(0,0,b)$, we need the $\frac{1}{b}$ term. It regularize the 1-form at $\eta = 0$ and make Stroke's theorem continue to work.
We have that the flux of ${\bf F}(x,y)=(F_x(x,y),F_y(x,y))$ through a given closed curve $\gamma$ is $$\int_{\gamma} ({\bf F} \cdot {\bf n}) ds$$ where ${\bf n}$ is the outward unit normal vector field to $\gamma$.
If $\gamma$ is the circle of radius $r$ centred at the origin then the outer normal is given by ${\bf n}=(\cos(t),\sin(t))$, hence $$\int_{\gamma} ({\bf F} \cdot {\bf n}) ds=\int_0^{2\pi}[F_x(r\cos(t),r\sin(t))\cos(t)+F_y(r\cos(t),r\sin(t))\sin(t)] (rdt).$$
In your case, $r=1$, $F_x(x,y)=a_{11}x+a_{12}y$ and $F_y(x,y)=a_{21}x+a_{22}y$. Now you can compute the above integral and find a formula in terms of the elements of the matrix $A$.
By the Divergence Theorem you will find that the flux is
$$\int_D \nabla \cdot {\bf F} \, dA = \int_D (a_{11}+a_{22}) \, dA=(a_{11}+a_{22})|D|=\pi(a_{11}+a_{22})$$ where $D$ is the interior of $\gamma$.
Best Answer
The flux is given by
$$\iint_{\text{square}} dA \, \mathbf{F} \cdot \hat{\mathbf{n}} $$
where $\hat{\mathbf{n}}$ is the unit normal to the square, which in this case is simply the $z$ direction. Thus, $\mathbf{F} \cdot \hat{\mathbf{n}}$ is the $z$ component of $\mathbf{F}$, or $x y z$. Thus, the flux is
$$a \int_{-a/2}^{a/2} dx \, x \, \int_{-a/2}^{a/2} dy \, y = 0$$