matrix-calculusordinary differential equationsvector-spacesvectors
For a two-by-two matrix $A$, find the flux of a linear vector field ${\bf F}(x,y)=A\begin{bmatrix}x\\ y\end{bmatrix}$ in $\mathbb{R}^2$ through a unit circle with the center at the origin in terms of entries of $A$.
I don't quite understand what to put inside the integration here I'm confused as to how ${\bf F}$ acts on the unit circle. And should I use $x= r \cos t$ and $y = r \sin t$ for the circle or a function of radius $r$?
From the following figure it can immediately been read off that when the point of contact $Q$ moves counterclockwise by the angle $\phi$ then the inner circle (as a rigid structure) rotates by the amount $2\phi$ clockwise. Therefore after a full turn of $Q$ the point $P$ has made $2$ turns.
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
$$a \int_{-a/2}^{a/2} dx \, x \, \int_{-a/2}^{a/2} dy \, y = 0$$
Best Answer
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 $\gamma$.