Flux Through Triangle

Consider a line passing through the points $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ . Any point on the line passing through these points and between them can be parameterised as such:

$x = t(b_1-a_1) +a_1$ $y = t(b_2-a_2) +a_2$ $z = t(b_3-a_3) +a_3$

Here, $t$ varies from 0 to 1. This position vector of this point is:

$\vec{r} = x \ \hat{i} + y \ \hat{j} + z \ \hat{k}$ $\implies d\vec{r} = \left((b_1-a_1) \ \hat{i} + (b_2-a_2) \ \hat{j} + (b_3-a_3) \ \hat{k}\right) \ dt$

Now putting this parameterization aside, in this problem we are provided with the vertices of a triangle and the magnetic field. To compute the flux through the given region, I made use of the 'gauge invariance' principle that @Steven Chase spoke of in a solution to an earlier problem. This means, first, the magnetic vector potential is derived. We know that:

$\nabla \times \vec{A} = \vec{B}$

Let:

$\vec{A} = A_x \ \hat{i} + A_y \ \hat{j} + A_z \ \hat{k}$ $\vec{B} = B_x \ \hat{i} + B_y \ \hat{j} + B_z \ \hat{k}$

This implies:

$B_x = \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} = 0$ $B_y = \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} = 0$ $B_z = \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = -2xy$

These are three partial differential equations. By careful inspection, one can see that the vector:

$\vec{A} = 2xy^2 \hat{i} + x^2y \ \hat{j} + 0 \ \hat{k}$

Satisfies the above equations. Now, recall that:

$\phi = \int_{S} \vec{B} \cdot d\vec{S}$

Applying Stokes theorem gives:

$\phi = \int_{S} \vec{B} \cdot d\vec{S} = \int_{S} (\nabla \times \vec{A}) \cdot d\vec{S}= \oint_{C} \vec{A} \cdot d\vec{r}$

Here, $C$ is the closed contour which is composed of the sides of the triangle. From here, the closed line integral can be split into three integrals as such:

$\oint_{C} \vec{A} \cdot d\vec{r} = \int_{1 \to 2}\vec{A} \cdot d\vec{r} + \int_{2 \to 3}\vec{A} \cdot d\vec{r} + \int_{3 \to 1}\vec{A} \cdot d\vec{r}$

Making use of the initial line parameterization, the integrands can be simplified. These simplifications are left out for now. The final integrand is:

$\phi = \int_{0}^{1} (135t^3 - 414t^2 + 405t - 90)dt$

$\boxed{\phi = \frac{33}{4}}$