Magnetic Flux (5-25-2021)

First, the magnetic vector potential is computed using the relation:

$\nabla \times \vec{A} = \vec{B}$ $\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}=0$ $\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}=0$ $\frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y}=3x^2$

From the above relations, with a little bit of trial and error, it is seen that the following vector satisfies the above relations.

$\vec{A} = 0 \ \hat{i} + x^3 \ \hat{j} + 0 \ \hat{k}$

Now, by applying Stokes' theorem

$\Phi= \int \int_{S}\vec{B} \cdot d\vec{S} =\int \int_{S} (\nabla \times \vec{A}) \cdot d\vec{S} = \oint_{C} \vec{A} \cdot d\vec{R}$

$\vec{R} = r(\theta) \cos{\theta} \ \hat{i} + r(\theta) \sin{\theta} \ \hat{j}$ $\frac{d\vec{R}}{d\theta} = \frac{d}{d\theta} \left(r(\theta) \cos{\theta}\right) \ \hat{i} + \frac{d}{d\theta} \left( r(\theta) \sin{\theta}\right) \ \hat{j}$

Plugging in the given parameterization:

$\implies\Phi= \oint_{C} \vec{A} \cdot d\vec{R} = \int_{0}^{2 \pi}\left(x^3(\theta) \ \frac{d}{d\theta} \left( r(\theta) \sin{\theta}\right) \right) \ d\theta$

Simplifying the integrand leads to:

$\Phi=\oint_{C} \vec{A} \cdot d\vec{R}=\int_{0}^{2 \pi} \left(1 + 0.5 \sin{4\theta}\right)^3 \cos^3{\theta} \left(2 \sin{\theta} \cos{4\theta} + \cos{\theta} \left(1 + 0.5 \sin{4\theta}\right) \right) \ d\theta$

$\boxed{\Phi= \frac{681 \pi}{512} \approx 4.178563666200487}$

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Shorter method:

$\vec{B} = 3r^2\cos^2{\theta} \ \hat{k}$ $d\vec{S} = r \ dr \ d\theta \ \hat{k}$

$\Phi = \int \int_{S}\vec{B} \cdot d\vec{S} = \int_{0}^{2\pi} \int_{0}^{1+0.5\sin{4\theta}} \left(3r^2\cos^2{\theta} \right) \ r \ dr \ d\theta \approx 4.178563666200487$