Mutual Inductance - Wavy Loop 2

Consider a point on loop 1:

$\vec{r}_1 = (3+\cos{\theta_1}) \ \hat{i} + (2+\sin{\theta_1}) \ \hat{j} + 0 \ \hat{k}$

And that on loop 2:

$\vec{r}_2 = \cos{\theta_2} \ \hat{i} + \sin{\theta_2} \ \hat{j} + (5+5\sin{\theta_2}) \ \hat{k}$

The arc length elements on each loop can be computed by evaluating the total differentials:

$d\vec{r}_1 = \left(-\sin{\theta_1} \ \hat{i} + \cos{\theta_1} \ \hat{j} + 0 \ \hat{k} \right) \ d\theta_1$ $d\vec{r}_2 = \left(-\sin{\theta_2} \ \hat{i} + \cos{\theta_2} \ \hat{j} + 5\cos{\theta_2} \ \hat{k} \right) \ d\theta_2$

Finally, we can compute the elementary magnetic vector potential at any general point due to loop 1:

$d\vec{A} = \frac{\mu_o I}{4 \pi}\left(\frac{d\vec{r}_1}{\lvert \vec{r} - \vec{r}_1\rvert}\right)$

The total magnetic potential vector due to loop at any point can be evaluated as such:

$\vec{A} = \frac{\mu_o I}{4 \pi} \oint_{L1} \frac{d\vec{r}_1}{\lvert \vec{r} - \vec{r}_1\rvert}$

Now, we know that the magnetic field at that point can be computed as such:

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

The elementary surface area through the loop is say: $d\vec{S}$ . This means that elementary flux is:

$d\Phi = \vec{B} \cdot d\vec{S}$

The total flux can be computed by integrating over the surface area of loop 2 as such:

$\Phi = \oint_{S2}\vec{B} \cdot d\vec{S}$ $\implies \Phi = \oint_{S2} \left(\nabla \times \vec{A}\right) \cdot d\vec{S}$

Applying Stoke's theorem leads to:

$\Phi = \oint_{S2} \left(\nabla \times \vec{A}\right) \cdot d\vec{S} = \oint_{L2} \vec{A} \cdot d\vec{r}_2$ $\Phi = \oint_{L2} \oint_{L1} \frac{\mu_o I}{4 \pi} \frac{d\vec{r}_1}{\lvert \vec{r}_2 - \vec{r}_1\rvert}\cdot d\vec{r}_2$

Substituting expressions and simplifying the integrand gives a function $f(\theta_1,\theta_2)$ and the integral looks as such:

$\Phi = \int_{0}^{2 \pi} \int_{0}^{2 \pi} f(\theta_1,\theta_2) d\theta_1 \ d\theta_2$

I performed this entire calculation process using a script of code and have only provided an outline. The answer evaluates to:

$\boxed{\Phi \approx 0.017}$