How do I calculate the relation between angle parameter($t$) and arc lenth parameter($s$)?

Question

Calculate the flux of the vector field $$\vec F=\left(2\pi x+\frac{2x^2y^2}{\pi} \right) \vec{\imath}+\left(2\pi xy-\frac{4y}{\pi}\right)\vec{\jmath}$$ along the outward normal across the ellipse $x^2+16y^2=4$

My attempt

How do I calculate the relation between angle parameter($t$) and arc lenth parameter($s$)? Help me to complete the question

Answer 1

It would be best to use the divergence theorem instead to calculate this integral:

$$\int_{\partial D} \vec{F}\cdot \hat{n} \:ds = \iint_D \nabla\cdot\vec{F}\:dA$$

which means we get

$$I = \iint_{x^2+16y^2\leq 4} 2\pi(1+x)+\frac{4xy^2-4}{\pi}\:dA$$

Now we could try to come up with a change of variables to evaluate this integral, or we can be clever. $2\pi x$ and $\frac{4xy^2}{\pi}$ are both odd functions of $x$ and the region has $x$ symmetry i.e. $-x \in D \iff x \in D$.

Therefore, these integrals evaluate to $0$, leaving us with

$$I = \int_{x^2+16y^2\leq 4} 2\pi - \frac{4}{\pi} \:dA = \left(2\pi - \frac{4}{\pi}\right)\operatorname{Area}(x^2+16y^2\leq 4) = 2\pi^2-4$$