The right way of using Green's theorem when x and y are given parametrized

Question

I am solving a line integral that is from $2nd \ kind$ and it can be solved by direct integration or by green's theorem so i already solved it using the direct integration but i have tried different ways to set it up with green's theorem but i am not understanding the following: I have $$\int_0^{2 \pi}(2x) \ dx + (3yx) \ dy,$$ and the line given by this parameters $$C: x = 4\cos(2t), y = 3\sin(2t)$$ so i want to apply the green's theorem this time, but all the examples i have are made without parametrization and i cannot figure out how to do it so i have tried : $$I_1=\iint_D(3yx)^{\prime}dx-(2x)^{\prime}dy$$ So when i take the partial derivatives i am left only with : $$\iint_D(3y-2)dxdy$$ and this isn't seems to be right. When i try to put $x=\cos(2t) \ \text{and} \ y=3\sin (2t)$ i am not sure how this can became double integral $dxdy$ because if i use the parameters from the equation i will have one limit [$0, 2 \pi$]. Is there a revert operation of the parameters and to take new limits etc ? Thank you for any help.

Answer 1

Green Theorem states that $$\int_C P(x, y)dx + Q(x, y)dy = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy$$

Where $C$ encloses the region $S$. Therefore we have

$$I = \iint_S \left(\frac{\partial(2x)}{\partial y} + \frac{\partial(3yx)}{\partial x} \right) dydx$$ $$= \iint_S 3y dydx = 0$$

Because the integrand is an odd function and $S$ is a complete ellipse which is symmetrical about both the x and y axes.

Answer 2

The parametrization of the line integral doesn't really help you calculate the corresponding surface integral. After all, the whole point of line integrals is that they're independent of the parametrization; if I changed to $t = u^3$ to parametrize the curve, I should still get the same result when I compute the line integral.

In general, you have to find a new parametrization for the surface $D$ that you're trying to integrate over. Sometimes this "parametrization" is just the coordinates $x$ and $y$, while other times it can be helpful to switch into other coordinates (rescaling $x$ and $y$ linearly, or changing to polar coordinates) via the Jacobian.