Help for verifying approach for executing a line integral of kind 2

Question

I have a line integral of kind 2. I want to use Green's theorem to solve it. I am not sure if I am setting it right so I want to ask for help to verify if I set up the integral correct. So the integral is given as follows : $$\iint (2x)dx+3(yx)dy$$ with parameters : $$C:x = 4\cos(2t) \ , \ y=3\sin(2t)$$ I found the region and its an ellipse: $$\frac{x^2}{4}+\frac{y^2}{3}=1$$ And I use polar coordinates parameters : $$x = 4r\cos 2t\\ y = 3r\sin 2t\\ dy\ dx = 12r \ dr\ dt$$ So I apply the greens theorem $$\int_C P(x, y)dx + Q(x, y)dy = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy$$ and I get $$\iint_S 3y dydx$$ $$\int_0^{2 \pi} \int_0^1(9r\sin(2t))(12r)drd \theta $$ Is this the right way to evaluate this integral with the Green's theorem ?

Answer 1

There are three things you have done here wrong, they are:

1. Equation of the ellipse would be $$ \cfrac{x^2}{16} + \cfrac{y^2}{9} = 1$$ and it can be verified by substituting $x = 4 \cos(2t), y = 3 \sin(2t)$ in the equation.

2. As you taken $x = 4r cos(2t), y = 3r sin(2t)$. Since$$ dx = \cfrac{\partial x}{\partial r} dr + \cfrac{\partial x}{\partial t}dt $$ and same for $dy$, $$\begin{align} &\Rightarrow dx = 4 \cos(2t) dr - 8r \sin(2t) dt \\ & \Rightarrow dy = 3 \sin(2t) dr + 6r \cos(2t) dt \end{align} $$ whose exterior product is $$dx dy = 24 r dr dt$$

3. Limit of $t$ would be from $ 0$ to $ \pi$ , you can check that by placing $\pi$ in $x = 4r \cos(2t) $, it reverts back to $4r$ i.e. it has completed one round.

4. In the last integral, $d \theta$ is not defined.