Help for evaluating a line integral with Green's theorem

Question

I have the following line integral of kind 2 $$\iint (2x)dx+3(yx)dy$$ and the region $$C:4\cos(2t) \ , \ y=3\sin(2t)$$ I sketch the region and its an elipse: $$\frac{x^2}{4}+\frac{y^2}{3}=1$$i am applying the greens theorem which is : $$\int_C P(x, y)dx + Q(x, y)dy = \iint_S \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dxdy$$ And i get $$\iint_S \frac{d}{dx} (3yx) -\frac{d}{dy}(2x)$$ So i take the partials and i get $$\iint_S 3y dydx$$ So my approach was to substitute $y=3 \sin(2t)$ in the integral and after integrating it i get $$\int_0^{2 \pi}9\sin(2t)=\frac{-9\cos2t}{2} \ \Bigg|_0^{2 \pi}= \frac{1}{2} - \frac{1}{2}= \color{blue}0$$ So is this the right approach i mean to substitute in the parameters and evaluate the integral this way or i am making an mitake. Really i will appriciate any help because i am having hard time understand what is the algorithm for solving Line integrals.

Answer 1

I get 0, too.

Note if $(x,y) = (4\cos 2t, 3\sin 2t)$ you will traverse the contour twice if you integrate from $0$ to $2\pi$.

Not that it matters. If it is $0$ twice around it is 0 the first time around.

At this point $\iint 3y\ dy\ dx$

Your integrand is odd and your region is symmetric about across the $x$ axis. This suggests the integral will be 0.

Suppose we didn't use Green's theorem, hopefully, we get the same result

$\int_0^{\pi} 2(4\cos 2t)(-8\sin 2t) + 3(3\sin 2t)(4\cos 2t)(6\cos 2t) \ dt\\ \int_0^{\pi}-64\cos 2t\sin 2t + 216\sin 2t\cos^2 2t \ dt\\ 0$

Did you do the integration correctly?

If you set up the integral in Cartesian:

$\frac{x^2}{16} + \frac{y^2}{9} = 1$

Note that you do not have the denominators squared in the OP.

$\int_{-4}^4\int_{-\frac 34 \sqrt{16-x^2}}^{\frac 34 \sqrt{16 - x^2}} 3y \ dy\ dx$

Clearly this evaluates to 0.

If you wanted to put the parameterization to use.

$x = 4r\cos 2t\\ y = 3r\sin 2t\\ dy\ dx = 24 r \ dr\ dt$

The last line is your jacobain.

$\int_0^{\pi}\int_0^1 (3r\sin 2t)(24 r)\ dr\ dt = 0$

Did you set it up correctly?