# Help for evaluating a line integral with Green's theorem

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.

## 1 Answer

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?

• I think you meant twice if you integrate $0$ to $2\pi$, your integration of $0$ to $\pi$ is once and is correct. – mathematics2x2life Jun 18 '19 at 7:02
• @Doug M Yes i am getting zero when i am integrating without the green's theorem but the question was if i evaluate the equation right with the greens theorem i mean like steps ? – Boris Borovski Jun 18 '19 at 7:06
• @mathematics2x2life Yes the limits are $t \in [0, \ 2 \pi]$ – Boris Borovski Jun 18 '19 at 7:10
• The final integral is clearly $0$ because $\int_0^\pi \sin 2t~dt =0$. – Robert Shore Jun 18 '19 at 7:28
• @Doug M Nowhere it says i must use polar coordinates, ah i now see what it should be - ok for the direct integration we don't need to go for polar coordinates we just evaluate the first equation that you wrote and for greens theorem we need to go into polar coordinates if i get it right. Thank you – Boris Borovski Jun 18 '19 at 7:30