Applying Green's theorem to this integral

Question

Let the area outside the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ bounded by the curves $\frac{x^2}{25}+\frac{y^2}{16}=1(y\leq 0)$,$x+y=5$ and $y-x=5$ be $R$, $C$ is the outer bound of $R$ and $C_1$ be the inner bound (the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$) apply Green's theorem to the integral $$\oint_C\frac{-y}{4x^2+9y^2}dx+\frac{x}{4x^2+9y^2}dy$$.

Now what I am getting is zero as the integrand is equal to the grad of some function f and the integral is over a closed curve, the problem is this question is going to be on the exam and my answer is too obvious to be true, I feel like there is a twist or a trick in there and the answer is not supposed to be 0.

Answer 1

The problem here is the singularity at the origin, which ends up making it so that the function is not the gradient of some function. You can even try finding such a function, and see that you can't (however you can locally, but this is not sufficient). What we can do, however, is observe that

$$I=\oint_C\left(\frac{-y}{4x^2+9y^2}~\mathrm{d}x+\frac{x}{4x^2+9y^2}~\mathrm{d}y\right)=\iint_R\left(\frac{\partial}{\partial x}\frac{x}{4x^2+9y^2}-\frac{\partial}{\partial y}\frac{-y}{4x^2+9y^2}\right)~\mathrm{d}x~\mathrm{d}y-\oint_{C_1}\left(\frac{-y}{4x^2+9y^2}~\mathrm{d}x+\frac{x}{4x^2+9y^2}~\mathrm{d}y\right)$$

by using Green's theorem. But notice that

$$\frac{\partial}{\partial x}\frac{x}{4x^2+9y^2}=\frac{-4x^2+9y^2}{(4x^2+9y^2)^2},$$

$$\frac{\partial}{\partial y}\frac{-y}{4x^2+9y^2}=\frac{-4x^2+9y^2}{(4x^2+9y^2)^2},$$

and so the double integral in the above vanishes, leaving us with

$$I=-\oint_{C_1}\left(\frac{-y}{4x^2+9y^2}~\mathrm{d}x+\frac{x}{4x^2+9y^2}~\mathrm{d}y\right).$$

But this is especially nice due to how $C_1$ is defined. In particular, we can parameterize it by the function $\gamma:[0,2\pi)\to C_1$ defined by

$$\gamma(\theta)=(3\cos\theta,2\sin\theta),$$

which has derivative given by

$$\gamma'(\theta)=(-3\sin\theta,2\cos\theta).$$

Applying the definition of the line integral we thus get that

$$I=-\int_0^{2\pi}\left(\frac{-2\sin\theta}{4(3\cos\theta)^2+9(2\sin\theta)^2}(-3\sin\theta)+\frac{3\cos\theta}{4(3\cos\theta)^2+9(2\sin\theta)^2}2\cos\theta\right)~\mathrm{d}\theta=-\int_0^{2\pi}\frac{6(\cos^2\theta+\sin^2\theta)}{9(\cos^2\theta+\sin^2\theta)}~\mathrm{d}\theta=-\frac{2}{3}\int_0^{2\pi}\mathrm{d}\theta=-\frac{4\pi}{3}.$$

Answer 2

   0 ==D[-y/(4 x ^2 + 9 y ^2), y] - D[x/(4 x ^2 + 9 y ^2), x] // Simplify   

So your function is indeed a gradient locally for

   {x,y}!=0.  

But the two partial integrals seem to be different

{ Integrate[-y/(4 x ^2 + 9y ^2),x] ,Integrate[x/(4 x ^2 + 9y ^2),y]}   
-> {-(1/6) ArcTan[(2 x)/(3 y)],1/6 ArcTan[(3 y)/(2 x)]}.

Introduce polar coordinates. The differentials are r-independent by inspection. The the arc integral is a constant.

     (-y dx + x dy)/(4 x^2 + 9 y^2)/.
        {x->r Cos[\[Phi]], y -> r Sin[\[Phi]],
         dx-> dr Cos[\[Phi]]- r Sin[\[Phi]] d\[Phi],
         dy -> dr Sin[\[Phi]]+ r Cos[\[Phi]] d\[Phi] }//FullSimplify


       Integrate[2  /(-13+5 Cos[2 \[Phi]]),{\[Phi],0,2\[Pi]}]

       -(Pi/3)