how to calculate $\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2}$ using green theorm or directly

Question

Calculate $$\int_C \frac{2xy^2dx-2yx^2dy}{x^2+y^2},$$ where $C$ is the ellipse $3x^2 +5y^2 = 1$ taken in the positive direction.

I tried to calculate the integral using green theorm.

now i need to build enclosier that doesn't enclose $(0,0)$ i am having hard time guessing what to build.

a circle and ellipse might be perfect but then the domain is not easy to write. can i have hint please ?

Answer 1

We have

$$\int_\mathcal C \frac{2xy^2\mathrm dx-2x^2y\mathrm dy}{x^2+y^2}=\int_{3x^2+5y^2=1}\frac{\mathrm d(x^2)y^2-x^2\mathrm d(y^2)}{x^2+y^2}\tag1$$

Let $u=x^2, v=y^2$. Then we have

$$\int_{3u+5v=1,\quad u,v\ge0}\left(\frac{v}{u+v}\mathrm du-\frac{u}{u+v}\mathrm dv\right)\tag2$$

We compute the first integral. The path is $3u+5v=1,\quad u,v\ge0$, so $v=\frac15-\frac35 u$ and $u\in[0,\frac13]$.

$$\int_\mathcal C\frac v{u+v}\mathrm du=\int_0^\frac13\frac{1-3u}{1+2u}=\cdots=\frac14\left(5\log\frac53-2\right)$$

We compute the second integral. The path is $u=\frac13-\frac53 v$ with $v\in[0,\frac15]$.

$$\int_\mathcal C\frac u{u+v}\mathrm dv=\int_0^\frac15\frac{1-5v}{1-2v}\mathrm dv=\cdots=\frac14\left(2-3\log\frac53\right)$$

Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$\frac14\left(5\log\frac53-2-2+3\log\frac53\right)=2\log\frac53-1$$

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Edit:

As pointed out in the comments, this change of co-ordinates was not a bijective map, so we have not included all the points on the ellipse by doing this calculation, we have only done it for $x,y>0$. For $x,y<0$, we would have the same path, but traversed backwards (i.e. the integration limits will be swapped), which would directly cancel out what we have computed here.

For $x>0,y<0$, we have the same path as in $(2)$, while for $x<0,y>0$, we have this path traversed backwards (you can see this by considering the ellipse traversed clockwise, seeing whether $x$, and correspondingly $u$, is increasing in magnitude or not). So again, these two contributions cancel. This is what gives $0$.

Answer 2

You can evaluate the line integral directly by taking $\mathbf r(t) = (\cos(t)/\sqrt 3, \sin(t)/\sqrt 5)$: $$I = \int_C \mathbf F \cdot d\mathbf r = -\int_0^{2 \pi} \frac {\sin 2 t} {4 + \cos 2 t} dt = -\int_0^\pi \frac {\sin 2 t} {4 + \cos 2 t} + \int_0^\pi \frac {\sin 2 t} {4 + \cos 2 t} dt = 0.$$ Green's theorem still holds for $\mathbf F$ even though $\mathbf F$ doesn't have continuous partial derivatives at $(0, 0)$: $$I = -\iint_{3 x^2 + 5 y^2 \leq 1} \frac {4 x y} {x^2 + y^2} dx dy.$$ This form makes it clearer that the result is zero because of the symmetries wrt the coordinate axes.