Jtdylktuy

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 ?

We have

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

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

$$int_{3u+5v=1,quad u,vge0}left(frac{v}{u+v}mathrm du-frac{u}{u+v}mathrm dvright)tag2$$

We compute the first integral. The path is $3u+5v=1,quad u,vge0$, so $v=frac15-frac35 u$ and $uin[0,frac13]$.

$$int_mathcal Cfrac v{u+v}mathrm du=int_0^frac13frac{1-3u}{1+2u}=cdots=frac14left(5logfrac53-2right)$$

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

$$int_mathcal Cfrac u{u+v}mathrm dv=int_0^frac15frac{1-5v}{1-2v}mathrm dv=cdots=frac14left(2-3logfrac53right)$$

Hence the amount that $(2)$ contributes to the total integral $(1)$ is$$frac14left(5logfrac53-2-2+3logfrac53right)=2logfrac53-1$$

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 $0$.

You can evaluate the line integral directly by taking $mathbf r(t) = (cos(t)/sqrt 3, sin(t)/sqrt 5)$:
$$I = int_C mathbf F cdot dmathbf r =
-int_0^{2 pi} frac {sin 2 t} {4 + cos 2 t} dt =
-int_0^pi frac {sin 2 t} {4 + cos 2 t} +
int_0^pi frac {sin 2 t} {4 + cos 2 t} dt = 0.$$
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.