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

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 ?

• Are you sure it is a minus between the two terms in the numerator? Green's theorem doesn't seem to work if it is a minus, since $$P_y=\frac{4xy}{x^2+y^2}\\Q_x=-\frac{4xy}{x^2+y^2}$$So $P_y\ne Q_x$ – John Doe Jan 7 at 22:19
• oh right , anyway can i apply green therom here ? even if its not $0$ or the integral is too complex to calculate ? – Mather Jan 7 at 22:22
• i have also tried to calculate it with basic parametrization of the ellipse but i got harsh integrand – Mather Jan 7 at 22:23
• @JohnDoe are you sure? Shouldn't there be a $(x^2+y^2)^2$ in the denominator from quotient rule. I don't think it cancels. – AHusain Jan 7 at 22:24
• @AHusain Oh whoops, I didn't differentiate that properly, you''re right! However, even if you do differentiate it properly, it still doesn't give $P_y=Q_x$, the minus sign will still mess this up – John Doe Jan 7 at 22:34

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$$\frac14\left(5\log\frac53-2-2+3\log\frac53\right)=2\log\frac53-1$$

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 $$0$$.

• shouldn't it be $0$ ? – Mather Jan 8 at 9:40
• @JohnDoe This isn't entirely correct. Your new coordinate system only maps a quarter of the ellipse. To go all the way around the ellipse, you must go forwards and backwards on the line in $(u,v)$ twice, making the integral $0$. – Dylan Jan 8 at 12:38
• @mather see the above comment ^ I'll edit my answer in a little while – John Doe Jan 8 at 13:20
• @Dylan I have edited the question, thanks for pointing that out – John Doe Jan 8 at 15:40

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.