Evaluate $\int_{\Gamma}\frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy$

Question

Evalaute $$\int_{\Gamma}\frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy$$

Where $\Gamma$ is the rectangle ABCD where $A=(1,0), B= (0,1), C=(-1,0), D=(0,-1)$

I have used:

$AB = (1-t,t), t\in[0,1]$

$BC = (-t,1-t), t\in[0,1]$

$CD = (-1+t,-t), t\in[0,1]$

$DA = (t,-1+t), t\in[0,1]$

I evaluate $\int_{AB} + \int_{BC} + \int_{CD} +\int_{DA}$ and got all zeros expect for $\int_{BC}= \frac{\pi}{2}$

a. is it correct?

b. is there a way to solve it using symmetry?

Answer 1

You can use Green's theorem to turn it into an integral on the unit circle

$$\int\limits_{\text{unit circle}} F\cdot dr- \int\limits_{\text{diamond}}F\cdot dr = \iint\limits_{\text{unit circle - diamond}}\operatorname{curl}F\:dA = 0$$

$$\implies \int\limits_{\text{unit circle}} F\cdot dr = \int\limits_{\text{diamond}}F\cdot dr$$

And on the unit circle $(x,y) = (\cos t, \sin t)$

$$\implies \int\limits_{\text{unit circle}} F\cdot dr = \int_0^{2\pi}-\cos t\sin t + \sin t \cos t\:dt = 0$$

which means the original integral also evaluates to $\boxed{0}$. This also proves that the vector field was conservative on $\Bbb{R}^2 - \{0\}$

Answer 2

Using symmetry, the total is zero. Keep the $x$ and $y$ integrals separate. For $x$, the AB and BC integrals are equal in magnitude and opposite in sign. Similarly CD with DA. For $y$, the pairings are AB with DA and BC with CD.