Find $\int F.dr$ where $F= \frac{y i – x j}{x^{2}+y^{2}}$ and $C$ is the circular path $x^{2} + y^{2} = 1$ described in counter clockwise sense.
The formula for flux is $\int {F_1 dy – F_2 dx}$ where $$F_1 = \frac{y}{(x^{2}+y^{2})},$$ $$F_2 = \frac{-x}{x^{2} + y^{2}}$$
$$\int F.dr = \int (\frac{y}{x^{2}+y^{2}}dy – \frac{-x}{x^{2}+y^{2}}dx$$
$$=\int_{-1}^{1} ydy + xdx$$
$($since $(x^{2}+y^{2} = 1$ and in given circle both x and y vary from -1 to 1 $)$
so the flux I get is
$2$ which is not right answer.
the right answer is $-2\pi$ which can be found by using polar coordinates and applying same formula of flux , i used in above method.
What is wrong with the method I applied to evaluate flux? How to find flux using cartesion coordinates?
Best Answer
It's hard to understand what you really meant, but first parametrize the unit canonical circle (positive direction):
$$r(t)=(\cos t,\,\sin t)\;,\;\;t\in[0,2\pi]\implies r'(t)=(-\sin t,\,\cos t)$$
so
$$\int_c\vec F\cdot d\vec r=\int_0^{2\pi}(\sin t,\,-\cos t)\cdot(-\sin t,\,\cos t)dt=\int_0^{2\pi}-dt=-2\pi$$
In cartesian coordinates:
With $\;y=\pm\sqrt{1-x^2}\;$ (upper and lower semicircle, resp.), we can do the parametrizations:
$$\begin{cases}r_1(x)=(x,\,\sqrt{1-x^2})\implies r_1'(x)=\left(1,\,-\frac x{\sqrt{1-x^2}}\right)\\{}\\ r_2(x)=(x,\,-\sqrt{1-x^2})\implies r_2'(x)=\left(1,\,\frac x{\sqrt{1-x^2}}\right)\end{cases}\;\;,\;\;\;-1\le x\le 1$$
and from here (be very careful with the direction!!):
$$\text{Over the upper semicircle $C_1$}: -\int_{C_1}\frac{\sqrt{1-x^2}}{x^2+y^2}dx-\frac x{x^2+y^2}dy =$$
$$=-\int_{-1}^1\left(\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}\right)\,dx=-\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}=-\int_{-\pi/2}^{\pi/2}dt=-\pi$$
I'll let you try to complete the calculation on the lower semicircle.