Consider: $$\overrightarrow { F } (x,y)=\left( \frac { -y }{ x²+y² } ,\frac { x }{ x²+y² } \right) $$ Show that $$\oint { \overrightarrow { F } \cdot \overrightarrow { dr } =2\pi } $$ For every closed simple curve C positively oriented around the origin
How i did:
$$\begin{align}\oint { \frac { -ydx }{ x²+y² } +\frac { xdy }{ x²+y² } } =\\\iint { \left( \frac { x²+y²-x(2x) }{ (x²+y²)² } \right) -\left( \frac { -(x²+y²)-(-y)(2y) }{ (x²+y²)² } \right) } dA=\\2\iint { \frac { x²dA }{ (x²+y²)² } } =\\2\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ r }{ \frac { r²cos²(\theta )rdrd\theta }{ { r }^{ 4 } } } } =2\int _{ 0 }^{ 2\pi }{ \int _{ 0 }^{ r }{ \frac { cos²(\theta ) }{ r } drd\theta }=2\pi* ln(r) } \end{align}$$
Questions: 1) I did the conversion from retangular to polar right? 2) Can I consider a closed simple curve with the integration limits: $$0<\theta <2\pi \\0<r<r$$ Obs: The answer should be 2pi instead 2*pi*lnr
Assuming you meant "closed simple curve around the origin": we can assume $\;C\;$ is a canonical circle with positive orientation (why?) , so $\;C:\;\;(r\cos t,\,r\sin t)\;,\;\;0\le t\le 2\pi\;$ , and
$$\int_C\vec F\cdot d\vec r=\int_0^{2\pi}\left(-\frac{r\sin t}{r^2}\,,\;\frac{r\cos t}{r^2}\right)\cdot\left(-r\sin t,\,r\cos t\right)=$$
$$=\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt=2\pi$$
If the curve passes through the origin then the integral isn't well defined (it doesn't exist), and if the curve doesn't enclose the origin then the integral is zero as $\;\vec F\;$ is a conservative vector field (why?)