Compute $\int \limits_{C} F.dr$ for $F(x,y)=(y,x)$ and $C$ is the curve given by $r=1+\theta$ for $\theta \in [0,2\pi]$
My Attempt
Am I correct in saying that $F$ is a conservative vector field therefore path independent. Using this property I can calculate the integral by parameterising the line segment from $(1,0)$ to $(1+2\pi,0)$ and "ignoring" the curve the question gives me, as these are the start and end points of the curve.
$x = r\cos(\theta), y = r\sin(\theta)$.
Then, $\theta = 0 \Rightarrow r = 1,$ and $\theta = 2\pi \Rightarrow r= 1 +2\pi$.
Then, the integral is $f(2\pi,1+2\pi) – f(0,1)$, where $f = r^2sin\theta cos\theta$ (which is the potential function)
Therefore the answer is 0. Is this correct?
Best Answer
You are correct that $F$ is conservative with potential $f(x,y) = xy$, which in polar coordinates can also be written as $\tilde{f}(\theta,r)=r^2\sin\theta\cos\theta$. So, then $\int_C F\cdot dr$ does only depend on the start and end points. I suppose if you wanted to be very clear, you could show that the function you give is the potential of $F$, which is easier to do in Cartesian coordinates.