Question Compute $\int \limits_{C} F\cdot dr$ for $F(x,y)=(y,x)$ and $C$ is the curve given by $r=1+\theta$ for $\theta \in [0,2\pi]$.
My Attempt
$F$ is conservative and a potential function is $f=xy$. First convert into polar coordinates, $f=r^2\sin(\theta)\cos(\theta)$
Start point: $(r, \theta) = (1,0)$
End point: $(r, \theta) = (1+2\pi,2\pi)$ Therefore by the properties of conservative vector fields:
$$f(b) - f(a) = (1+2\pi)^2\sin(2\pi)\cos(2\pi) - (1)^2\sin(0)\cos(0) = 0 - 0 = 0. $$
Is this the correct method and answer? Thank you!
$\endgroup$ 31 Answer
$\begingroup$More generally: if the field is conservative and the potential is the same in the endpoints of the curve, the integral will be zero. Let be $\gamma:[a,b]\longrightarrow\Bbb R^2$ a parametrization of $C$: $$\int_C F\cdot dr = \int_C \nabla f\cdot dr = f(\gamma(b)) - f(\gamma(a)) = 0.$$
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