Consider the vector field $\mathbb{R}^2\setminus\{0\}$,
$F = \frac{2x}{\sqrt{(x^2 + y^2)}}\hat{i} + \frac{2y}{\sqrt{(x^2 + y^2)}}\hat{j}$
Compute the integral:
$\int_{C}^{}F\cdot{dr}$
for the curve:
$x(t) = t^3 + 1, y(t) = (1 - t^2)e^{2t}$ , $t\in{[-1, 1]}$
So, the vector field is conservative, except at the point $(0,0)$ , but this curve contains the point $(0,0)$. Does that mean that I have to use brute force to do this and cannot use the fundamental theorem of line integrals?
Thanks!
You can consider the limit of the line integral as its start goes to $t=-1$. Define $r(t)=(x(t),y(t))$. Then $$\int_CF\cdot dr=\lim_{a\to-1^+}\int_a^1F(r(t))\cdot r'(t)\,dt=\lim_{a\to-1^+}(G(r(1))-G(r(a))=2\sqrt{2^2+0^2}-2\sqrt{0^2+0^2}=4$$ where $G$ is the scalar field $2\sqrt{x^2+y^2}$ whose gradient is the stated vector field.