I have a vector field
$$F(x,y) = \langle x^2 + y^2, 2xy\rangle$$
which is conservative. I also have a curve $C$ defined as the boundary of the region between the two curves $y = x+2$ and $y = x^2$. Since $F$ is conservative, I know that the circulation of $F$ over $C$ must be zero. However, if I wanted arrive at this conclusion using strict computation (without Green's theorem), I could try the following.
Break $C$ into pieces using curves $C_1$ and $C_2$; here, $C_1$ has the parameterization $r(t) = \langle t, t^2\rangle$ where $t$ flows from $-1$ to $2$, and $C_2$ has the parameterization $s(t) = \langle t, t+2\rangle$ where $t$ flows from $2$ to $-1$. I'll compute the circulation as:
$$\oint_C F = \int_{-1}^2 [F(r(t))\cdot r'(t)]~dt + \int_2^{-1} [F(s(t))\cdot s'(t)]~dt.$$
I find that evaluating $F$ at these parameterizations results in
$$F(r(t)) = \langle t^2 + t^4, 2t^3\rangle; \quad F(s(t)) = \langle 2t^2 + 2t + 4, 2t^2 + 4t\rangle.$$
Since the derivative of my parameterizations are $r'(t) = \langle 1,2t\rangle$ and $s'(t) = \langle 1, 1\rangle$, I can compute the corresponding dot products and I get the expression
$$\oint_C F = \int_{-1}^2 (5t^4 + t^2)~dt - \int_{-1}^2 (4t^2 + 6t + 4)~dt = \int_{-1}^2 (5t^4 - 3t^2 - 6t - 4)~dt.$$
Curiously enough, this evaluates to $3$ and not $0$. What am I doing wrong here?
