$$ \ F(x,y) = ( e^{x+y^2}, \ 2y \ e^{x+y^2} ) \ $$
$$ C: x^{\frac{2}{3}} + y^{\frac{2}{3}} =1 \text{ from } (1,0) \text{ to } (-1,0) $$
Then $ C: r(t)=( \cos^3(t),\sin^3(t) \ ) \text{ with } 0\leq t \leq \pi $
$$ r'(t) = ( \ -3 \sin(t) \cos^2(t), \ 3 \sin^2(t) \cos(t) \ ) $$
$$ F(r(t)) = F( \cos^3(t), \sin^3(t) \ ) = ( \ e^{\cos^3(t)+\sin^6(t)} , 2 \ \sin^3(t) \ e^{\cos^3(t)+\sin^6(t)} ) $$
$$ \langle F(r(t)) \ , \ r'(t) \rangle \ = \ e^{\cos^3(t)+\sin^6(t)} [ \ -3 \sin(t) \cos^2(t) + 6 \sin^5(t) \cos(t) \ ] $$
$$\int_C \langle F(r(t)) \ , \ r'(t) \rangle> \, dt = e^{-1} - e $$