We just learned about path integrals today, and have this on our online homework:
Let $\mathbf{y}(t)=(\cos(t),\sin(t))$ be the path defining the outside of the unit circle. Let $\mathbf{F}$ be the vector field $\mathbf{F}(x,y)=(x+e^{x^2}, -2xye^{x^2})$. Calculate the path integral of $\mathbf{F}$ along the path.
I've done path integrals before, and my first thought based on this chapter is that I need the norm of $\mathbf{F}$.
$$T_x=(2x^2e^{x^2}, -2x^2ye^{x^2})$$ $$T_y=(0, -2xe^{x^2})$$
The norm is their cross product divided by length of the cross product: $$T_x \times T_y = (0, 0, -4x^3e^{2x^2})$$ $$||T_x \times T_y||=\sqrt{(-4x^3e^{2x^2})^2}=4x^3e^{2x^2}$$ $$\mathbf{n}=\frac{(0, 0, -4x^3e^{2x^2})}{4x^3e^{2x^2}}=(0,0,-1)$$
Now is when I get confused. What exactly is the path, and where does $\mathbf{y}(t)$ come into play?