Could anyone tell me how to show the (a) following line integral is of independent of the path?
$$I= \int_C[(3y^2+\pi^3\cos x)i+(6xy-3\pi^3\sin y)j]\cdot dr,$$ where $C$ is a curve connecting the point $P(-\frac{\pi}{2}, \pi)$ to $Q(\pi,\pi)$
(b) How to find a potential function and use that to find line integral?
(c) Choose the integration path $C$ to be formed from the straight line segments $C_1$ and $C_2$ where $C_1$ joins $P(-\frac{\pi}{2}, \pi)$ to $Q(\pi,\pi)$ and $C_2$ joins $Q(\pi,\pi)$ to $R(\pi,\frac{\pi}{2})$
parametrise $C_1$ and $C_2$ and evaluate the integral $I$.
Thanks so much for the help!
Best Answer
You have to find a potential function $\phi$ such that $\nabla \phi = (3y^{2}+\pi^{3}\cos(x),6xy-3\pi^{3}\sin(y))$. This implies $$\frac{\partial \phi}{\partial x} = 3y^{2}+\pi^{3}\cos(x) \Rightarrow \phi(x,y) = 3y^{2}x+\pi^{3}\sin(x)+c(y)$$ where $c(y)$ is a differentiable function which might depend only on the $y$-variable, since we integrated on $x$. Now, we expect that $$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(3y^{2}x+\pi^{3}\sin(x)+c(y)) = 6xy-3\pi^{3}\sin(y)\Rightarrow c'(y) = -3\pi^{3}\sin(y) \Rightarrow c(y) = 3\pi^{3}\cos(y) + K $$ with $K$ being a constant of integration. Thus, we conclude $\phi(x,y) = 3y^{2}x+\pi^{3}\sin(x)+3\pi^{3}\cos(y)+K$. Now, to your second question, use your own integral together with the parametrisations $-\frac{\pi}{2}\le x \le \pi$ and $y=\pi$ for $C_{1}$ and $x= \pi$ and $\pi \le y \le \frac{\pi}{2}$ for $C_{2}$.