Line Integral of Vector Calculus

36 Views Asked by At

this question comes from my practice exam for vector calculus and I just have no clue on how to do it. Can somebody provide me with some hints on how to do it? Thanks! Here is the link to the screenshot of the questionline integral

1

There are 1 best solutions below

0
On BEST ANSWER

First observe the vector field \begin{align} \vec{F}(x, y) = e^{\frac{1}{2}\cos^3(x+y)}\vec{i}+e^{\frac{1}{2}\cos^3(x+y)}\vec{j} =: M(x, y)\vec{i}+N(x, y)\vec{j} \end{align} is globally defined on the plane. Next, observe $\partial_y M = \partial_x N$ which means $\vec{F}$ is a conservative vector field. Hence there exists $f:\mathbb{R}^2\rightarrow \mathbb{R}$ such that $\vec{F} =\nabla f$. More importantly, since \begin{align} \frac{\partial f}{\partial x}-\frac{\partial f}{\partial y} = 0 \end{align} then we see that $f(x, y) = g(x+y)$ for some $g:\mathbb{R}\rightarrow \mathbb{R}$.

Finally, by the fundamental theorem of line integral, we get that \begin{align} \int_C \vec{F}\cdot d\vec{r} = f(1, -1)-f(0, 0) = g(1-1)-g(0+0) = 0. \end{align} Likewise, for the other curve we see that \begin{align} \int_C \vec{F}\cdot d\vec{r} = f(-\pi, 2\pi)-f(2\pi, -\pi)= g(-\pi+2\pi)-g(2\pi-\pi) = 0. \end{align}