Question: Find a simple closed curve C with counterclockwise orientation that maximizes the value of $$\int_C\frac{1}{3}y^3dx+\left(x-\frac{1}{3}x^3\right)dy$$ and explain your reasoning.
My approach: First I check the vector field as it was a conservative field or not. Because if it is then we have path-independence.
\begin{align*}
\mathbf{F}=\frac{1}{3}y^3\vec{i}+\left(x-\frac{1}{3}x^3\right)\vec{j}\\
f(x,y)=\frac{1}{3}y^3 \text{ and } g(x,y)=\left(x-\frac{1}{3}x^3\right)\\
f_y(x,y)=y^2 \text{ and } g_x(x,y)=1-x^2\\
\end{align*}
$f_y(x,y)\ne g_x(x,y)$. Hence, $\mathbf{F}$ isn't conservative field.
Because $f_y(x,y)=g_x(x,y) \implies\nabla\times\mathbf{F}=0\implies \mathbf{F}\text{ is conservative }$.
So now I stuck how to find the curve and Is there any general idea$(1)$ to solve similar kind of question$?$ and If the orientation$(2)$ isn't told then how to determine which orientation give us maximum value.
Thanks for your time.
Thanks in advance .
Best Answer
Use Green's theorem: $$\int_{\partial D} f(x,y)dx+g(x,y)dy=\int_D (g_x-f_y)\>{\rm d}(x,y)=\int_D(1-x^2-y^2)\>{\rm d}(x,y)\ .\tag{1}$$This is valid for any reasonable domain $D\subset{\mathbb R}^2$ and its boundary $\partial D$. Now we should choose $D$ such that the RHS of $(1)$ becomes maximal. This is the case when $D$ includes all points $(x,y)$ for which $1-x^2-y^2\geq0$, hence $D$ should be the unit disc. It follows that the maximal value of your path integral is given by $$2\pi\int_0^1(1-r^2)\>r\>dr={\pi\over2}\ .$$