Let $F=(xy^2, 3z-xy^2, 4y-x^2y)$ . Find the maximum value of the line integral of F over a simply closed curve $C$ in the plane $x+y+z=1$. What is the curve that maximises it?
I am a bit confused on how to approach this question. I tried parameterising $C$ which clearly didn't help as I needed two parameters to do so.
I then tried using Stokes theorem. I took out the dot product of the curl and the normal vector which came out to be $1-x^2+2x-y^2-2xy $ but I don't know how to proceed further. What substitution would we make?
This question is actually solvable using Stoke's theorem. You made a mistake when calculating the curl.
The curl should be $<1-x^2, 2xy, -y^2-2xy>$. Considering the plane $x+y+z=1$, we take $<\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$ as the unit normal vector.
Then, $$\oint_{C}{\vec{F} \cdot d\vec{r}} = \iint_{S}{(\triangledown\times \vec{F})\cdot \hat{n} dS} = \iint_{S}{\frac{1-x^2-y^2}{\sqrt{3}}dS}$$
There's another post talking about maximizing a line integral. You may check for more ideas here. Basically the idea is using polar coordinates and come to a conclusion that $C$ should be a circle with radius 1. Hope this helps!