If C is any simple closed piecewise-smooth curve in $\mathbb{R}^3$, what is the value of the line integral?

Question

If C is any simple closed piecewise-smooth curve in $\mathbb{R}^3$, what is the value of the line integral $\oint\limits_{C} x^4 dx +y^4 dy+z^4 dz$

What I know so far:

I think the answer is $0$ because the integral around any close path is $0$.

Answer 1

Consider the function $f(x,y,z):={1\over5}(x^5+y^5+z^5)$ with gradient $$\nabla f(x,y,z)=(x^4,y^4,z^4)\ .$$ It follows that for any smooth curve $\gamma$ that connects the point ${\bf r}_0$ with ${\bf r}_1$ one has $$\int_\gamma x^4\>dx+y^4\>dy+z^4\>dz=\int_\gamma\nabla f({\bf r})\cdot d{\bf r}=f({\bf r}_1)-f({\bf r}_0)\ .$$ A piecewise smooth closed curve $C$ is a finite sum of such $\gamma$, with pairwise matching start and end points. It follows that $$\int_C x^4\>dx+y^4\>dy+z^4\>dz=0\ .$$