Use Green’s Theorem to evaluate $$\int_C \langle \sin \sqrt{{1+x^3}},\ 7x \rangle \cdot d\textbf{r}$$ where $C$ is the boundary of the region $K(4)$. $K(3)$ is a triangle with $3 \cdot 4^2$ of length $\frac{1}{27}$. $K(4)$ is a Koch Snowflake, with $3 \cdot 4^3$ triangles of length $\frac{1}{81}$ added. It has dimension $\frac{\log(4)}{\log(3)}$.
First and foremost, we can calculate the curl of the vector field, which is just $7$. Then, we just have to find the area of $K(3)$ and multiply by $7$. However, I am having issues trying to find the area of this. Can anyone help?
The Koch snowflake is an equilateral triangle with side length L
$A = \frac {\sqrt {3}}{4} L^2$
plus 3 triangles with side length $\frac {L}{3}$
$+ 3\frac {\sqrt {3}}{4} (\frac {L}{3})^2$
plus 12 triangles with side length $\frac {L}{9}$
$+ 12\frac {\sqrt {3}}{4} (\frac {L}{9})^2$
plus 48 triangles with side length $\frac {L}{27}$
$+ 48\frac {\sqrt {3}}{4} (\frac {L}{27})^2$
It grows regularly after this.
$\frac {\sqrt {3}}{4} L^2(1 + \frac 13\sum_\limits{n=0}^{\infty} (\frac {4}{9})^n)$
And that is a geometric series...
$\frac {\sqrt {3}}{4} L^2(1 + \frac 13 \frac {1}{1-\frac 49})$
Actually re-reading the problem, it looks like this one is finite...
$\frac {\sqrt {3}}{4} L^2(1 + \frac 13\sum_\limits{n=0}^{3} (\frac {4}{9})^n)$
$\frac {\sqrt {3}}{4} L^2(1 + \frac 13 \frac {1-\frac {4}{9}^4}{1-\frac 49})$