# Flux through the boundary of a rectangle

given the rectangle described by:

$0\leq x \leq 6$, $0 \leq y \leq 2$

And a fluid following the vector field:

$F = \langle x^3+2, y \cos(6x) \rangle$

We get the 2D curl:

$curl(F) = -6y\sin(6x)-0=-6y\sin(6x)$

By green's theorem, the flux through the boundary is thus:

$\int^6_0\int^2_0 -6y\sin(6x) dydx = -4\sin^2(18)$ (by wolfram, manually I got $-2\cos(36)$)

Neither of these answers is correct and I am not sure where the problem is.

• $\int \cos(6x) dx = \frac{1}{6} \sin(6x) + C$ – infinitylord Dec 11 '17 at 3:44
• I believe the flux is related to the divergence, not the curl – Alex Pavellas Dec 11 '17 at 4:53

\begin{align} \int^6_0\int^2_0 -6y\sin(6x) dydx &= \int^6_0\sin(6x)dx\int^2_0 -6y dy \\ &= \dfrac{-\cos6x}{6}\Big|^6_0\dfrac{-6y^2}{2}\Big|^2_0 \\ &= \dfrac{1-\cos36}{6}\dfrac{-24}{2} \\ &= \color{blue}{-4\sin^218} \end{align}

• Which is not the correct answer as I said – Makogan Dec 11 '17 at 4:09
• The integral gives this answer, so your fault comes from other part! – Nosrati Dec 11 '17 at 4:11
• Hence why I described the entire problem and then gave my solution, so that people could identify what error I made – Makogan Dec 11 '17 at 4:18
• I thought you want to verify the integral answer. Anyway I will check your solution. How did you find it with wolfram? – Nosrati Dec 11 '17 at 4:21
• To simplify my query, I need the answer (and know how to get the answer) to the problem: Find the flux of $F=\langle x^3+2, ycos(6x) \rangle$ through the boundary of the rectangle $0 \leq x \leq 6$; $0 \leq 2 \leq 2$ – Makogan Dec 11 '17 at 4:26

I attempted to solve this.

$\frac{\partial F1}{\partial y}=0\ \ \ \frac{\partial F2}{\partial x}=-6ysin(6x)$

$\int^6_0 \int^0_2 -6ysin(6x) \ dydx$

$\int^6_0 -6ysin(6x) \ dx ]^2_0 = \int^6_0 -12sin(6x) \ dx$

$\frac{12}{6}cos(6x)]^6_0 = 2[cos(36)-1]$

Which is basically the same answer. What is the answer you are trying to attain?