Problem:
Use Stokes’ theorem to evaluate the integral
$I = \int\limits_C \textbf{F} \centerdot \textbf{ds}$
when $\textbf{F}$ is the vector field
$\textbf{F} = 3zx\textbf{i} + 3xy\textbf{j} + yz\textbf{k}$
and C is the path consisting of the three edges
of the triangle $∆ABC$ shown in
formed by the portion of the plane
$x + y + z = 1$
in the first octant of 3-space, oriented as
shown.
Attempt at Solution:
I know that by Stokes' theorem, $\int\limits_C \textbf{F} \centerdot \textbf{ds} = \iint\limits_S (\nabla \times \textbf{F}) \centerdot \textbf{dS}$, where $S$ is the surface closed by $C$. So, evaluating $\nabla \times \textbf{F}$, I found it to be equal to $z\textbf{i} + 3x\textbf{j} + 3y\textbf{k}$. Substituting $z = 1 – x – y$, $\nabla \times \textbf{F} = (1 – x – y)\textbf{i} + 3x\textbf{j} + 3y\textbf{k}$.
Next, to find $\textbf{dS}$, I took the gradient of $x + y + z = 1$, which turned out to be equal to $<1, 1, 1>$. Thus, $(\nabla \times \textbf{F}) \centerdot \textbf{dS} = 1 + 2x + 2y$.
For the limits of integration, I used the projection of the surface onto the $xy$-plane, which produced the image of a triangle bounded by $x = 0$, $y = 0$, and $y = -x$. Accordingly, I chose the limits of integration as follows: $\int_0^1 \int_0^{-x} (1 + 2x + 2y)dydx$. After performing integration, I found the result to be $-\frac{5}{6}$. However, this answer was incorrect. Where did I go wrong?
Best Answer
As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $\int_0^1\int_0^{1-x} (1 + 2x + 2y)dydx = \frac{7}{6}$. Thank you!