Using Stokes theorem to evaluate $\int_C \textbf{F} \cdot d\textbf{r}$ given $\textbf{F}(x,y,z)=2y\textbf{i}+3z\textbf{j}+x\textbf{k}$

Question

I am asked to use Stokes theorem to evaluate $$\textbf{F}(x,y,z)=2y\textbf{i}+3z\textbf{j}+x\textbf{k}$$ given that $C$ is the triangle with edges $(2,0,0)$, $(0,2,0)$ and $(0,0,2)$.

My solution

The equation for the plane is $z = 2-x-y$ and $\textbf{F} = (2y , 6-3x-3y, x)$ (here I'm using only my parameters $x$ and $y$).

Since I'm parametrizing using $x$ and $y$, I get

\begin{align*} r_x = (1,0,-1) \quad r_y = (0,1,-1) \quad \Rightarrow \quad r_x \times r_y = (1,1,1)\\ \end{align*}

\begin{align*} \int_C \textbf{F} \cdot d\textbf{r} = \iint_S (2y , 6-3x-3y, x) \cdot (1,1,1) \ dS = \int_{0}^{2} \int_{0}^{2-x} (6-2x-y) \ dydx = 8 \end{align*}

But the answer should be $-12$. Where did I go wrong?

Answer 1

You forgot to compute the curl of the vector field. Note that $\operatorname{curl}(\textbf{F})=-(3,1,2)$ and $$\int_C \textbf{F} \cdot d\textbf{r} = \iint_S \operatorname{curl}(\textbf{F})\cdot d\textbf{S}=-\iint_S (3,1,2)\cdot \frac{(1,1,1)dS}{\sqrt{3}}=-\frac{6}{\sqrt{3}}\cdot \frac{(2\sqrt{2})^2\sqrt{3}}{4}=-12.$$