Flux integral over triangle

Question

Compute the flux of the field $$ F=(x-y+xy, \, -2x+y, \, xz)$$ over the plane triangle with cornerns in $(1,0,0), \, (0,1,0), \, (0,0,1)$.

My method: $\\$

Parametrize the triangle by $r=(x, \, y, \, 1-x-y), \, (x,y) \in D$ where $D: 0 \leq x \leq 1, \, 0 \leq y \leq 1$. Then $r'_x = (1,0,-1), \, r'_y=(0,1,-1)$ and we get the normal vector of the triangle by the cross product $$r'_x \times r'_y = (1,1,1)$$ In my naive understanding, this should be divided by $|r'_x \times r'_y| = \sqrt{3}$. We should be ready to easily compute the double integral given by $$ \iint_{D} F \cdot \frac{r'_x \times r'_y}{|r'_x \times r'_y|} dxdy $$ We may express $F$ in terms of only $x$ and $y$ as $$F=(x-y+xy, \, -2x+y, \, x(1-x-y)$$ and the dot product is calculated to $$(x-y+xy) + (-2x+y) + (x(1-x-y)) = x-y+xy-2x+y+x-x^2-xy = -x^2$$. So the double integral is $$\frac{1}{\sqrt{3}} \int_{y=0}^1 (\int_{x=0}^1 - x^2 dx) dy = -\frac{1}{3\sqrt{3}}$$ This is very wrong though. The correct answer is $1/12$ downwards in the direction of the $z$-axis. Where did I go wrong? How can I do it correctly?

Answer 1

You forgot to compute $dS$ ; since the transformation to the $xy$-triangle to the parametrized triangle is linear, this will change the answer by a factor. Also, you integrated over a square ($0 \le x,y \le 1$), not a triangle. That doesn't help either! I suggest you correct that and comment if you need more help.

Hope that helps,

Answer 2

The given triangle surface can be parametrized by $$S: \vec r=\langle x, y, 1-x-y\rangle,\,\,\,0\leq y\leq 1-x,\, 0\leq x\leq 1.$$ The scalar surface area element $dS$ is given in the comments by Lozansky. Here, $\vec r_x=\langle 1, 0, -1\rangle$ and $\vec r_y=\langle 0, 1, -1\rangle$. Hence, the vectoral surface are element is $$d\vec S=\vec r_x\times \vec r_y\, dxdy=\langle 1,1,1\rangle dxdy.$$ Then, the flux of $F=\langle x-y+xy, \, -2x+y, \, xz\rangle$ through $S$ can be computed by $$\int\int_S\vec F\cdot d\vec S=\int_0^1\int_0^{1-x}(-x^2)dydx=\int_0^1(x^3-x^2)dx=\frac14-\frac13=-\frac1{12}.$$ The sign of the answer depends on the orientation. If we define the surface area element by $d\vec S=\vec r_y\times \vec r_x\, dxdy$ then the answer will be positive.