0

Let $S$ denote the plane surface whose boundary is the triangle with vertices $(1,0,0), (0, 1, 0), (0, 0, 1),$ and let $F(x, y, z) = xi + yj + zk$. Let $n$ be the unit normal to S where $z \ge 0$. Evaluate the integral $\iint_S F \cdot dS$ using $r(u, v) = (u + v)i + (u − v)j + (1 − 2u)k.$

I am trying to find the bounds on $u$ and $v$. I get that $0 \le u \le \frac 12$ from the fact that $0 \le z \le 1,$ but I am having trouble finding the strictest bounds on $v$ in terms of $u$. I am also confused about how to determine whether the bounds I find are the strictest possible.

Euclid
  • 45
  • 7

1 Answers1

1

$$\begin{align*} 0\le z\le1&\implies0\le1-x-y\le1\\[1ex] &\implies-1\le x+y-1\le0\\[1ex] &\implies0\le x+y\le1 \end{align*}$$

On the surface $S$, you know that both $0\le x\le1$ and $0\le y\le 1$, so you get that

$$\begin{align*} 0\le x\le x+y&\implies0\le x\le1-(1-x-y)\\[1ex] &\implies0\le x\le 2u\\[1ex] &\implies0\le u+v\le2u\\[1ex] &\implies-u\le v\le u \end{align*}$$

So with $0\le u\le1$, you need to have $v$ range from $-u$ to $u$. Then take the normal vector to $S$ to be

$$\vec r_v\times\vec r_u=2\,\vec\imath+2\,\vec\jmath+2\,\vec k$$

so the integral is

$$\iint_S\vec F\cdot\mathrm d\vec S=2\int_0^{1/2}\int_{-u}^u\bigg((u+v)\,\vec\imath+(u-v)\,\vec\jmath+(1-2u)\,\vec k\bigg)\cdot\bigg(\vec\imath+\vec\jmath+\vec k\bigg)\,\mathrm dv\,\mathrm du=2$$

user170231
  • 19,334