Problem statement: Calculate the flux of $\mathbf{F}=(x-y+xy,-2x+y,xz)$ through the flat triangle with corners in $(1,0,0), (0,1,0),(0,0,1)$.

My progress:

The triangle surface Y is a part of the plane $x+y+z=1$ and a parameterization with $x,y$ as parameters is given by $\boldsymbol{r}=\boldsymbol{r}(s,t)=(s,t,1-s-t)$, $s\in \left [ 0,1 \right ], t\in \left [ 0,1 \right ]$

We know that the flow is given by $\iint_{Y}^{.} \boldsymbol{F}\bullet \boldsymbol{N}dS$ where $\boldsymbol{N}=\boldsymbol{r'_{s}}\times \boldsymbol{r'_{t}}=(1,0,-1)\times (0,1,-1)=(1,1,1)$ which is pointing outward.

Hence $\iint_{Y}^{.} \boldsymbol{F}\bullet \boldsymbol{N}dS=\iint_{D}^{.}\boldsymbol{F}\bullet (\boldsymbol{r'_{s}\times \mathbf{r'_{s}}})dsdt=\iint_{D}^{.}(s-t+st,-2s+t,s-s^2-st)\bullet (1,1,1)dsdt=\iint_{D}^{.}-s^2dsdt=\frac{1}{3}$

Correct answer is $\frac{1}{12}$

  • 2
    $\begingroup$ Do you mean flux? $\endgroup$ – MITjanitor Mar 11 '13 at 20:07
  • $\begingroup$ Yes. Sorry, not familiar with the correct translations yet. I'll edit. $\endgroup$ – EricAm Mar 11 '13 at 20:11

I think you just may have integrated with the wrong bounds, this worked for me:

$$\int_0^1\int_0^{1-t}-s^2dsdt=-\frac{1}{3}\int_0^1 (1-t)^3dt$$ then with $u=1-t$ and $du=-dt$ we get:

$$\frac{1}{12}(1-t)^4 |_{0}^1=\frac{1}{12}$$

heres a crude picture to show the region of integration once everything is parameterized:

enter image description here

  • $\begingroup$ Hmm..but why is $s\in \left [ 0,1-t \right ]$ instead of $s\in \left [ 0,1 \right ]?$ $\endgroup$ – EricAm Mar 11 '13 at 20:42
  • 2
    $\begingroup$ Because you're still integrating the triangle, and not a cube $\endgroup$ – MITjanitor Mar 11 '13 at 20:43
  • $\begingroup$ Yes I see, that makes sense actually. $\endgroup$ – EricAm Mar 11 '13 at 20:46

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