# Surface Integral (Flux?)

Evaluate the surface integral:

$$\iint_S \mathbf{\vec F} \cdot d\mathbf{\vec S}$$

for the vector field

$$\mathbf{\vec F}(x,y,z) = xze^y \mathbf{ \hat i} - xze^y \mathbf{\hat j} + z\mathbf{\hat k}$$

where $S$ is part of the plane $x + y + z = 1$ in the first octant and has a downward orientation.

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$d\mathbf{S}$ is parallel to $\left<1,1,1\right>$ so the first two components of $\mathbf{F}$ will cancel in the integral. Therefore you need only evaluate $\int_S z\mathbf{k}\cdot\,d\mathbf{S}$. –  kiwi Aug 16 '12 at 19:58
So just the integral of z...? WHat would be the limits of the integral? –  Nick Aug 16 '12 at 20:17

In other words, we have to integrate the $2$-form associated with the vector field $\vec{\mathbf{F}}(x,y,z)=[xze^y,-xze^y,z]$, namely $$\omega=xze^y\ dy\wedge dz+(-xze^y)\ dz\wedge dx+z\ dx\wedge dy.$$ In order to evaluate $\int_S \omega$, we have to choose a parametrisation of S. The natural one seems to be a good choice.
Let $\Psi\colon \{(s,t)\in\mathbb{R}^2\mid s+t\lt 1,\, s,t\gt0\}\longrightarrow S$ be a parametrisation given by $$\left\{\begin{array}{} x=s\\ y=t\\ z=1-s-t \end{array}\right. .$$ Since its Jacobian matrix $\mathrm{D}\Psi$ is of the form $$\left[\begin{array}{cc}1&0\\0&1\\-1&-1 \end{array}\right],$$ we could easily see that $\Psi$ is orientation-reversing. In fact, $$\vec{\mathfrak{n}}=\left[\begin{array}{c}1\\0\\-1\end{array}\right]\times\left[\begin{array}{c}0\\1\\-1\end{array}\right]=\left[\begin{array}{c} 0\cdot(-1)-(-1)\cdot1\\-(1\cdot(-1)-(-1)\cdot0)\\1\cdot1-0\cdot0 \end{array}\right]=\left[\begin{array}{c}1\\1\\1\end{array}\right]$$ is pointing to the inside of the surface.
Therefore, \begin{align} \iint_S \vec{\mathbf{F}}\ dS & =\int_S \omega=-\iint_{\Psi^{-1}(S)}\vec{\mathbf{F}}(\Psi(s,t))\cdot\vec{\mathfrak{n}}\ dsdt \\& =-\iint_{\Psi^{-1}(S)}(s(1-s-t)e^t\cdot1+(-s(1-s-t)e^t)\cdot1+(1-s-t)\cdot1)\ dsdt\\&=-\int_0^1\left(\int_0^{1-t}(1-s-t)\ ds\right)\ dt=-\int_0^1\frac{(1-t)^2}{2}\ dt=\left.\frac{(1-t)^3}{6}\right|_{t=0}^{1}\\&=-\frac{1}{6}. \end{align} The domain of the parametrisation (hence the limits of integration) could be obtained from rewriting $\textit{the first octant}$ condition. Actually, $0\le z=1-x-y\le1$ means that $0\le x+y\le1$. To come up with $1-t$ in upper limit of the $ds$ integral, try fixing $t$ and simply determine the possible range for $s$ in the triangle $\Psi^{-1}(S).$

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I'll give my idea here:

A normal to our surface is $\,(1,1,1)\,$ , but we're asked in downward orientation, which I translate as towards the inside of the tetrahedron determined by the plane and the three planes in the first octant, thus we need the normal $$n:=(-1,-1,-1)\Longrightarrow\,\, \stackrel \wedge n:=\frac{n}{||n||}=\frac{n}{\sqrt 3}$$ and putting $\,D:=\{(x,y)\;:\;0\leq x\leq 1\,,\,0\leq y\leq 1-x\}\,\,,\,dS=\sqrt 3\, d\mathbf A=\sqrt 3\,dxdy\,$ , and since $\,z=1-x-y\,$ , we get

$$\int\int_S\mathbf{\vec F}\cdot d\mathbf{\vec S}=\int\int_D\mathbf{\vec F}\cdot \mathbf {\stackrel {\wedge} n}\cdot d\mathbf S=\int_0^1dx\int_0^{1-x}(xze^y\,,\,-xze^y\,,\,z)\cdot (-1,-1,-1)dy=$$ $$=\int_0^1dx\int_0^{1-x}-(1-x-y)\,dy=\int_0^1\left[-(x-1)^2\left.+\frac{y^2}{2}\right|_0^{1-x}\right]=$$ $$\int_0^1\left(-(x-1)^2+\frac{(x-1)^2}{2}\right)dx=-\left.\frac{1}{6}(x-1)^3\right|^1_0=-\frac{1}{6}$$

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I believe you have a typo in the exponent in the last line, since $\int_0^1\left(-\frac{(x-1)^2}{2}\right)\ dx=-\left.\frac{1}{6}(x-1)^{\mathbf{3}}\right|^1_0=-\frac{1}{6}$. –  Kuba Helsztyński Aug 17 '12 at 20:47
Rats, you're right @KubaHelsztyński ! That typo really fooled me and, in fact, I get the same answer as you did. Thanks a lot. –  DonAntonio Aug 17 '12 at 23:22

We need only evaluate $\int\!\!\int_S z\mathbf{k}\cdot d\mathbf{S}$. Consider the tetrahedron with vertices at $(0,0,0),(1,0,0),(0,1,0),(0,0,1)$. This has four faces - the surface $S$ over which we want the integral and three others which are part of the coordinate planes. It's clear that the integral over these other three faces is zero --- on two of them $\mathbf{k}\perp d\mathbf{S}$ and on the third $z=0$. By the divergence theorem $$\int\!\!\!\int_S z\mathbf{k}\cdot d\mathbf{S} = -\int\!\!\!\int\!\!\!\int_V \text{div}(z\mathbf{k})\,dV = -\int\!\!\!\int\!\!\!\int_V dV$$ (the $-$ is because the chosen normal is inward). This last integral is just the volume of the tetrahedron $V=\frac13($area of base$\times$height$)=\frac16$.

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