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Let $S$ be the surface where $x + 2y + 3z = 0$ and $-1 \leq y \leq 1, 0 \leq z \leq 1$. Compute the surface integral $$\int \int_S (2x,3y-x,1-2y) \cdot \mathbf{\hat{N}}dS$$ where the unit normal vector $\mathbf{\hat{N}}$ points upwards.

My attempt: Using that $f = z = -x/3 -2y/3$ we find that $\mathbf{\hat{N}}$ points in the direction $(-f_x,-f_y,1) = (1/3,2/3,1)$. Further, by looking at the projection of $f$ onto the $xy$-plane, we set $z = 0$ and see that $y = -x/2$, which means $-2 \leq x \leq 2$ when $-1 \leq y \leq 1$as given.

We therefore obtain $\mathbf{F} \cdot \mathbf{\hat{N}}dS = 1dydx$. Further, when integrating over the region in the $xy$-plane given by $y = -x/2$ for $-2 \leq x \leq 2$ and $-1 \leq y \leq 1$ we obtain $0$, since there is as much area above and below the $y$-axis. Hence, I obtain the answer zero for the surface integral, however the correct answer is $6$.

Is anyone able to see where I'm wrong?

Pame
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  • I get $$ \mathbf{\hat N}=\frac1{\sqrt{14}}\begin{pmatrix}1\2\3\end{pmatrix},, \quad\mathbf{F}\cdot\mathbf{\hat N}=\frac{2x+4y-2x+3-6y}{\sqrt{14}}= \frac{-2y+3}{\sqrt{14}},. $$ – Kurt G. Dec 12 '23 at 08:35
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    @KurtG. Should be $6y$ instead of $4y$ which gives $3/\sqrt{14}$ – Pame Dec 12 '23 at 08:39
  • I also get zero. Why do you say the correct answer is 6? – DonAntonio Dec 12 '23 at 09:34
  • @DonAntonio It's what the solution sheet says. They parametrize the surface $S$ as $\mathbf{r}(y,z) = (-2y-3z,y,z)$ which then gives $$\mathbf{F}(\mathbf{r}) \cdot \mathbf{\hat{N}}dS = 3dzdy$$ and then integrate over $0 \leq z \leq 1, -1 \leq y \leq 1$. They used $$\mathbf{\hat{N}}dS = (1,2,3)dzdy.$$ – Pame Dec 12 '23 at 09:52

1 Answers1

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A parametrization of the surface is $$ [-1,1]\times [0,1]\ni (y,z)\mapsto \boldsymbol{\phi}(y,z)=\begin{pmatrix}-2y-3z\\y\\z\end{pmatrix}\,. $$ I have shown in this answer that $$ \boldsymbol{\phi}_y\times \boldsymbol{\phi}_z=\begin{pmatrix}-2\\1\\0\end{pmatrix}\times \begin{pmatrix}-3\\0\\1\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix} $$ is the unnormalized normal vector $\mathbf{N}$ to the surface which we use to integrate in this parametrization: \begin{align} &\int_S\mathbf{F}\cdot\mathbf{\hat N}\,dS=\int_{-1}^1\int_0^1\mathbf{F}\cdot\begin{pmatrix}1\\2\\3\end{pmatrix}\,dz\,dy=\int_{-1}^1\int_0^1(2x+6y-2x+3-6y)\,dz\,dy\\ &=\int_{-1}^1\int_0^13\,dz\,dy=6\,. \end{align}

Another parametrization of the surface is $$ [-2y-3,-2y]\times [-1,1]\ni (x,y)\mapsto \boldsymbol{\psi}(x,y)=\begin{pmatrix}x\\y\\\frac{-2y-x}3\end{pmatrix}\,. $$ Note that $(x,y)$ are not from a rectangle anymore. Therefore a slightly sloppy notation.

How to find the bounds on $x$: we want
$$ z=\frac{-2y-x}{3} $$ to be in $[0,1]\,.$ The two inequalities that are equivalent to this are $$ -2y-3\le x\le -2y\,. $$ The normal vector to the surface is now $$ \boldsymbol{\psi}_x\times \boldsymbol{\psi}_y=\begin{pmatrix}1\\0\\-\frac13\end{pmatrix}\times \begin{pmatrix}0\\1\\-\frac23\end{pmatrix}=\frac13\begin{pmatrix}1\\2\\3\end{pmatrix} $$and the integral is \begin{align} &\int_S\mathbf{F}\cdot\mathbf{\hat N}\,dS=\frac13\int_{-1}^1\int_{-2y-3}^{-2y}\mathbf{F}\cdot\begin{pmatrix}1\\2\\3\end{pmatrix}\,dx\,dy=\frac13\int_{-1}^1\int_{-2y-3}^{-2y}3\,dx\,dy=6\,. \end{align}

Kurt G.
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  • The answer is correct, but is it possible to solve the surface integral by integrating over the $xy$-plane instead? I don't see what is wrong with my approach other than that it gives an incorrect answer. – Pame Dec 13 '23 at 10:41
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    @Pame Must be possible. I do not want to be the circus horse jumping over that bar also. Instead you should write out the steps detail by detail to perform that approach. – Kurt G. Dec 13 '23 at 10:55
  • I already explained my approach in the original post. Should be enough details to follow. In any case the approach to finding the unit normal vector should be the same as in your answer, the only thing changing is the region of integration. – Pame Dec 13 '23 at 11:08
  • @Pame You messed up the region of integration. – Kurt G. Dec 13 '23 at 11:49
  • Can I ask how you found the bounds for integration in the $xy$-plane? Setting $z = 0$ I get $x = -2y$. I don't see how you get $-3+2y$ and $+2y$ as the bounds for $x$. – Pame Dec 14 '23 at 07:56
  • I had a sign error. But the reasoning and the end result are the same. – Kurt G. Dec 14 '23 at 08:13