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I'm a bit lost as to how to work through this problem. I started out by taking the curl of $yi+2j$ and got a result of $-\hat{k}$ normalizes to $-1$. I decided to use Stokes theorem $$\int_{\partial \sigma} \vec{V} \cdot d \vec{l}.$$ But I'm confused as to what $\sigma$ signifies; I read that the first octant is where $x,y,z$ are all positive so I'm assuming that just means what's in the diagram, but I don't know where to go with this problem or what to do with the plane and triangles. If someone could help set me up with the directions on how to work out the problem I can do the calculations, I just need help knowing what is being asked.

Mark Fantini
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anonymous
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  • $\sigma$ is the surface to which you are applying Stokes' theorem. $\partial\sigma$ is the boundary of that surface. So in your case $\partial\sigma$ would be the triangle in the $xy$-plane consisting of part of the $x$-axis, $y$-axis, and the intersection of your plane with the $xy$-plane. – Bill Cook Apr 28 '14 at 01:44

2 Answers2

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Here sigma is the 2-dimensional surface you have drawn with an orientation, that is, an up and down side which the normal vector n describes. One the line integral side of Stokes theorem one integrates around the boundary of sigma, traversing it so that your upright position is in agreement with n, that is, one walks around the boundary of sigma so that one's head is in the direction of n, while the surface is constantly on your left. This gives the direction of the line integrals.

In the case above, if n is pointed outward to the plane shown then the line integrals will be 1) from (6,0,0) to (0,4,0) and then 2) from (0,4,0) to (0,0,3) and finally 3) from (0,0,3) back to (6,0,0).

Also remember that a simple parametrization of the line from A to B is A + t (B - A), 0 <= t <= 1, so for example one can parametrize the line from (6,0,0) to (0,4,0) by (6,0,0) + t(-6,4,0) = (6 - 6 t, 4 t, 0 ) = (x, y, z). That is,

x = 6 - 6 t, y = 4 t, z = 0.

This should be enough for you to carry out the calculations required.

berkeleychocolate
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$\sigma$ consists of the triangle in the tilted plane shown, together with the triangles in the $xz$- and $yz$-planes with one edge on the tilted plane and the other two edges along the coordinate axes. If we included the triangle $$T = \{2x+3y\le 12, \quad x,y\ge 0\}$$ in the $xy$-plane as well, we'd have a closed surface. The textbook is a bit remiss here, as they should specify which direction $\mathbf n$ is pointing. Perhaps they've made a universal convention that it's always "outward," where we're supposed to understand here that that means that $\mathbf n$ has a positive $z$-component along the tilted plane.

One way to apply Stokes's Theorem is to observe that if we likewise orient $T$ with $\mathbf n = \mathbf k$, then $\partial T = \partial\sigma$. It follows then that $$\iint_\sigma \text{curl}\,\mathbf F\cdot\mathbf n\,d\sigma = \int_{\partial\sigma} \mathbf F\cdot d\mathbf r = \int_{\partial T} \mathbf F\cdot d\mathbf r = \iint_T \text{curl}\,\mathbf F\cdot\mathbf n\,d\sigma\,.$$ But this last integral is super easy!

Ted Shifrin
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