# stokes theorem application

-edit, i really need to know which part i got wrong as i'm having trouble proceeding, please help me, thanks-

Applying Stokes' Theorem, evaluate the integral

Curve C is the intersection of the boundary surface of the cube $0≤x,y,z≤a$, with the plane $x+y+z=3a/2$ where a = 1.17. The contour C is oriented counterclockwise if seen from the positive direction of the Ox-axes.

my workings:

Let S be the hexagon that lies in the cube 0<=x,y,z <=a and on the plane x + y + z = 3a/2

then we can parameterize it using x,y obtaining z = 3a/2 - x - y.

normal vector to plane = | {i j k}, {1 0 -1}, {0 1 -1} | = i + j + k

curl F = |{i j k}, {dx dy dz}, {$y^{2}-z^{2}$ $z^{2}-x^{2}$ $x^{2}-y^{2}$}| = ($-2y-2z$)i + ($-2z-2x$)j + ($-2x-2y$)k

using stroke's theorem

the integral = $\int \int_{S}$ curl F.N dS

= $\int \int_{0<=x,y<=a} -4y - 4x - 4(\frac{3a}{2} - x - y) dx dy$

=$\int \int_{0<=x,y<=a} 6a$ dx dy = 6a $\int \int_{0<=x,y<=a}1 dx dy$ = 6a * area = 6a^3.

which part of my workings is wrong? thanks for looking through

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How do you know this answer is wrong? The only thing that jumps out at me is the sign error in the last step. –  user7530 Nov 17 '11 at 8:43
because i have the final numerical answer and its different from mine. i'm really at a lost here. you're right, i think my orientation is wrong, but that only changes the sign. –  adsisco Nov 17 '11 at 10:56
The projection of the hexagon to the $(x,y)$-plane is not the full square, but only ${3\over4}$ of it. See my answer below. –  Christian Blatter Nov 17 '11 at 12:59

Given a force field ${\bf F}=(P,Q,R)$ and the oriented boundary $\partial S$ of a piece of surface $S$ Stokes' theorem says that

$$W:=\int_{\partial S}\ {\bf F}\cdot d{\bf x}\ =\ \int_S\ {\bf rot\ F}\cdot{\bf n}\ {\rm d}\omega\ ,$$

where ${\rm d}\omega$ denotes the scalar surface element. In your case we have

$${\bf rot\ F}\cdot{\bf n}=(-2y-2z, -2z-2x,-2x-2y)\cdot\Bigl({1\over\sqrt{3}},{1\over\sqrt{3}},{1\over\sqrt{3}}\Bigl)=-{4\over\sqrt{3}}(x+y+z)=-2\sqrt{3}\ a$$

at all points of $S$, whence $W=-2\sqrt{3} a\ \omega(S)$. Now $S$ is a regular hexagon with edge length $a/\sqrt{2}$. Therefore $\omega(S)=6\cdot{\sqrt{3}\over 8}a^2$, so that we get definitively

$$W=-2\sqrt{3}a\cdot{3\sqrt{3}\over 4}a^2=-{9\over2}\ a^3\ .$$

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THANKS! i really couldn't thanks you enough. –  adsisco Nov 17 '11 at 14:07