Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

-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.

the problem

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

share|cite|improve this question
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
up vote 2 down vote accepted

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

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

share|cite|improve this answer
THANKS! i really couldn't thanks you enough. – adsisco Nov 17 '11 at 14:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.