I am a bit unsure the way Stoke's theorem is applied in this case.

Evaluate $\oint\limits_C {xydx + yzdy + zxdz} $ around the triangle with vertices $(1,0,0), (0,1,0), and (0,0,1)$, oriented clockwise as seen from the point $(1,1,1)$

My reasoning

Since Stokes's theorem says that we can sum up individual microscopic rotations, i.e. we can turn a line integral into a double integral:

$$\oint\limits_C {xydx + yzdy + zxdz} = \iint\limits_D {\nabla \times \vec F \cdot \hat NdS} = - \iint\limits_D {(y,z,x) \cdot \hat NdS}$$

Then, I used parametrization $r = (u,v,1 - u)$.

The surface element turns out to be $$d{\mathbf{S}} = \frac{{\partial (x,y,z)}}{{\partial (u,v)}}dudv = (1,0,1)dudv$$

So, the integral is $$\iint\limits_{D'} {u + vdudv} = \int\limits_0^1 {\int\limits_0^{1 - v} {u + vdudv = \int\limits_0^1 {(1 - v)v + \frac{{{{(1 - v)}^2}}}{2}dv} = \frac{1}{2}\int\limits_0^1 {1 - {v^2}dv} = \frac{1}{2}\int\limits_0^1 {1 - {v^2}dv} = \frac{1}{3}} } $$

According to the answer, it should be $1/2$. Is this correct application of Stokes's theorem?

  • $\begingroup$ Your normal vector is incorrect. $\endgroup$
    – Paul
    Commented May 27, 2015 at 19:10
  • $\begingroup$ As is the parameterization of the triangle. $\endgroup$
    – Paul
    Commented May 27, 2015 at 19:14
  • $\begingroup$ @Paul, is the normal vector supposed to be $(1,1,1)$? Feels to be the case. $\endgroup$
    – Artem
    Commented May 27, 2015 at 19:20
  • $\begingroup$ I read it as (1,0,1) in your problem statememt. If I misread it I apologize. But even so the parameterization of the triangle should be (u,v,1-u-v). $\endgroup$
    – Paul
    Commented May 27, 2015 at 20:11
  • $\begingroup$ @Paul, You read it the right way. It's just that I was a bit unsure whether it would be (1,0,1) or (1,1,1). Thank you for describing the parameterization! $\endgroup$
    – Artem
    Commented May 27, 2015 at 20:39

1 Answer 1


Thanks to @Paul's comment, I was able to put all the pieces together to finally obtain the right answer.

The application of Stokes's theorem is correct, so everything is fine up to (and including): $$\oint\limits_C {xydx + yzdy + zxdz} = \iint\limits_D {\nabla \times \vec F \cdot \hat NdS} = - \iint\limits_D {(y,z,x) \cdot \hat NdS}$$

After this, the hard job is to find the right parametrization for the surface. I.e. we want to express $x,y,z$ in terms of $u,v$.

enter image description here

First, let's illustrate this in 3 space.

By projecting the pyramid like object onto $xy$ plane, we get a triangle. We could keep $x$ and $y$ as they are, i.e. by letting $x=u,\quad y=v$. Now the tricky part, the $z$-axis. If we look at this, we see a tilted triangle (from $(1,1,1)$). This is a plane with a normal $$n=(1,1,1)\quad\implies \quad 1=x+y+z$$ The final parametrization is therefore $$r=(u,v,1-u-v)$$

By finding the $$\left\| {\frac{{\partial r}}{{du}} \times \frac{{\partial r}}{{dv}}} \right\|$$ or the way I expressed in my question $$ \frac{{\partial (x,y,z)}}{{\partial (u,v)}} = (1,1,1)$$

If we perform the dot product of $\nabla \times F$ with the normal vector, we obtain $$\int_0^1 {\int_0^{1 - v} {dudv} } = \frac{1}{2}$$

Note, the reason why we have $1-v$ as the upper bound is because the domain (the body projected onto xy plane) has a boundary $1=u+v$.


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