# Stokes theorem

1. Aug 9, 2005

### daftjaxx1

I'm trying to solve this problem:

Compute $$\oint_c(y+z)dx + (z-x)dy + (x-y)dz$$ using Stoke's theorem, where c is the ellipse $$x(t) = asin^2t, \ y(t) = 2asintcost, z(t) = acos^2t, 0\leq t \leq \pi$$

The version of stoke's theorem I learned is:
$$\int_c \overrightarrow{F} \cdot d\overrightarrow{r} = \int_s curl \overrightarrow{F} \cdot d\overrightarrow{S} =\iint_s curl \overrightarrow{F}\cdot \overrightarrow{n} \cdot dS$$

where S is the elliptical surface bounded by the curve c, F is a vector field and n is the unit vector pointing out at that point.

In this case, $$F = <y+z, z-x, x-y>$$, and I calculated curl F to be $$<-2, 0, -2>$$.

So we have to find

$$\iint_s <-2, 0, -2> \cdot \overrightarrow{n} \cdot dS$$

How would I find $$\overrightarrow {n}$$ and dS, and also the bounds of integration for the double integral?

2. Aug 9, 2005

### HallsofIvy

Staff Emeritus
Fortunately, since the curl is a constant vector, you don't have to do any integration! Just find $$curl \overrightarrow{F}\cdot \overrightarrow{n}$$ which will be a constant, of course, and multiply by the area of the ellipse.

I don't know that this is the easiest way but here's what I did off the top of my head: When t= 0, a point on the ellipse is (0, 0, a). When t= $$\frac{\pi}{4}$$, a point on the ellipse is (a/2, a, a/2). When t= $$\frac{\pi}{2}$$, a point on the ellipse is (a, 0, 0). Since the interior of the ellipse is a plane, the vectors from one of those points to the other 2 lie in that plane and so perpendicular to any normal to the plane: Take the cross product of the two vectors to find a perpendicular (and then, of course, divide by its length to get a unit perpendicular).
Take the dot product of that unit perpendicular with the curl you have already calculated and "integrate" that over the ellipse. Since, as I said before, the integrand is a constant, that is just the constant times the area of the ellipse. It should be easy to see, by looking at the points for t= 0, t= $$\frac{\pi}{4}$$, t= $$\frac{\pi}{2}$$ , and t= $$\frac{3\pi}{4}$$ that the ellipse has semi-axes of length a and a/2. You can calculate the area of the ellipse from that.