Solving the Elliptical Integral Using Stoke's Theorem

[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:
<br /> \int_c \overrightarrow{F} \cdot d\overrightarrow{r} <br /> = \int_s curl \overrightarrow{F} \cdot d\overrightarrow{S}<br /> =\iint_s curl \overrightarrow{F}\cdot \overrightarrow{n} \cdot dS<br />

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 = &lt;y+z, z-x, x-y&gt;, and I calculated curl F to be &lt;-2, 0, -2&gt;.

So we have to find

\iint_s &lt;-2, 0, -2&gt; \cdot \overrightarrow{n} \cdot dS

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

 

[HallsofIvy]

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.

 

[M98Ranger]

To solve this problem using Stoke's theorem, we first need to find the normal vector n and the surface element dS for the elliptical surface S. The normal vector n can be found by taking the cross product of the two tangent vectors to the ellipse, which can be calculated using the parametric equations given for x(t), y(t), and z(t). The surface element dS can be calculated using the formula dS = ||\overrightarrow{T_1} \times \overrightarrow{T_2}|| dt, where \overrightarrow{T_1} and \overrightarrow{T_2} are the tangent vectors at a given point on the ellipse and dt is the infinitesimal element of arc length along the ellipse.

Once we have found n and dS, we can then evaluate the double integral using the bounds of integration 0 \leq t \leq \pi. This will give us the value of the surface integral, which is equal to the line integral along the curve c by Stoke's theorem. We can then solve for the line integral by setting up the parametric equations for x(t), y(t), and z(t) and evaluating the integral over the given range of t. This will give us the final solution to the problem.

In summary, to solve this problem using Stoke's theorem, we need to first find the normal vector and surface element for the elliptical surface, then evaluate the double integral over the given bounds to find the value of the surface integral, and finally use this value to evaluate the line integral along the curve c. This approach may be more efficient and simpler than directly evaluating the line integral, especially for more complex curves and surfaces.