INSTRUCTOR: Verify that Stokes's theorem is true for vector field F of x comma y comma z, equal to angle bracket y comma x comma negative z, end angle bracket, and surface S, where S is the upwardly oriented portion of the graph of f of x comma y, equal to x squared times y, over a triangle in the xy plane with vertices of points 0 comma 0, the point 2 comma 0, and the point a 0 comma 2.

Stokes's theorem says that the integral-- the line integral over the curve C of F times dr is equal to the double integral over the surface S of the curl of F dotted with dS. For that purpose, we need to find each of these and show that they are in fact equal.

Now, the curve C that we are concerned with is traced out by a triangle. However, this curve is composed of three paths. I'm going to call these C sub 1, that's the path from the point 0,0 to the point 2 comma 0. C sub 2, which is the path from the point 2 comma 0 to the point 0 comma 2. And to complete this, we have C sub 3, which is the path from 0 comma 2 to the point 0 comma 0.

Going to have to parameter each of these. And then, as we know, we'll sum those to find the complete line integral for this curve C. So the parameterization for C sub 1 would be r of t is equal to open angle bracket t comma 0 end angle bracket, with 0 less than or equal to t less than or equal to 2.

Curve C sub 2, will have a parameterization of r of t equals open angle bracket 2 minus t comma t end angle bracket, with 0 less than or equal to t less than or equal to 2. And C3, C sub 3 will have a parameterization of r of t equal to angle bracket 0 comma 2 minus t end angle bracket, with 0 less than or equal to t less than or equal to 2.

Convince yourself for a moment that these in fact do trace out this triangle and they do have the correct beginning and ending points. Now, let's go ahead and find the integral of over the curve C sub 1 of F dotted with dr. This is equal to the integral from 0 to 2 of open angle bracket t comma 2 minus t end angle bracket, dotted with angle bracket-- oh, actually, that's for the wrong curve.

Sorry. The integral over the curve C sub 1 of F dotted with dr is equal to the integral from 0 to 2 of open angle bracket 0 comma t end angle bracket, dotted with angle bracket 1 comma 0 end angle bracket with respect to t. Now, with this dot product, this becomes the integral from 0 to 2 with respect to t of the function 0, which is going to be equal to 0.

Now, for the integral over the curve C sub 2 of F dotted with dr, this will be equal to the integral from 0 to 2 of open angle bracket t comma 2 minus t end angle bracket, dotted with angle bracket negative 1 comma 1 end angle bracket dt.

Finding this dot product, this becomes the integral from 0 to 2 of the function 2 minus 2 t, dt. And we can find that value with the antiderivative and evaluating that is also 0.

Finally, you can find the integral over the curve C sub 3 of F dotted with dr. This would be equal to integral from 0 to 2 with respect to t of open angle bracket to minus t comma 0 end angle bracket, dotted with angle bracket 0 comma negative 1 end angle bracket.

Finding that dot product, this becomes the integral from 0 to 2 with respect to t of 0, which is equal to 0. From this, we can say the integral over the curve C of F dotted with dr is equal to 0 because we sum those three terms.

To verify Stokes's theorem, we need to show that the surface integral over the surface S of curl of F dotted with dS is also equal to 0. So the double integral over S of curl of F dotted with dS.

And let's first actually go ahead and find what the curl of that is. And actually, I think I'll save that for you. There are a few definitions.

But the curl of F is actually equal to angle bracket 0 comma 0 comma 0 end angle bracket. So that our integral is the double integral over the surface S of angle bracket 0 comma 0 comma 0 end angle bracket, dotted with dS. Which no matter what that parameterization is, this will be equal to 0. So in fact, Stokes's theorem is true for this vector field and surface.