Please help with Stoke's Theorem?

Question

Use Stoke's Theorem to evaluate the integral F.dr where F(x,y,z) = 3z i + 5x j - 2y k and is the ellipse in which the plane z=y+3 intersects the cylinder x^2 + y^2 = 1 and the ellipse is oriented counterclockwise as viewed from the above.

Any help would be greatly appreciated.

Anyone? Please?

Answer 1

In order to use Stokes' theorem, we convert the line integral of the field around the curve into the surface integral of the curl of the field over the surface bounded by the curve. First, we need to find the curl of the vector field:

curl (F) = ∇×F =

|i ........ j ....... k|
|∂/∂x ∂/∂y ∂/∂z|
|3z ... 5x .. -2y|

= <-2, 3, 5>

So now me must find the unit vector normal to the surface. Since the ellipse is bounded entirely within the plane -y+z=3, we know the normal vector to that ellipse is simply the vector normal to the plane, which is <0, -1, 1>. Note that this vector points upwards, which is good because the curve is positively oriented, so we will obtain the correct sign. Of course, there is a small problem, in that we need a unit normal vector, and this is not a unit vector. We fix this by dividing this vector by its magnitude, which is √2. So we have:

∬curl (F) · n dS = ∬<-2, 3, 5>·<0, -1, 1>/√2 dS = ∬2/√2 dS

Much nicer than that ugly line integral. Now, we must find dS. Since the surface is given by a function of x and y (namely, z=y+3), we can use the formula that dS = √((∂z/∂x)² + (∂z/∂y)² + 1) dA. This gives us that dS = √2 dA. So substituting:

∬2 dA

And now all that remains is to find the region of integration. Since the projection of the surface onto the xy-plane is the region bounded by the circle x²+y²=1, this means (switching to polar coordinates) that the region of integration is θ∈[0, 2π], r∈[0, 1]. So we have:

[0, 2π]∫ [0, 1]∫2r dr dθ

And this integrates almost trivially as:

2π