Stoke's Theorem - Example 1

curl(\vec F) =\left|  {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ x&{x + z}&z \end{array}} \right|= <-1,0,1> and

C is the intersection of the sphere and the plane. So C is a circle and S is the disk within C. 

So \vec n is the unit vector normal to the plane x+y=1. This means \vec n=\frac{1\vec i+1\vec j+0\vec k}{\sqrt{1^2+1^2+0^2}}=<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0>

(Notice that with this, we have assumed that the \vec n is pointing into the x0y plane, that is towards the smaller part of the "cut" sphere.)

\int_C {\vec Fd\vec s}=\iint_S{(\frac{-1}{\sqrt{2}}+0 +0)dS} =(\frac{-1}{\sqrt{2}}+0 +0)\iint_S{dS}=-\frac{1}{\sqrt{2}} \cdot \text{Area of S}

Area of S = Area of the Circle = \pi \cdot r^2

So we need the radius of C. We pick an arbitrary point on the circle, say (1,0,0). We need the center point O of C.

The sphere is centered at (0,0,0). The plane is symmetric with respect to x and y and normal to the plane x0y. So the x- and y-coordinates of O are equal and the z-coordinate of O is 0. O=(\frac{1}{2},\frac{1}{2},0) and r^2=\frac{1}{2}

\int_C {\vec Fd\vec s} =\frac{-1}{\sqrt{2}} \cdot \pi \cdot \frac{1}{2}=\frac{-\pi}{4\sqrt{2}}. This means a particle being carried by \vec F is being negatively rotated through S about \vec n.

If we spin the 3d plot, we can see that \vec F has rotation through S about \vec{n}. (I would like to be able to draw only the vectors of F on the disk itself to make this clearer.)

If the plane is x+z=1, then the \int_C {\vec Fd\vec s}=\iint_S{(\frac{-1}{\sqrt{2}}+0 +\frac{1}{\sqrt{2}})dS} =\iint_S{0\,dS}=0 so a particle being carried by the vector field F is not being rotated through this S about this \vec{n}.