3918.sagews

In this worksheet we will evaluate the line integral: $\int_C \vec{F} d\vec{s}$ where $\vec{F}=<x,x+z,z>$ and C is the closed curve defined by the intersection of the sphere $x^2+y^2+z^2=1$ and the plane $x+y=1$.

We use Stoke's theorem: $\int_C {\vec Fd\vec s}  = \iint_S{curl(\vec F)\cdot \vec n\,dS}$ where $\vec n$ is the unit vector normal to the surface S.

Here:

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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.)

It follows that:

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

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 $\vec n$.

var('x y z')
var('t')
plane1=implicit_plot3d(x+y==1,(x, -1, 1), (y, -1,1), (z, -1,1),color='blue', opacity=.3)
sphere1=implicit_plot3d(x^2+y^2+z^2==1,(x, -1, 1), (y, -1,1), (z, -1,1), color='lightgrey', opacity=.3)
circle1a=parametric_plot3d((t,1-t,sqrt(2)*sqrt(t-t^2)),(0,1), color='yellow',thickness=10)
circle1b=parametric_plot3d((t,1-t,-sqrt(2)*sqrt(t-t^2)),(0,1), color='pink',thickness=10)
n1=arrow((0,0,0),(1/sqrt(2),1/sqrt(2),0))
show(plane1+sphere1+circle1a+circle1b+n1,figsize=5)

F=plot_vector_field3d((x,x+z,z), (x,0,1), (y,0,1), (z,-1/sqrt(2),1/sqrt(2)),plot_points=[5,5,7])
plane1a=implicit_plot3d(x+y==1,(x, -1, 1), (y, -1,1), (z, -1/sqrt(2),1/sqrt(2)),color='white')
show(F+plane1a+circle1a+circle1b+n1)

Thoughts:

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 $\vec{n}$.

var('x y z'); var('t')
plane2=implicit_plot3d(x+z==1,(x, -1, 1), (y, -1,1), (z, -1,1),color='blue', opacity=.3)
circle2a=parametric_plot3d((t,sqrt(2)*sqrt(t-t^2),1-t),(0,1), color='red',thickness=10)
circle2b=parametric_plot3d((t,-sqrt(2)*sqrt(t-t^2),1-t),(0,1), color='pink',thickness=10)
n2=arrow((0,0,0),(1/sqrt(2),0,1/sqrt(2)))
show(plane2+sphere1+circle2a+circle2b+n2,figsize=5)

F2=plot_vector_field3d((x,x+z,z), (x,0,1), (y,-1/sqrt(2),1/sqrt(2)), (z,0,1),plot_points=[5,7,5])
plane2a=implicit_plot3d(x+z==1,(x, -1, 1), (y, -1/sqrt(2),1/sqrt(2)),(z,-1,1),color='white')
show(F2+sphere1+plane2a+circle2a+circle2b+n2)

No rotation in above.