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\centerline
{
\bf Solution to the ``QUIZ'' for Lecture 24
}

\bigskip


By using Stokes' Theorem, or otherwise, evaluate 
$\int_C {\bf F} \cdot d {\bf r}$, where
$$
F(x,y,z)= (yz+2y+3z){\bf i} + (xz+2x+4z) {\bf j} + (xy+3x+4y) {\bf k} \quad ,
$$
where $C$ is the curve of intersection of the plane $x+y+z=1$
and the cylinder $x^2+y^2=1$, oriented counterclockwise as viewed
from above. Be sure to explain everything.


{\bf Sol.}
Stokes's theorem says
$$
\int_C {\bf F} \cdot d{\bf r}= \int\int_S curl({\bf F}) \cdot d \, {\bf S} \quad,
$$
where $S$ is {\it any} surface that bounds $C$. This would be very hard to compute {\it unless}
we are lucky and $curl {\bf F}$ would happen to be the zero vector $\langle \, 0,0,0 \, \rangle$.
In that case we would be done, since if the vector field that is being integrated is the {\bf zero}
vector-field, and answer is auomatically $0$!

So let's compute $curl ({\bf F})$:

$$
curl( {\bf F})=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
yz+2y+3z & xz+2x+4z & xy+3x+4y 
} \right \vert
$$
$$
{\bf i} \left ( {{\partial} \over {\partial y}} (xy+3x+4y) - {{\partial} \over {\partial z}} (xz+2x+4z) \right )
$$
$$
-{\bf j} \left ( {{\partial} \over {\partial x}} (xy+3x+4y) - {{\partial} \over {\partial z}} (yz+2y+3z) \right )
$$
$$
+{\bf k} \left ( {{\partial} \over {\partial x}} (xz+2x+4z) - {{\partial} \over {\partial y}} (yz+2y+3z) \right )
$$
$$
{\bf i} \left ( x+4 - (x+4) \right )
-{\bf j} \left ( y+3 - (y+3) \right )
+{\bf k} \left ( z+2 - (z+2) \right )
=
{\bf i} 0
-{\bf j} 0
+{\bf k} 0 =
\langle 0, 0, 0 \rangle  \quad .
$$
So we are lucky and the curl is indeed the zero-vector and we have
$$
\int_C {\bf F} \cdot d{\bf r}= \int\int_S curl({\bf F}) \cdot d \, {\bf S} =
\int\int_S \langle 0,0,0 \rangle \cdot d \, {\bf S} = 0 \quad .
$$

{\bf Ans.:} $0$.

Comments: 

1. About \%60 of the people got it right. 

2. Some people did the ``otherwise'' method, but didn't finish it. No wonder, doing it directly is much more complicated. 

3. Some people said that the answer is $0$ because the surface 
$S$ is {\bf closed}. This is {\bf wrong!}. The {\it surface} $S$ (whatever it is) is not a closed surface. The {\bf curve}
$C$ is closed, but the surface is open of course. If a surface is closed, than it has {\bf no boundary}, and
in this problem we are told that the boundary of our surface is $C$.
You can never use the argument $S$ is closed when you are asked to find a {\it line-integral}, $\int_C \bf F \cdot d \, {\bf s}$,
because $S$ is {\bf never} closed (for that scenario). Your only hope at an easy way out is to compute 
$curl {\bf F}$ and see that it is the zero-vector $\langle 0,0,0 \rangle$ .
The ``closed surface'' argument works when you have to computer a {\bf surface integral} $\int\int_S curl({\bf F}) \cdot d {\bf S}$,
where $S$ is the {\it whole} sphere or the {\it whole} ellipsoid or whatever, then it is automatically zero.



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