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\centerline
{
\bf Dr. Z's Math251 Handout \#16.5 [Curl and Divergence]
}

By Doron Zeilberger

{\bf Problem Type 16.5a}: 
Find (a) the curl and (b) the divergence
of the vector field 
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}
+ Q(x,y,z) \, {\bf j}
+ R(x,y,z) {\bf k} \quad .
$$
{\bf Example Problem 16.5a}: 
Find (a) the curl and (b) the divergence
of the vector field 
$$
{\bf F}(x,y,z)= 
2 e^x \sin y \, {\bf i}+ 3 e^x \cos y \, {\bf j}+ (4 z^2+x+y) \, {\bf k}
\quad .
$$



\bigskip
\hrule
\bigskip
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\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} $curl \, {\bf F}$ equals
$$
\g \times {\bf F}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
P & Q & R 
} \right \vert \quad.
$$
Set it up for the specific $P,Q,R$.
&
{\bf 1.} 
$$
\g \times {\bf F}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
2 e^x \sin y & 3 e^x \cos y  & 4 z^2+x+y
} \right \vert \quad .
$$
\cr


\bigskip & \bigskip \cr
{\bf 2.} Evaluate the `determinant'.

&
{\bf 2.}
 $$
{\bf i} \, 
\left ( {{\partial} \over {\partial y}} (4z^2+x+y) -
{{\partial} \over {\partial z}} (3e^x \cos y) \right )
$$
$$
-{\bf j} \, 
\left ( {{\partial} \over {\partial x}} (4z^2+x+y) -
{{\partial} \over {\partial z}} (2e^x \sin y) \right )
$$
$$
+{\bf k} \, 
\left ( {{\partial} \over {\partial x}} (3 e^x \cos y) -
{{\partial} \over {\partial y}} (2 e^x \sin y)   \right )
$$
$$
={\bf i} \, 
(1-0)
-{\bf j} \, 
( 1 -0)
+{\bf k} \, 
(3 e^x \cos y -2 e^x \cos y)
$$
$$
={\bf i} -{\bf j} + e^x \cos y \, {\bf k} \quad .
$$
{\bf Ans. to (a)}: $curl \, {\bf F}={\bf i} -{\bf j} + e^x \cos y \, {\bf k}$.
\cr
\bigskip & \bigskip \cr

{\bf 3.} Set-up the formula for the divergence
$$
div \, {\bf F}= {{\partial P} \over {\partial x}}
+{{\partial Q} \over {\partial y}}
+{{\partial R} \over {\partial z}} \quad .
$$
Then compute it.

&
{\bf 3.} 
$$
div \, {\bf F}= {{\partial } \over {\partial x}} (2 e^x \sin y)
+{{\partial } \over {\partial y}} (3 e^x \cos y)
+{{\partial } \over {\partial z}}(4 z^2+x+y)
$$
$$
= 2 e^x \sin y-3 e^x \sin y+8z =
-e^x \sin y +8z \quad .
$$
{\bf Ans. to (b):}  $div \, {\bf F}= -e^x \sin y +8z$ .
\cr}

{\bf Problem Type 16.5b}: 
Determine whether or not the vector field is conservative.
If it is, find a function $f$ such that ${\bf F}=\g f$.
$$
{\bf F}(x,y,z)= P(x,y,z) \, {\bf i}
+ Q(x,y,z) \, {\bf j}
+ R(x,y,z) {\bf k} \quad .
$$
{\bf Example Problem 16.5b}: 
Determine whether or not the vector field is conservative.
If it is, find a function $f$ such that ${\bf F}=\g f$.
$$
{\bf F}(x,y,z)= 
(y^2z+2xyz)\, {\bf i}+ (2xyz+x^2z) \, {\bf j}+ (xy^2+x^2y+2z) \, {\bf k}
$$

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\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute $curl \, {\bf F}$ 
$$
\g \times {\bf F}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
P & Q & R 
} \right \vert \quad.
$$
If it is the {\bf zero} vector (i.e. {\bf all} compoents are zero)
then the vector field ${\bf F}$ is conservative.
Otherwise not. If it is not, end of story.
If it is, go on.
&
{\bf 1.} 
$$
\g \times {\bf F}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
{{\partial} \over {\partial x}} 
& {{\partial} \over {\partial y}} 
& {{\partial} \over {\partial z}} 
\cr
y^2z+2xyz & 2xyz+x^2z  & xy^2+x^2y+2z
} \right \vert \quad .
$$
This equals
 $$
{\bf i} \, 
\left ( {{\partial} \over {\partial y}} (xy^2+x^2y+2z) -
{{\partial} \over {\partial z}} (2xyz+x^2z) \right )
$$
$$
-{\bf j} \, 
\left ( {{\partial} \over {\partial x}} (xy^2+x^2y+2z) -
{{\partial} \over {\partial z}} (y^2z+2xyz) \right )
$$
$$
+{\bf k} \, 
\left ( {{\partial} \over {\partial x}} (2xyz+x^2z) -
{{\partial} \over {\partial y}} (y^2z+2xyz)   \right )
$$
$$
={\bf i} \, 
(2xy+x^2-2xy-x^2)
-{\bf j} \, 
( y^2 +2xy-y^2-2xy)
$$
$$
+{\bf k} \, 
(2yz+2xz-2yz-2xz)
$$
$$
=0 \, {\bf i} - 0 \, {\bf j} + 0 {\bf k} = {\bf 0} \quad.
$$
Since the curl of ${\bf F}$ is ${\bf 0}$, the vector field 
${\bf F}$ is conservative, and we must go on.
\cr

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{\bf 2.} Find a function $f(x,y,z)$ such that
$\g F=f$, in other words
$$
f_x=P \quad , \quad f_y=Q  \quad , \quad f_z=R \quad.
$$
You first integrate $P$ w.r.t. to $x$ getting
that $f$ equals something {\bf plus} a function
$g(y,z)$. Then you plug that expression for $f$ and
use it in $f_y=Q$ getting that $g(y,z)$ equals
something explicit {\bf plus} a function $h(z)$.
Plug-it back into $f$, and use $f_z=Q$ to get
what $h(z)$ is, and plug it back into $f$.
&
{\bf 2.}
$f_x=y^2z+2xyz$, means that
$$
f= \int (y^2z+2xyz) \, dx= xy^2z+x^2yz+ g(y,z) \quad.
$$
$f_y=2xyz+x^2z$ means that
$$
2xyz+x^2z+g_y=2xyz+x^2z \quad,
$$
so $g_y=0$ and $g(y,z)=h(z)$, for some function, $h(z)$, of $z$.
So now
$$
f= xy^2z+x^2yz+ h(z)
$$
$f_z=xy^2+x^2y+2z$ means
that
$$
xy^2+x^2y+h'(z)=xy^2+x^2y+2z \quad 
$$
So $h'(z)=2z$ and $h(z)=z^2$.
It follows that
$$
f= xy^2z+x^2yz+z^2 \quad .
$$

{\bf Ans.}: ${\bf F}$ is conservative, and the potential function
$f$ such that $\g f={\bf F}$ is
$f= xy^2z+x^2yz+z^2$.
\cr}

\end