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\centerline
{
\bf Dr. Z's Math251 Handout \#17.1a (2nd ed.) [Green's Theorem]
}

By Doron Zeilberger

{\bf Problem Type 17.1aa}: 
Use Green's Theorem to evaluate the
line integral along the given positively oriented curve.
$$
\int_{C} P(x,y) \, dx+ Q(x,y) \, dy \quad ,
$$
where $C$ is a given curve.

{\bf Example Problem 17.1aa}:  
Use Green's Theorem to evaluate the
line integral along the given positively oriented curve.
$$
\int_{C} {{2x^2 y^3} \over {3}} \, dx+ 4xy^3 \, dy \quad ,
$$
where $C$ is the triangle with vertices $(0,0)$, $(1,3)$, and
$(0,3)$.

\bigskip
\hrule
\bigskip
%\halign{#\hfill  & \quad #\hfill \cr
\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Set-up Green's Theorem
$$
\int_{C} P(x,y) \, dx+ Q(x,y) \, dy \quad 
$$
$$
=\int\int_{D} \left (
{{\partial Q} \over {\partial x}}
-
{{\partial P} \over {\partial y}}
\right ) \, dA \quad ,
$$
where $D$ is the region enclosed by $C$. 


&
{\bf 1.} Here $P={{2x^2 y^3} \over {3}}$, $Q=4xy^3$, so
$$
{{\partial Q} \over {\partial x}}
-
{{\partial P} \over {\partial y}}
={{\partial } \over {\partial x}} (4xy^3)
-
{{\partial } \over {\partial y}} ({{2x^2 y^3} \over {3}})
$$
$$
= 4y^3-2x^2y^2 \quad .
$$
We have to evaluate the area integral
$$
\int\int_{D} (4y^3-2x^2y^2) \, dA \quad,
$$
where $D$ is the region inside our triangle.
\cr

\bigskip & \bigskip \cr
{\bf 2.} Draw $D$, and write it as a type I or type II region.
Set up our area integral as an iterated integral.
&
{\bf 2.} Our triangle has one side along the $y$-axis,
so it is more convenient to express it as a type II
region. The hypotenuse is the line $y=3x$, that should
be written as $x=y/3$, since {\bf now} $y$ is the boss!
So
$$
D= \{ (x,y) \, \vert \, 0 \leq y \leq 3 \, , \, 0 \leq x \leq y/3 \, \}
\quad .
$$
The iterated integral is thus:
$$
\int_{0}^{3}\int_{0}^{y/3} (4y^3-2x^2y^2) \, dx \, dy \quad .
$$
\cr
\bigskip & \bigskip \cr

{\bf 3.} Evalute this iterated integral.

&
{\bf 3.} The inner integral is
$$
\int_{0}^{y/3} (4y^3-2x^2y^2) \, dx =
4y^3x- 2y^2{{x^3} \over {3}} \, \Bigl \vert_{0}^{y/3}
$$
$$
=
4y^3 (y/3)- {{2} \over {3}} y^2 (y/3)^3=
{{4} \over {3}}y^4-{{2} \over {81}}y^5 \quad .
$$
The whole thing is:
$$
\int_{0}^{3} \left [
\int_{0}^{y/3} (4y^3-2x^2y^2) \, dx  \right ]
\, dy \quad,
$$
$$
=\int_{0}^{3} 
({{4} \over {3}}y^4-{{2} \over {81}}y^5)\, dy \quad
=
{{4} \over {3}}{{y^5} \over {5}} -
{{2} \over {81}}{{y^6} \over {6}} \, \Bigl \vert_{0}^{3}
$$
$$
={{4} \over {3}} \cdot {{3^5} \over {5}} -{{2} \over {81}}
\cdot {{3^6} \over {6}} -0=
{{309} \over {5}} \quad .
$$
{\bf Ans.:} ${{309} \over {5}}$.
\cr}

{\bf Problem Type 17.1ab }:Use Green's Theorem to evaluate
$\int_C {\bf F} \cdot d {\bf r}$,
(Check orientation of the curve before applying the theorem).
$$
{\bf F} (x,y) = \langle P(x,y) \, , Q(x,y) \rangle \quad,
$$
where $C$ is a certain given closed curve.

{\bf Example Problem 17.1ab}:  Use Green's Theorem to evaluate
$\int_C {\bf F} \cdot d {\bf r}$,
(Check orientation of the curve before applying the theorem).
$$
{\bf F} (x,y) = \langle \sin x + y^2 \, , \, x+ \cos^3 y \rangle \quad,
$$
where $C$ consists of the arc of the curve $y=\sin x$ from
$(0,0)$ to $(\pi,0)$ and the line segment from $(\pi,0)$
to $(0,0)$.

\bigskip
\hrule
\bigskip
%\halign{#\hfill  & \quad #\hfill \cr
\halign{           
\vtop{\hsize=15pc\noindent #\strut} \qquad
&
\vtop{\hsize=15pc\noindent #\strut} \cr
{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Set-up Green's Theorem
$$
\int_{C} P(x,y) \, dx+ Q(x,y) \, dy \quad 
$$
$$
=\int\int_{D} \left (
{{\partial Q} \over {\partial x}}
-
{{\partial P} \over {\partial y}}
\right ) \, dA \quad ,
$$
where $D$ is the region enclosed by $C$. 
Also decide whether the specified description of
the curve is in the {\bf positive} direction
({\bf counterclockwise}) or in the {\bf negative}
direction ({\bf clockwise}).
&
{\bf 1.} Here $P=\sin x + y^2$, $Q=x+ \cos^3 y$, so
$$
{{\partial Q} \over {\partial x}}
-
{{\partial P} \over {\partial y}}
={{\partial } \over {\partial x}} (x+ \cos^3 y)
-
{{\partial } \over {\partial y}} (\sin x + y^2)
$$
$$
= 1-2y \quad .
$$
We have to evaluate the area integral
$$
\int\int_{D} (1-2y) \, dA \quad,
$$
where $D$ is the region inside our closed curve.
The way $C$ is described is {\bf clockwise} so it
is in the negative direction. So at the end, we have
to take the {\bf negative} of the result.
So multiplying by $(-1)$, we really have to evaluate
$$
\int\int_{D} (2y-1) \, dA \quad .
$$

\cr

\bigskip & \bigskip \cr
{\bf 2.} Draw $D$, and write it as a type I or type II region.
Set up our area integral as an iterated integral.
&
{\bf 2.} Our region has one side along the $x$-axis,
from $x=0$ to $x=\pi$, and the other part along
the curve $y=\sin x$.
It is more convenient to express it as a type I
region. 
$$
D= \{ (x,y) \, \vert \, 0 \leq x \leq \pi \, , \, 0 \leq y \leq \sin x \, \}
\quad .
$$
The iterated integral is thus:
$$
\int_{0}^{\pi}\int_{0}^{\sin x} (2y-1) \, dy \, dx \quad .
$$
\cr
\bigskip & \bigskip \cr

{\bf 3.} Evaluate this iterated integral.

&
{\bf 3.} The inner integral is
$$
\int_{0}^{\sin x} (2y-1) \, dy =
y^2-y \, \Bigl \vert_{0}^{\sin x}=
\sin^2 x - \sin x \quad .
$$
The whole thing is:
$$
\int_{0}^{\pi}
\left [ \int_{0}^{\sin x} (2y-1) \, dy 
\right ]
\, dx \quad,
$$
$$
=\int_{0}^{\pi} (\sin^2 x - \sin x ) \,  dx 
$$
$$
\int_{0}^{\pi} ({{1- \cos 2x} \over {2}} - \sin x ) \,  dx 
$$
$$
={{x} \over {2}} - {{\sin 2x} \over {4}} +\cos x \, \Bigl \vert_{0}^{\pi}
$$
$$
=
 ({{\pi} \over {2}} - {{\sin 2\pi} \over {4}} +\cos \pi )
- ({{0} \over {2}} - {{\sin (2 \cdot 0)} \over {4}} +\cos 0  )
$$
$$
=
{{\pi} \over {2}} - 0 -1-(0- 0 +1)=
{{\pi} \over {2}} - 2 \quad .
$$
{\bf Ans.:} ${{\pi} \over {2}} - 2$.
\cr}

{\bf A Problem from a previous Final}

Evalute
$$
\int_C \,\, (5y-\sin(e^x)) \, dx + (10x - e^{\cos^2 y}) \, dy \quad ,
$$
where $C$ is the closed curve consisting of the boundary of the rectangle
$$
\{ \, (x,y) \, | \, 0 \leq x \leq 4 \quad , \quad 0 \leq y \leq 3 \, \} .
$$

{\bf Ans.} $60$.

\end




  

