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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<meta
name="viewport"
content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no"
/>
<title>APMA E2000 - Green's Theorem</title>
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</head>
<body>
<div class="reveal">
<div class="slides">
<section>
<section>
<h3 class="framelabel">Lecture 19</h3>
<h1>Green's Theorem</h1>
<h2>APMA E2000</h2>
<div class="r-stretch"></div>
<p style="text-align: right">
Drew Youngren <code>[email protected]</code>
<span style="font-size: xx-small"
><a
href="https://onlineboard.eu/b/eOtbAEk4"
target="_blank"
rel="noopener noreferrer"
>bd</a
></span
>
</p>
</section>
<section>
<!-- Set up standard LaTeX macros -->
$\gdef\RR{\mathbb{R}}$ $\gdef\vec{\mathbf}$
$\gdef\bv#1{\begin{bmatrix} #1 \end{bmatrix}}$
$\gdef\proj{\operatorname{proj}}$ $\gdef\comp{\operatorname{comp}}$
<h2>Announcements</h2>
<ul>
<li class="fragment">HW 10 due Tuesday</li>
<li class="fragment">No quiz this week</li>
<li class="fragment">Recitation (11) as usual</li>
</ul>
</section>
</section>
<section>
<section>
<h1>1-minute review</h1>
</section>
<section>
<h3 class="framelabel">Fundamental Theorem of Line Integrals</h3>
\[ \int\limits_C \nabla f \cdot d\vec r = f(C_\text{end}) -
f(C_\text{start}) \]
</section>
<section>
<h6 class="framelabel">Potential Difference</h6>
<p style="font-size: smaller">
Compute the path integral $\int\limits_C \left(x -
\frac{y^3}{3}\right)\,dx - xy^2\,dy$ over the the path below.
</p>
<img
src="assets/spiral_vector_field.png"
alt="Spiral path in conservative vector field"
srcset=""
class="r-stretch"
/>
</section>
</section>
<section>
<section>
<h1>Green's Theorem</h1>
</section>
<section
data-background-image="assets/comma_region.png"
data-background-size="contain "
data-background-opacity="0.5"
>
<h3 class="framelabel">Simple Closed Curve</h3>
<p>
A <strong>simple closed curve</strong> $C$ is a curve with a
parametrization $\vec r(t), a\leq t \leq b$ such that
</p>
<ul>
<li>$\vec r(a) = \vec r(b)$, and</li>
<li>it does not cross itself.</li>
</ul>
<p class="fragment">
In $\RR^2$, such a curve surrounds a region $D$. We write \[C =
\partial D\] meaning "$C$ is the <em>boundary</em> of $D$."
</p>
</section>
<section
data-background-image="assets/comma_region.png"
data-background-size="contain "
data-background-opacity="0.5"
>
<h3 class="framelabel">Green's Theorem</h3>
\[\oint\limits_{\partial D} P\,dx + Q\,dy = \iint\limits_{D}
\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}
\right)\,dA \]
<!-- <img src="../img/swirlyD.png" width="33%" style="float: right; right: 100px;" /> -->
<div style="font-size: smaller">
where:
<ul>
<li class="thm fragment">
$D$ is a (simply connected) region in the plane,
</li>
<li class="thm fragment">
$\partial D$ is an counterclockwise-oriented, simple, closed
curve forming the boundary of $D$, and
</li>
<li class="thm fragment">
$\vec F(x,y) = \langle P(x,y),Q(x,y) \rangle$ is a
continuously differentiable vector field.
</li>
</ul>
</div>
</section>
<section>
<h6 class="framelabel">Example</h6>
<div class="container">
<div class="col">
<p style="font-size: smaller">
Compute the line integral of \[\vec F(x,y) =\langle xy - y^2 ,
x^2\rangle\] on the triangular curve connecting $(0,0)$,
$(2,0)$, and $(3,2)$ counterclockwise.
</p>
</div>
<div class="col">
<img
src="assets/tri_circ_int.png"
alt="Triangular path in vector field"
srcset=""
class="r-stretch"
/>
</div>
</div>
</section>
<section
data-background-image="assets/tri_circ_int.png"
data-background-opacity="0.5"
>
<h6 class="framelabel">Solution</h6>
<p>
Say, $ \langle P, Q \rangle = \langle xy - y^2 , x^2\rangle$.
Then,
</p>
<p class="fragment">
\[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} =
2x - (x - 2y) = x + 2y \]
</p>
<p class="fragment">
\[ \oint_T P\,dx + Q\,dy = \int_0^2 \int_{\frac32 y}^{2 + \frac12
y} (x + 2y)\,dx\,dy = 6 \]
</p>
<div class="r-stretch"></div>
</section>
<section
data-background-image="assets/tri_circ_int.png"
data-background-opacity="0.5"
>
<h6 class="framelabel">Solution</h6>
<p style="font-size: smaller">Check by integrating each piece.</p>
<p class="fragment" style="font-size: smaller">
Bot $\langle 2t, 0 \rangle$: $\displaystyle \int_0^1 ((2t)0 -
0^2)2\,dt + (2t)^2(0)\,dt + = 0$
</p>
<p class="fragment" style="font-size: smaller">
R $\langle 2 + t, 2t \rangle $: $\displaystyle \int_0^1 (2+t)(2t)
- (2t)^2 + (2 + t)^2 2\,dt = 14$
</p>
<p class="fragment" style="font-size: smaller">
L $\langle 3 - 3t,2 - 2t \rangle $: $\displaystyle \int_0^1 ((3 -
3t)(2 - 2t) - (2 - 2t)^2)(-3) + (3 - 3t)^2 (-2)\,dt = -8$
</p>
</section>
<section>
<h6 class="framelabel">Justification</h6>
<div class="container">
<div class="col">
<p class="r-stretch">
<!-- It is sufficient to show the two identities -->
\[\begin{align*}\oint_{\partial D} P\,dx &= \iint_D
-\partial_y P \,dA \\ \oint_{\partial D} Q\,dy &= \iint_D
\partial_x Q \,dA \end{align*}\]
</p>
<p class="fragment r-stretch" style="font-size: smaller">
Consider only the first case and that $D$ is the region \[
\begin{align*} g(x) &\leq y \leq f(x) \\ a &\leq x \leq b \\
\end{align*}\]
</p>
</div>
<div class="col">
<img
src="assets/bounded_by_functions.png"
alt="Region bounded by 2 functions of 1 variable"
class="r-stretch"
/>
</div>
</div>
</section>
<!-- <section>
Step through proof argument.
</section> -->
</section>
<section>
<section>
<h1>Areas</h1>
</section>
<section
data-background-image="assets/reservoir_plain.png"
data-background-opacity="0.7"
>
<h6 class="framelabel">Question</h6>
<p>
Can we determine the surface area of the Jackie Onassis Reservoir
if it's fenced off?
</p>
<div class="r-stretch"></div>
</section>
<section
data-background-image="assets/reservoir.png"
data-background-opacity="0.6"
>
<h6 class="framelabel">Answer</h6>
<p class="fragment">Yes, with a GPS watch.</p>
<p class="fragment">
Consider the vector field $\vec F(x,y) = \langle 0, x \rangle$.
</p>
<p class="fragment">
\[ \oint\limits_{\partial R} \vec F\cdot d\vec r =
\oint\limits_{\partial R} x\,dy = \iint\limits_R 1\,dA =
\text{Area}(R) \]
</p>
</section>
<section>
<h6 class="framelabel">Alternatives</h6>
\[\text{Area}(D) = \iint\limits_D \,dA \]
<div class="fragment">\[ = \oint\limits_{\partial D} x\,dy \]</div>
<div class="fragment">
\[ = \oint\limits_{\partial D} (-y)\,dx \]
</div>
<div class="fragment">
\[ = \oint\limits_{\partial D} \frac{x}{2}\,dy - \frac{y}{2}\,dx
\]
</div>
</section>
</section>
<section>
<section>
<h1>Notes on a Theorem</h1>
</section>
<section>
<h6 class="framelabel">Compare to FTLI</h6>
<p>
In what case do both Green's and the FTLI apply to a line integral
$\int_C \vec F\cdot d\vec r$?
</p>
<p class="fragment">$C$ is closed, and $\vec F$ is conservative.</p>
<div class="fragment">
\[\oint_{\partial D} \nabla f\cdot d\vec r = 0 \]
</div>
</section>
<section data-auto-animate>
<h3 class="framelabel">
What is $\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}$?
</h3>
<p style="font-size: smaller">
Let $B_r$ be a small ball of radius $r$ around a point $(a,b)$.
For a continuous function, \[ h(a,b) = \lim_{r\to 0}
\frac{\iint_{B_r} h\,dA}{\text{Area}(B_r)}\]
</p>
<p style="font-size: smaller" class="fragment">
That is, over smaller and smaller regions, the average value
converges to the value at that point.
</p>
</section>
<section data-auto-animate>
<h3 class="framelabel">
What is $\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}$?
</h3>
<p style="font-size: smaller">
Let $B_r$ be a small ball of radius $r$ around a point $(a,b)$.
Then,
</p>
<p>
\[ \left(\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}\right)(a,b) = \lim_{r\to 0} \frac{\iint_{B_r}
\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial
y}\right)\,dA}{\text{Area}(B_r)}\]
</p>
</section>
<section data-auto-animate>
<h3 class="framelabel">
What is $\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}$?
</h3>
<p style="font-size: smaller">
Let $B_r$ be a small ball of radius $r$ around a point $(a,b)$.
Then,
</p>
<p>
\[ \left(\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}\right)(a,b) = \lim_{r\to 0} \frac{\oint_{C_r} P\,dx
+ Q\,dy}{\text{Area}(B_r)}\]
</p>
</section>
<section data-auto-animate>
<h3 class="framelabel">
What is $\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}$?
</h3>
<p style="font-size: smaller">
Let $B_r$ be a small ball of radius $r$ around a point $(a,b)$.
Then,
</p>
<p>
\[ \left(\frac{\partial Q}{\partial x} - \frac{\partial
P}{\partial y}\right)(a,b) = \lim_{r\to 0} \frac{\oint_{C_r} P\,dx
+ Q\,dy}{\text{Area}(B_r)}\]
</p>
<p style="font-size: smaller" class="">
That is, the integrand of Green's Theorem is the "circulation
density" of the vector field $\langle P, Q \rangle$.
<span class="fragment"
>This is known as the <strong>scalar curl</strong>.</span
>
</p>
</section>
<section
data-background-image="assets/holey_domain.png"
data-background-opacity="0.3"
>
<h6 class="framelabel">Hole-y Domains</h6>
<p>
Green's Theorem works for non-simply-connected domains, provided
the boundary components are oriented correctly.
</p>
<div class="r-stretch"></div>
</section>
<section>
<h6 class="framelabel">Flux Form</h6>
<p>
Let $\vec N$ be the outward-oriented unit normal vector to a
closed curve.
</p>
<p class="fragment">
\[ \begin{align*}\vec T\,ds &= \langle x'(t), y'(t) \rangle \,dt
\\ \vec N\,ds &= \langle y'(t), -x'(t) \rangle\,dt \end{align*} \]
</p>
<p class="fragment">
\[ \oint_{\partial D} \vec F \cdot \vec N\,ds = \oint_{\partial D}
(-Q)\,dx + P\,dy \]
</p>
<p class="fragment">\[ = \iint\limits_D (P_x + Q_y)\,dA \]</p>
</section>
<section>
<h6 class="framelabel">Exercise</h6>
<div class="container">
<div class="col">
<p style="font-size: smaller">
Compare the methods of computing the integral \[
\oint\limits_C \left \langle 3, - \frac{x^2}{2} \right \rangle
\cdot d\vec r \] around the boundary of the
<a
href="https://3demos.ctl.columbia.edu/?Y3VycmVudENoYXB0ZXI9SG93K1RvJnNjYWxlPTAmc2hvd1BhbmVsPWZhbHNlJmdyaWQ9dHJ1ZSZvYmowX2tpbmQ9ZmllbGQmb2JqMF9jb2xvcj0lMjMwZDA4ODcmb2JqMF9hbmltYXRpb249dHJ1ZSZvYmowX3BhcmFtc19wPS4zJm9iajBfcGFyYW1zX3E9LXglNUUyJTJGMiZvYmowX3BhcmFtc19yPTAmb2JqMF9wYXJhbXNfblZlYz02Jm9iajFfa2luZD1ncmFwaCZvYmoxX2NvbG9yPSUyM2RjNWU2NiZvYmoxX3BhcmFtc19hPTAmb2JqMV9wYXJhbXNfYj0xJm9iajFfcGFyYW1zX2M9MCZvYmoxX3BhcmFtc19kPTEmb2JqMV9wYXJhbXNfej0wJm9iajFfcGFyYW1zX3QwPTAmb2JqMV9wYXJhbXNfdDE9MQ=="
target="_blank"
rel="noopener noreferrer"
>
unit square
</a>
$0\leq x \leq 1$, $0 \leq y \leq 1$.
</p>
</div>
<div class="col">
<img
src="assets/square_lineint.png"
alt="Square curve in vector field"
/>
</div>
</div>
</section>
</section>
<section>
<section><h1>Learning Outcomes</h1></section>
<section id="learning-outcomes">
<h6 class="framelabel">You should be able to...</h6>
<ul>
<li>Verify the hypotheses of Green's Theorem.</li>
<li>Use Green's Theorem to compute areas in the plane.</li>
<li>Interpret the scalar curl as the spin of a vector field.</li>
<li>Utilize the flux form of Green's when appropriate.</li>
</ul>
</section>
</section>
</div>
</div>
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