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6.4 H - Green's theorem problems
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<h2>
Green's theorem problems
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<ol>
<li>Let $C$ be the triangle path $(0,0)\to(1,1)\to(0,1)\to(0,0)$.
<img src="0g/13.4.triangle-less.jpg"><br>
<p id="1"><a class="toggle" onclick="toggle_visibility('a1');">show / hide</a> solution
<p id="a1" class="result toggle">
The orientation of the path is negative.  So, multiplying by $-1$, The vector integral around the outside is equal to the positive orientation:
$$\begineq \oint_{-C}(-2y\,dx+3x\,dy)&=&\\
\oint  P\,dx+Q\,dy &=&\iint\left(\frac{\del Q}{\del x}-\frac{\del P}{\del y}\right)\,dA\\
&=&\iint\left(3-(-2)\right)=5\iint dA=5*\frac 12.
\endeq$$
<span class="answer">=5/2<span>.

<li style="page-break-before:always">
Use Green's Theorem to calculate $\oint_C(y-x)\,dx
+(2x-y)\,dy$ where $C$ is the boundary of the rectangle shown.<br>
<img src="0g/13.4.3x4.jpg"><br>

<p id="2"><a class="toggle" onclick="toggle_visibility('a2');">show / hide</a> solution
<p id="a2" class="result toggle">
$$Q_x-P_y=2-1=1$$
so
$$\oint_C P\,dx+Q\,dy=\int_{-4}^{4}\int_{-3}^3 (1) \,dx\,dy=8\times 6.$$
<span class="answer">=48</span>


<li style="margin-top: 6ex">
Compute $\oint_C\left(-\frac{xy^4}{2}\right)\,dx +
\left(x^2y^3\right)\,dy$ where $C$ is the curve shown below.<br>
<img src="0g/13.4.shield.jpg"><br>

<p id="3"><a class="toggle" onclick="toggle_visibility('a3');">show / hide</a> solution
<p id="a3" class="result toggle">
$$Q_x-P_y=2xy^3-(-4xy^3/2)=4xy^3$$
then
$$\begineq\oint_C...&=&\int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^1 4xy^3\,dy\,dx\\
&=&\int_{x=-1}^1 4x \left(\int_{y=-\sqrt{1-x^2}}^1 y^3\,dy\right)\,dx\\
&=&\int_{x=-1}^1 4x \left(
\left.\frac{y^4}{4}\right|_{-\sqrt{1-x^2}}^1 \,dy\right)\,dx\\
&=&\int_{x=-1}^1 \frac{4x}{4} \left(1-(1-x^2)^2
\right)\,dx\\

&=&\int_{x=-1}^1 x (2x^2-x^4)\,dx=\int_{x=-1}^1  (2x^3-x^5)
\,dx\\

&=&\left[\frac{2x^4}{4}-\frac{x^6}{6}\right]_{x=-1}^1=\left[\frac24-\frac24-\left(\frac16-\frac16\right)\right]


\endeq$$
<span class="answer">=0</span>

<li style="page-break-before:always">
Use Green's Theorem to evaluate the following line integrals:
<ol type="a">
<li>$\oint_C\left(\arctan(x^2)-y^2\right)\,dx+
\left(x^2y-\ln(y^2+1)\right)\,dy$ where $C$ is the semicircle $y=\sqrt{4-x^2}$ together with the line segment $(-2,0)\to(2,0)$ as shown.<br>
<img src="0g/13.4.semic.jpg">
<br>
<p id="4"><a class="toggle" onclick="toggle_visibility('a4');">show / hide</a> solution
<p id="a4" class="result toggle">
$$Q_x-P_y=2xy-(-2y)=2y(x+1).$$
So,
$$\begineq \oint_C...&=&\int_{x=-2}^2\int^{\sqrt{4-x^2}}_{y=0} 2y(x+1)\,dy\,dx\\
&=&\int_{x=-2}^2 2(x+1)\int_0^{\sqrt{4-x^2}} y\,dy\,dx\\
&=&\int_{x=-2}^2 2(x+1)\left.\frac{y^2}{2}\right|_0^{\sqrt{4-x^2}}\,dx\\
&=&\int_{x=-2}^2 2(x+1)\frac12 (4-x^2)\,dx\\
&=&\int_{x=-2}^2 (x+1)(4-x^2)\,dx=\int_{x=-2}^2 4x-x^3+4-x^2\,dx\\
&=&
\left.\left(\frac{4x^2}2-\frac{x^4}{4}+4x-\frac{x^3}{2}\right)\right|_{x=-2}^2\\
&=& 0 - 0 + 16-8
\\



\endeq$$
<span class="answer">=8</span>
<br>

<li style="margin-top:20ex">
$\oint_C xy\,dx+(x^2+y^2)\,dy$ where $C$ is this triangle.<br>
<img src="0g/13.4.tri2.jpg"><br>

</ol>

<li style="page-break-before:always">
Consider the <i><b>non-closed</b></i> curve $C$, $(3,0)\to(0,2)\to (1,0)$ as shown.  Figure out a way to use Green's Theorem to help you compute
$\int_C(x+y)\,dx+(3x-y)\,dy$.<br>
<img src="0g/13.4.open.jpg"><br>
Hint:<br>
<img src="0g/greenhint.jpg">

<p id="5"><a class="toggle" onclick="toggle_visibility('a5');">show / hide</a> solution
<div id="a5" class="result toggle">
<p>$\myv F=P\uv i+Q\uv j$ With $P=x+y$ and $Q=3x-y$, <br>

<img src="0g/06-Green5a.png"><br>
The vector integral $\oint_\text{triangle}\myv F\cdot\,d\myv r$ <i>all the way around</i> the triangle above is, according to Green, given by the double integral:
$$\nonumber\iint Q_x-P_y\,dA=\iint (3-1)\,dA=2*A=2*(\text{base}*\text{height}/2)=4.$$

<img src="0g/06-Green5b.png"><br>
The vector line integral along the path above ($d\myv r=dx$ and $y=0$) is
$$\nonumber\int \myv F\cdot\,d\myv r=\int_3^1P(x,y=0)\,dx=\int_3^1x\,dx=\left.\frac{x^2}2\right|_3^1=\frac 12 -\frac92=-4$$

<p>Adding the two contributions together: 4-4<span class="answer">=0</span>.
</div>
</ol>
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