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13.4 - Green's theorem problems
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<h2>
[13.4] - 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 class="result">
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 class="result">
$$Q_x-P_y=2-1=1$$
so
$$\oint_C P\,dx+Q\,dy=\int_{-4}^{4}\int_{-3}^3 (1) \,dx\,dy=8\times 6=48.$$


<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 class="result">
$$Q_x-P_y=2xy^3-(-4xy^3/2)=4xy^3$$
then
$$\oint_C...=\int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^1 4xy^3\,dy\,dx =...=0$$


<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 class="result">
$$Q_x-P_y=2xy-(-2y)=2y(x-1).$$
So,
$$\oint_C...=\int_{x=-2}^2\int^{\sqrt{4-x^2}}_{y=0} 2y(x-1)\,dy\,dx =...=-32/3$$
<br><img src="0g/13.4.4.jpg">

<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">



</ol>
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