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Problem of the Week #89 - December 9th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Find the positively oriented simple closed curve $C$ for which the value of the line integral
\[\int_C (y^3-y)\,dx - 2x^3\,dy\]
is a maximum.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by MarkFL and mathbalarka. You can find Mark's answer below.

To begin, I would consider the following theorem:

Green's Theorem in the Plane

Suppose that $C$ is a piecewise smooth simple closed curve bounding a region R. If $P$, $Q$, \(\displaystyle \frac{\partial P}{\partial y}\) and \(\displaystyle \frac{\partial Q}{\partial x}\) are continuous on $R$, then:

\(\displaystyle \oint_{C}P\,dx+Q\,dy=\underset{R}\iint \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)\,dA\)

In the given problem, we have:

\(\displaystyle P=y^3-y\implies \frac{\partial P}{\partial y}=3y^2-1\)

\(\displaystyle Q=-2x^3\implies \frac{\partial Q}{\partial x}=-6x^2\)

And so using Green's theorem, the given line integral may be expressed as the double integral:

\(\displaystyle \oint_{C}\left(y^3-y \right)\,dx+\left(-2x^3 \right)\,dy=\underset{R}\iint \left(-6x^2-\left(3y^2-1 \right) \right)\,dA=\underset{R}\iint \left(1-3\left(2x^2+y^2 \right) \right)\,dA\)

Next, let's parametrize $x$ and $y$ as follows:

\(\displaystyle x(r,\theta)=\frac{r}{\sqrt{2}}\cos(\theta)\)

\(\displaystyle y(r,\theta)=r\sin(\theta)\)

We may make use of the simple closed curve (a circle) $r=a$ where $0<a$.

Thus, $0\le r\le a$ and $0\le\theta\le2\pi$ and our double integral may now be expressed as follows:

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{a} \left(1-3r^2 \right)\left|\frac{\partial(x,y)}{ \partial(r,\theta)} \right|\,dr\,d\theta\)

Calculating the Jacobian matrix, we find:

\(\displaystyle \left|\frac{\partial (x,y)}{\partial (r,\theta)} \right|=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}\\\end{vmatrix}=\begin{vmatrix} \frac{1}{\sqrt{2}}\cos(\theta)& -\frac{r}{\sqrt{2}}\sin(\theta)\\ \sin(\theta)& r\cos(\theta)\\\end{vmatrix}= \frac{r}{\sqrt{2}} \left(\cos^2(\theta)+\sin^2(\theta) \right)=\frac{r}{\sqrt{2}}\)

Hence, there results:

\(\displaystyle \frac{1}{\sqrt{2}}\int_{0}^{2\pi}\int_{0}^{a} \left(r-3r^3 \right)\,dr\,d\theta=\frac{1}{\sqrt{2}}\int_{0}^{2\pi}\left(\left[\frac{r^2}{2}-\frac{3r^4}{4} \right]_0^a \right)\,d\theta=\frac{1}{\sqrt{2}}\int_{0}^{2\pi}\left(\frac{a^2}{2}-\frac{3a^4}{4} \right)\,d\theta=\)

\(\displaystyle \frac{2a^2-3a^4}{4\sqrt{2}}\int_{0}^{2\pi}\,d\theta=\frac{\pi}{2\sqrt{2}}\left(2a^2-3a^4 \right)\)

Now, if we consider the following function for maximization:

\(\displaystyle f(a)=2a^2-3a^4\)

We find by differentiating with respect to $a$ and equating the result to zero:

\(\displaystyle f'(a)=4a-12a^3=4a\left(1-3a^2 \right)=0\)

We then have the critical value (observing 0<a):

\(\displaystyle a=\frac{1}{\sqrt{3}}\)

We may use the second derivative test to determine the nature of the extremum associated with this critical value:

\(\displaystyle f''(a)=4-36a^2\implies f\left(\frac{1}{\sqrt{3}} \right)=-8<0\)

Hence, this critical value is at a maximum. And so the parametric equations of the simple close curve which maximizes the given line integral are:

\(\displaystyle x(\theta)=\frac{1}{\sqrt{6}}\cos(\theta)\)

\(\displaystyle y(\theta)=\frac{1}{\sqrt{3}}\sin(\theta)\)

or:

\(\displaystyle \sqrt{6}x=\cos(\theta)\)

\(\displaystyle \sqrt{3}y=\sin(\theta)\)

Squaring both and adding (and applying a Pythagorean identity), we eliminate the parameter to obtain the ellipse:

\(\displaystyle 6x^2+3y^2=1\)

Thus, we may conclude that this ellipse is the simple closed curve which maximizes the value of the given line integral.
 
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