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

Status
Not open for further replies.

#### Chris L T521

##### Well-known member
Staff member
Thanks again to those who participated in last week's POTW! Here's this week's problem!

-----

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.

-----

#### Chris L T521

##### Well-known member
Staff member
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.

Status
Not open for further replies.