Question

greens theorem. find the max value of the line integral where f=(13x^2y+3y^3-y)i-12x^3j and C is any positively oriented closed curve. max=?

Answer 1

The line integral is given by

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int_C((13x^2y+3y^3-y)\,\mathrm dx-12x^3\,\mathrm dy)$

By Green's theorem, the line integral along $C$ is equivalent to the double integral over $R$ (the region bounded by $C$)

$\displaystyle\iint_R\left(\frac{\partial(-12x^3)}{\partial x}-\frac{\partial(13x^2y+3y^3-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$
$=\displaystyle\iint_R(-36x^2-(13x^2+9y^2-1))\,\mathrm dx\,\mathrm dy$
$=\displaystyle\iint_R(1-49x^2-9y^2)\,\mathrm dx\,\mathrm dy$

Now consider the function $g(x,y)=1-49x^2-9y^2$. We can think of the double integral above as a volume integral; namely, it's the volume of the region below $g(x,y)$ and above the region $R$ in the $x$-$y$ plane (i.e. $z=0$). This volume will be maximized if $C$ is taken to be the intersection of $g(x,y)$ with the plane, which means $C$ is the ellipse $49x^2+9y^2=1$.

For the double integral, we can convert to an augmented system of polar coordinates using

$\begin{cases}x=\frac17r\cos\theta\\\\y=\frac13r\sin\theta\end{cases}$

where $0\le r\le1$ and $0\le\theta\le2\pi$. We have the Jacobian determinant

$\det\mathbf J=\left|\dfrac{\partial(x,y)}{\partial(r,\theta)}\right|=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial\theta}\\\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial\theta}\end{vmatrix}$
$\det\mathbf J=\begin{vmatrix}\frac17\cos\theta&-\frac17r\sin\theta\\\\\frac13\sin\theta&\frac3r\cos\theta\end{vmatrix}=\dfrac r{21}$

So the double integral, upon converting to our polar coordinates, is equivalent to

$=\displaystyle\frac1{21}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\left(1-49\left(\frac r7\cos\theta)^2-9\left(\frac r3\sin\theta\right)^2\right)r\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac1{21}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}(1-r^2\cos^2\theta-r^2\sin^2\theta)r\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac1{21}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}(r-r^3)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{2\pi}{21}\int_{r=0}^{r=1}(r-r^3)\,\mathrm dr\,\mathrm d\theta$
$=\dfrac\pi{42}$