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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=?

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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((\partial(-12x^3))/(\partial x)-(\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|(\partial(x,y))/(\partial(r,\theta))\right|=\begin{vmatrix}(\partial x)/(\partial r)&(\partial x)/(\partial\theta)\\\\(\partial y)/(\partial r)&(\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}=\frac 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(2\pi)/(21)\int_(r=0)^(r=1)(r-r^3)\,\mathrm dr\,\mathrm d\theta

=\frac\pi{42}
User Aviv Ben Shabat
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