The area of the ellipse [tex]E[/tex] is given by
[tex]\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy[/tex]
To use Green's theorem, which says
[tex]\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy[/tex]
([tex]\partial E[/tex] denotes the boundary of [tex]E[/tex]), we want to find [tex]M(x,y)[/tex] and [tex]L(x,y)[/tex] such that
[tex]\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1[/tex]
and then we would simply compute the line integral. As the hint suggests, we can pick
[tex]\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1[/tex]
The line integral is then
[tex]\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy[/tex]
We parameterize the boundary by
[tex]\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}[/tex]
with [tex]0\le t\le2\pi[/tex]. Then the integral is
[tex]\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt[/tex]
[tex]=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi[/tex]
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Notice that [tex]x^{2/3}+y^{2/3}=4^{2/3}[/tex] kind of resembles the equation for a circle with radius 4, [tex]x^2+y^2=4^2[/tex]. We can change coordinates to what you might call "pseudo-polar":
[tex]\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}[/tex]
which gives
[tex]x(t)^{2/3}+y(t)^{2/3}=(4\cos^3t)^{2/3}+(4\sin^3t)^{2/3}=4^{2/3}(\cos^2t+\sin^2t)=4^{2/3}[/tex]
as needed. Then with [tex]0\le t\le2\pi[/tex], we compute the area via Green's theorem using the same setup as before:
[tex]\displaystyle\iint_E\mathrm dx\,\mathrm dy=\frac12\int_0^{2\pi}(-4\sin^3t(12\cos^2t(-\sin t))+4\cos^3t(12\sin^2t\cos t))\,\mathrm dt[/tex]
[tex]=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt[/tex]
[tex]=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt[/tex]
[tex]=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt[/tex]
[tex]=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt[/tex]
[tex]=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi[/tex]
