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shusha [124]
3 years ago
14

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

∂d−y dx+x dy . hint: x(t)=5cos(t). the area is 85pi .
b.find a parametrization of the curve x2/3+y2/3=42/3 and use it to compute the area of the interior. hint: x(t)=4cos3(t).
Mathematics
1 answer:
Pavel [41]3 years ago
3 0

The area of the ellipse E is given by

\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy

To use Green's theorem, which says

\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

(\partial E denotes the boundary of E), we want to find M(x,y) and L(x,y) such that

\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1

and then we would simply compute the line integral. As the hint suggests, we can pick

\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

The line integral is then

\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy

We parameterize the boundary by

\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}

with 0\le t\le2\pi. Then the integral is

\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt

=\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

###

Notice that x^{2/3}+y^{2/3}=4^{2/3} kind of resembles the equation for a circle with radius 4, x^2+y^2=4^2. We can change coordinates to what you might call "pseudo-polar":

\begin{cases}x(t)=4\cos^3t\\y(t)=4\sin^3t\end{cases}

which gives

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}

as needed. Then with 0\le t\le2\pi, we compute the area via Green's theorem using the same setup as before:

\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

=\displaystyle24\int_0^{2\pi}(\sin^4t\cos^2t+\cos^4t\sin^2t)\,\mathrm dt

=\displaystyle24\int_0^{2\pi}\sin^2t\cos^2t\,\mathrm dt

=\displaystyle6\int_0^{2\pi}(1-\cos2t)(1+\cos2t)\,\mathrm dt

=\displaystyle6\int_0^{2\pi}(1-\cos^22t)\,\mathrm dt

=\displaystyle3\int_0^{2\pi}(1-\cos4t)\,\mathrm dt=6\pi

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