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aleksklad [387]
3 years ago
14

b) Use Greens theorem to find∫x^2 ydx-xy^2 dy where ‘C’ is the circle x2 + y2 = 4 going counter clock wise.​

Mathematics
1 answer:
anygoal [31]3 years ago
4 0

It looks like the integral you want to find is

\displaystyle \int_C x^2y\,\mathrm dx - xy^2\,\mathrm dy

where <em>C</em> is the circle <em>x</em> ² + <em>y</em> ² = 4. By Green's theorem, the line integral is equivalent to a double integral over the disk <em>x</em> ² + <em>y</em> ² ≤ 4, namely

\displaystyle \iint\limits_{x^2+y^2\le4}\frac{\partial(-xy^2)}{\partial x}-\frac{\partial(x^2y)}{\partial y}\,\mathrm dx\,\mathrm dy = -\iint\limits_{x^2+y^2\le4}(x^2+y^2)\,\mathrm dx\,\mathrm dy

To compute the remaining integral, convert to polar coordinates. We take

<em>x</em> = <em>r</em> cos(<em>t</em> )

<em>y</em> = <em>r</em> sin(<em>t</em> )

<em>x</em> ² + <em>y</em> ² = <em>r</em> ²

d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>t</em>

Then

\displaystyle \int_C x^2y\,\mathrm dx - xy^2\,\mathrm dy = -\int_0^{2\pi}\int_0^2 r^3\,\mathrm dr\,\mathrm dt \\\\ = -2\pi\int_0^2 r^3\,\mathrm dr \\\\ = -\frac\pi2 r^4\bigg|_{r=0}^{r=2} \\\\ = \boxed{-8\pi}

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