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Over [174]
3 years ago
14

Consider the force field and circle defined below.

Mathematics
1 answer:
grin007 [14]3 years ago
4 0

By Green's theorem,

\displaystyle\int_{x^2+y^2=9}\vec F(x,y)\cdot\mathrm d\vec r=\iint_D\left(\frac{\partial(xy)}{\partial x}-\frac{\partial(x^2)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=\iint_Dy\,\mathrm dx\,\mathrm dy

where C is the circle x^2+y^2=9 and D is the interior of C, or the disk x^2+y^2\le1.

Convert to polar coordinates, taking

\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta

Then the work done by \vec F on the particle is

\displaystyle\iint_Dy\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^3(r\sin\theta)r\,\mathrm dr\,\mathrm d\theta=\left(\int_0^{2\pi}\sin\theta\,\mathrm d\theta\right)\left(\int_0^3r^2\,\mathrm dr\right)=\boxed0

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