find the area of double integral over R dxdy enclosed by the polar curve r(t) = 2 + sin (10t) where t ranges from 0 to 2pi.

i used the method integral (magnitude of r'(t)) over t= 0 to 2pi..

but it was wrong..may i know why?

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- Apr 22nd 2010, 03:30 PM #1

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- Apr 23rd 2010, 11:01 PM #2

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I think the correct formula is $\displaystyle \frac{1}{2}\int_0^{2\pi}\mid{r}\mid^2\ d\theta$.

To do the integral, you'll need the identity $\displaystyle \sin^2(x)=\frac{1}{2}(1-\cos{2x})$.

To check your answer, the result should be roughly the area of a circle with radius 2, which, of course, is $\displaystyle 4\pi$.

- Hollywood

- Apr 24th 2010, 03:43 AM #3

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- Apr 24th 2010, 05:16 PM #4

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If you look at a small piece of the graph, it is a sector of a circle (the shape of a slice of pie). The angle at the origin is $\displaystyle d\theta$, and the radius is $\displaystyle \mid{r}\mid$. Since the angle at the origin for a whole circle is $\displaystyle 2\pi$, our wedge shape is $\displaystyle d\theta/2\pi$ of a circle. Since the area of a circle is $\displaystyle \pi\mid{r}\mid^2$, the area of the wedge is $\displaystyle (d\theta/2\pi)(\pi\mid{r}\mid^2)=\frac{1}{2}\mid{r}\mid^2\ d\theta$. That explains the integrand.

Our curve $\displaystyle r=2+\sin{10\theta}$ is traced by $\displaystyle \theta$ going from 0 to $\displaystyle 2\pi$, so that explains the limits of integration.

- Hollywood