1. ## vol

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?

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

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

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

- Hollywood

3. may i know how you got that formula?

4. 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 $d\theta$, and the radius is $\mid{r}\mid$. Since the angle at the origin for a whole circle is $2\pi$, our wedge shape is $d\theta/2\pi$ of a circle. Since the area of a circle is $\pi\mid{r}\mid^2$, the area of the wedge is $(d\theta/2\pi)(\pi\mid{r}\mid^2)=\frac{1}{2}\mid{r}\mid^2\ d\theta$. That explains the integrand.

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

- Hollywood