Find the area inside the loop of the following limacon: r = 7 − 14 sin (θ).
Please help!
to find the limits of integration we set r =0
$\displaystyle 0=7-14\sin(\theta) \iff \sin(\theta)=\frac{1}{2}$
so we get $\displaystyle \frac{\pi}{6} \mbox{ and } \frac{5 \pi}{6}$
$\displaystyle \frac{1}{2}\int_{\pi/6}^{5\pi/6}(7-14\sin(\theta))^2d\theta$
After integrating I get $\displaystyle 49\pi-\frac{147\sqrt{3}}{2} \approx 26.63$
did you draw the graph?
now that we have drawn it, we see that the inner loop happens when the graph goes to the origin, that is, when $\displaystyle r = 0$. so let's solve for that:
$\displaystyle \Rightarrow 0 = 7 - 14 \sin \theta$
$\displaystyle \Rightarrow \sin \theta = \frac 12$
$\displaystyle \Rightarrow \theta = \frac {\pi}6,~\frac {5 \pi}6$ for $\displaystyle 0 \le \theta \le 2 \pi$
thus, the area is given by: $\displaystyle A = \frac 12 \int_{\pi /6}^{5 \pi /6}r^2~d \theta = \frac 12 \int_{\pi / 6}^{5 \pi / 6}(7 - 14 \sin \theta)^2 ~d \theta$
EDIT: Geez, too late. Thanks a lot EmptySet!
We can factor out the 7 and write it as $\displaystyle 7(1-2sin{\theta})$
$\displaystyle 1-2sin{\theta}=0, \;\ {\theta}=\frac{\pi}{6}, \;\ {\theta}=\frac{5\pi}{6}$
Then, we get $\displaystyle \frac{49}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}( 1-2sin{\theta})^{2}d{\theta}$
$\displaystyle \frac{49}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\ left[4sin^{2}{\theta}-4sin{\theta}+1\right]d{\theta}$
Oops, beat to the punch. At least we agree.