The diagram shows a sketch of the loop whose polar equation is
$$r=2(1-\sin\theta),\qquad -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$$
a)Show that the area enclosed by the loop is 16/3.
b)Show that the initial line divides the area enclosed by the loop in the ratio 1:7.
[Math] Find the area enclosed by the loop $r=2(1-\sin\theta)\sqrt{\cos\theta}$
calculuspolar coordinatestrigonometry
Related Solutions
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
The two circles are intersecting at $\theta=\dfrac{\pi}{4}$ and $r=0$
The area region can be divided into two parts.
Note that we are bounded by the region of the red circle $($the lower part of the half$)$ by $\theta=0\mbox{ to }\theta=\dfrac{\pi}{4}$
Similarly, the region of the purple circle $($the upper part of the half$)$ is bounded by $\theta=\dfrac{\pi}{4}$ to $\dfrac{\pi}{2}$
So, we get $$\dfrac12\int_{0}^{\dfrac{\pi}{4}}(3\sin\theta)^2d\theta+\dfrac12\int_{\frac{\pi}{4}}^{\dfrac{\pi}{2}}(3\cos\theta)^2d\theta$$
Best Answer
Fleshing out Aneesh's comment: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{2}r^2d\theta=-2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1-\sin\theta\right)^2\,\overbrace{d(1-\sin\theta)}^{-\cos\theta}=\left.-\frac{2}{3}\left(1-\sin\theta\right)^3\right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}=$$ $$=-\frac{2}{3}\left[(1-1)^3-(1+1)^3\right]=\frac{16}{3}$$