Question

Sketch the graph and find the area of the region that lies inside both

r = 1 and r = 2sinθ.

Answer 1

intersection of both between two curves $sin{\theta}=1$

${\theta}=\dfrac{\pi}{6}$

The area between these two curve can be easily found with the help of double integeration whose formula is given by =

$\iint{r}drd{\theta}$

since we can see there is similarity between the region so if we find area of right hand side of the region then we can easily get region of left hand side .

$\int_{0}^{\dfrac{\pi}{6}}\int_{r=1}^{r=2sin{\theta}}$ $rdrd{\theta}$

now integerating this double integeration function , we get -

$\int_{0}^{\dfrac{\pi}{6}}$ $\dfrac {|r^{2}|_{r=1}^{r=2sin{\theta}}}{2}d{\theta}$

$\dfrac{1}{2}\int_{0}^{\dfrac {\pi}{6}}(4sin^{2}{\theta}-1)d{\theta}$

$\dfrac{1}{2}[\int_{0}^{\dfrac{\pi}{6}} 4sin^{2}{\theta}d{\theta}-\int _{0}^{\dfrac{\pi}{6}}d{\theta}]$

$sin^{2}{\theta}=\dfrac {1-cos2{\theta}}{2}$

$=\dfrac{1}{2}[\int_{0}^\dfrac{\pi}{6}(2-2cos2{\theta})d{\theta}-\int_{0}^\dfrac{\pi}{6}d{\theta}]$

$=\dfrac{1}{2}[[2{\theta}-sin2{\theta}]_{0}^{\dfrac{\pi}{6}}-\dfrac{\pi}{6}]$

$=\dfrac{\pi}{12}-\dfrac{1}{4}$

2(area ABCA) , we get whole area =$\dfrac{\pi}{6}-\dfrac{1}{2}$

This is area covered by region r=1 and r=$2sin{\theta}$ .