Find the area of the region that lies inside both r=2sin(th) and r=sin(th).
I think I have to put it in the form of integral 1/2(r1)^2 - (r2)^2 d(th) but I'm not sure what the bounds would be. I set them equal to each other and I got the values pi/2 and 3pi/2.
I assumed the integral would be from pi/2 to 3pi/2 1/2(2sin(th)^2 - sin(th)^2)d(th), could someone confirm? I'm not sure about the bounds
On
Both the polar equations represent circles with second one inside the first. The first equation $r=2\sin\theta$ represents the circle $$x^2+(y-1)^2=1$$ with center $(0, 1)$ and radius $1$ and the second equation $r=\sin\theta$ represents the circle $$x^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{4}$$ with center $(0, \frac{1}{2})$ and radius $r=\frac{1}{2}$
So the area of the region inside both is the area of the second circle which is $\dfrac{\pi}{4}$.
Area = $$\frac{1}{2}\int_0^{2\pi}\big( 4\sin^2 \theta - \sin^2 \theta) d\theta = \frac{3}{2} \int_0^{2\pi} \sin^2 \theta d\theta$$.
Use the trig identity $\sin^2 \theta = \frac{1}{2}(1-cos2\theta)$ to help evaluate the integral. It helps to graph the two polar equations, this may give you a better view for your bounds of integration.