# Area between two polar curves $r = 2 \sin\theta$ and $r =2\cos\theta$

I am trying to find the area between the polar curves $r = 2 \sin θ$ and $r = 2 \cos θ$.

I set up the area equation as follows:

$$\frac12\int_0^{\pi/4}((2\sinθ)^2-(2\cosθ)^2)\,d\theta$$

I could not get the correct answer with this, which is $\frac\pi2-1$.

Any help with this problem would be appreciated :D

• i accidentally put the wrong correct answer, it is (pi/2)-1 – nshah Apr 7 '14 at 23:33

A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.

The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$\underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1.$$

this is just a hint and you have to do it by yourself so see below..... by using 12th method so we know by polar co-ordinates $x= r \sin(x)$ and $y=r \cos(x)$ here in question

$r=2\sin(x)$

multiply both side $r$

$r^2=2r\sin(x)$

$r^2 = 2x$ since $x=r \sin(x)$

since a point circle equation is $x^2+y^2=r^2$

therefore $x^2+y^2=2x$ implies that $y=\sqrt (2x-x^2)$ similarly for $r=2\cos(x)$, $x^2+y^2=2y$ implies that $y= \sqrt (2y-x^2)$ and integrate it by drawing the circle

get the answer. best of luck