Evaluating the area between the curves $r=2\sin\theta$ and $r=\sin\theta+\cos\theta$ 
So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad  r=\sin\theta +\cos\theta $$

I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder to me.
$$\begin{align*}
x&=r\cos\theta=\cos\theta(\sin\theta + \cos\theta)\\
y&=\sin\theta(\sin\theta+\cos\theta)
\end{align*}$$
so $x+y=1+2\cos\theta \sin\theta$....Which gives me no help.
I'm getting used to polar coordinates, but I need some help! Thank you

To clarify the common area, you need to plot two graphs. Both are circles. 
$\cos{\theta} + \sin{\theta} = \sqrt{2}\sin(\theta + \pi/4)$. So draw a circle $r = \sqrt{2} \sin {\theta}$, and rotate it $\pi/4$ clockwise. 
Then, it is the circle whose cernter is (1/2,1/2) and the radius is $\frac{\sqrt{2}}{2}$.
Purely algebraically, $r= \sqrt{x^2 + y^2}$, $\cos\theta = \frac{x}{\sqrt{x^2+y^2}}$, and $\sin\theta = \frac{y}{\sqrt{x^2+y^2}}$. 
Then $\sqrt{x^2 + y^2 } = \frac{x}{\sqrt{x^2+y^2}} + \frac{y}{\sqrt{x^2+y^2}}$, i.e. $x^2 + y^2 = x + y$. 
So $(x-\frac{1}{2})^2 + (y - \frac12)^2 = \frac12$.
