Integrals and area of circle I have no idea where to start this one guys: 
Write a definite integral whose value is the area of the region between the two circles: 
$$x^2+y^2=1$$
$$(x-1)^2+y^2=1$$
Do I need to find my intersection points and then the integrals? 
 A: 
Easy way

Area is given by $$A=4\int_{1/2}^{1}\sqrt{1-x^2}\,\mathrm dx$$


Hard way

By solving these two equations for $x$ we find that they Intersect at $x=\dfrac{1}{2}$
Area is given by $$A=2\left[\int_0^{1/2}\sqrt{1-(x-1)^2}\,\mathrm dx+\int_{1/2}^{1}\sqrt{1-x^2}\,\mathrm dx\right]$$
hope you can proceed from here

Here's graphical explanation of multiplying by $2$


A: It seems to me $\int dy$ might work better. 
So, start with sketching each circle, and find the points where they intersect (by solving the two equations simultaneously). One circle is centered at the origin, the other at $(1,0)$, each has radius 1. They intersect when $x=\frac12$, $y=\pm \frac{\sqrt{3}}2$ (look at the nice picture that @Integrator posted in their answer, $A(\frac12,\frac{-\sqrt{3}}2)$, $B(\frac12,\frac{\sqrt{3}}2)$). Instead of splitting the integrals as @Integrator did into two integrals (which is fine), you could get away with doing just one integral $\int\limits_{\frac{-\sqrt{3}}2}^{\frac{\sqrt{3}}2}...dy$, but then  you need to solve for $x$ in terms of $y$ each of the two equations, and pick a suitable branch in each case. From the first equation you get $x=\pm \sqrt{1-y^2}$, use the $+$ branch as the "curve at the right". From the second $x=1\pm \sqrt{1-y^2}$, use the $-$ branch as the "curve at the left". So then compute 
$\int\limits_{\frac{-\sqrt{3}}2}^{\frac{\sqrt{3}}2}\sqrt{1-y^2}-(1-\sqrt{1-y^2})\ dy= \int\limits_{\frac{-\sqrt{3}}2}^{\frac{\sqrt{3}}2} 2\sqrt{1-y^2}-1\ dy$. You should get the same answer both ways, using the integral in the answer by @integrator, or using the integral in my answer. 
Though ... @Integrator (with help from @abel) beat me, found a way to do just one integral even with $\int...dx$ :) And, just for the record, the integrals (either answer) evaluate to $\frac{2\pi}3-\frac{\sqrt{3}}2$ . 
A: Just make a sketch first if you cannot figure it out at first,after sketching you will see clearly what is going on.
