Finding Area between curves Find the area enclosed between the circles:
$x^2 + (y- 3.5)^2 = (3.5)^2$
and
$(x-3.5)^2 + y^2 = (3.5)^2$ 
I tried using definite Integrals but I was unable to frame the equation. Please help me
 A: HINT
Take a look to the graph 

and use that for a quarter


*

*$A_{circle}=\frac{\pi R^2}4$

*$A_{triangle}=\frac12R^2$

A: Sometimes graphing helps a lot. You can easily verify that the circles meet at the Origin and at $(3.5,3.5)$. You can also (with or without calculus) verify that at both points of intersection, the circles met at a right angle (Orthogonal intersection); that is, their respective tangents meet at a right angle. Now when you connect the points of intersection with a line segment, you "split up" the region they share into two equal regions. Now with some formulas of circle segment and circle sector, this is no longer a challenge
A: You need to find the points where they intersect, since you already fund that to be $(0,0)$ and $(3.5,3.5)$ I'll skip that step;
next  to find the area you find the definite integral ;
$A = \displaystyle\int_0^{3.5}\bigg(\sqrt{(3.5)^2-(x-3.5)^2}+\sqrt{(3.5)^2-x^2}-3.5\bigg)\,dx$
A: You can use simple geometry, like in the answers provided by other people. If you must use integrals, @gimusi provided a figure for you. Integrate $y_U(x)-y_L(x)$ for $x$ between $0$ and $3.5$. $y_U$ is the equation of the upper curve, $y_l$ is the equation of the lower curve.
