# Determine the area of such a portion of the surface

Suppose two spheres of radius $r > 0$ are positioned with centers at distance $2r\cos\theta$ from one another, where $0<\theta<\frac \pi2$. These spheres intersect in a circle, and a portion of each sphere lies in the interior of the other. Determine the area of such a portion of the surface.

I was wondering if I did this right: I started with writing the formulas of the 2 spheres:

1st sphere: $x^2 + y^2 + z^2 = r^2$

2nd sphere $(x+2r\cos \theta)^2 + y^2 + z^2 = r^2$

Then I equated the 2 equations and solved for $x$, and got $x = -r\cos\theta$

From this point on do I just plug $-r\cos\theta$ in and take the partial derivatives, then integrate with respect to $x$ and $y$ to get the surface area?

$$S=2(\theta r^2-\frac12r^2\sin2\theta)=r^2(2\theta-\sin2\theta)$$
I would suggest that you get the projection of the intersection on $zy$-plane to get the double integral limits. Do double integration on one of the sphere with that limit and multiply by 2.