Area of the part of a sphere Let $x \in \mathbb{R^3}$ and $t>0$. Also let $R>0$. Why is the area of the part of the sphere $\partial B(x,t)$ inside $B(0,R)$ smaller than the area of $\partial B(0,R)$:
$$|\partial B(x,t) \cap B(0,R)| \leq |\partial B(0,R)|$$
How can this be shown rigorously?   

If $t<R$ the result is obvious, let's then suppose $t>R$ and let the two spherical surfaces intersect along a circle of radius $r$ (if they don't intersect the result is once again trivial): the part of the larger sphere contained inside the smaller sphere is a spherical cap, the smaller of the two spherical caps on the larger sphere having the said circle as base.
The area of the smaller spherical cap of base radius $r$ is given by 
$S(\rho)=2\pi\rho(\rho-\sqrt{\rho^2-r^2})$, where $\rho$ is the radius of the sphere. This is a decreasing function of $\rho$ (when $r$ is fixed), so we have $S(t)<S(R)<4\pi R^2$. 
In other words: the area of the spherical cap of the larger sphere inside the smaller sphere is less than the spherical cap of the smaller sphere, which in turn is less than the area of the smaller sphere.
