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?   

Let $C = B(0, R)$ and $X = B(x, t)$ where we rotated the scene to have 
$x = (d,0,0)$.

The very easy case is $C \cap X = \emptyset$ (empty intersection). This happens if $d-t > R$, because the left-most point of $X$, which is $(d-t,0,0)$ is separated from the right-most point of $C$, which is $(R,0,0)$.

The easy case is $X \subseteq C$ ($X$ fits in $C$). To fit the left- and right-most points of $X$ within $C$ we need $-R \le d-t$ and $d+t\le R$.
This means $\pm d \le R-t$, so $0 \le R-t$ and $t \le R$, and 
$$
A_X = 4\pi t^2 \le 4\pi R^2 = A_C
$$
For reversed roles $C \subseteq X$, then $C \cap X = C$ and the inequality holds equal.

Now the general case: We attempt to calculate the area within the intersection. For the area calculation we will use spherical coordinates,
for this we rotate and translate the scene such that $X$
centers around the origin and $C$ around $(0,0,d)$.

We need the upper bound for the polar angle $\theta_i$.
For points in $I = \partial C\cap \partial X$ we have
$$
x^2 + y^2 + z^2 = t^2 \\
x^2 + y^2 + (z - d)^2 = R^2
$$
The difference equation is
$$
-2zd + d^2 = R^2 - t^2 \Rightarrow \\
z_i = \frac{d^2 - R^2 + t^2}{2d}
$$
where $z_i$ is the $z$-coordinate of the intersection $I$, which shows up as red line or red plane $z=z_i$ in the images above.
Then $I$ is given by
$$
x^2 + y^2 + z_i^2 = t^2
$$
which is a circle with radius $r = \sqrt{t^2-z_i^2}$.
This gives
$$
\cos\theta_i = z_i/t
$$
For the surface of $X$ within $C$ we get
$$
A = 
\int\limits_0^{2\pi}\!\!d\varphi
\int\limits_0^{\theta_i}\!\! d\theta \, t^2 \sin\theta
= 2\pi t^2 [-\cos\theta]_0^{\theta_i}
= 2\pi t^2 (1-\cos\theta_i)
= 2\pi t^2(1-z_i/t)
$$
From the sphere equation for $X$ we have $\lvert z\rvert \le t \iff z\in [-t,t]$, which limits $z_i$ as well and this means 
$$
0 \le A \le 4\pi t^2 = A_X
$$
as expected. Now we need to compare $t$ and $R$. 
If $t\le R$ we have $A \le A_X \le 4\pi R^2 = A_C$ and we are done.
