Consider the following surface in $\mathbb{R^3}$:
$$\Sigma = S_r(x) \cap B_R(0)$$
Where $S_r(x)$ is a sphere of radius $r$ centered at $x$ and $B_R(0)$ is an open ball of radius $R$ centered at the origin.
I'd like to show that:
$$\text{area}( S_r(x) \cap B_R(0)) \leq \text{area}(S_R(0))$$
I tried to evaluate the following surface integral:
$$\int_{\Sigma} dS$$
But it's not clear to me how the surface $\Sigma$ can be parameterized. I assumed that $x = (p,0,0)$ and then use spherical coordinates. The expression I got is not very helpful.
Is there another easier way to prove that inequality?
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as: $$ \mathbf{r}(t) = \langle r \cos t, r\sin t\rangle $$ where $|t| \leq \alpha = \arcsin(\frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So $$ \text{length}= \int_{-\alpha}^{\alpha} \|\mathbf{r}'(t)\| \; dt = 2r\arcsin(\frac{R'}{r}) \leq 2r \arcsin(\frac{R}{r}) $$ since $R'\leq R$ and $\arcsin$ is increasing. Lastly you can use convexity of $\arcsin$ on $[0,1]$ to show that $$ \arcsin(\frac{R}{r})\leq \frac{\pi}{2}\frac{R}{r} $$ (convexity implies that $\arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,\pi/2)$). So we end up with $$ \text{length} \leq \pi R \leq 2\pi R $$ (twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.