geometry
It is well known that for the area of a triangle $A$ we have
$$ A=r\cdot s,$$
where $s$ is the semiperimeter, and $r$ is the radius of the inscribed circle.
Is there an analogue for the higher-dimensional case. In other words, can I express the volume of a $d$-simplex in terms of the radius of its inscribed sphere and the volume of its boundary? If such a formula exists, what are the references?
For the sphere problem, if $x$ is the length of the side which is nearly $8$, the volume $V(x)$ of the sphere is $\dfrac{4\pi}{3}r^3$, where $r(x)=\dfrac{\sqrt{36+x^2}}{2}$.
Substituting in the formula for the volume of the sphere, and simplifying a bit, we obtain $$V(x)=\frac{\pi}{6}(36+x^2)^{3/2}.$$ We want to use the tangent line approximation to approximate $V(x)$ when $x$ is near $8$. Differentiate. We get $$V'(x)=\pi\frac{x}{2}(36+x^2)^{1/2}.$$ Here the Chain Rule was used.) Set $x=8$. At $x=8$, the derivative is $40\pi$.
The tangent line approximation says that $$V(a)\approx V(8)+40\pi(x-8).$$ If we know that "$x=8\pm h$," where $h$ is positive and small, then by the tangent line approximation we have approximated $V(8)$ within roughly $\pm 40\pi h$.
The area problem is technically easier, since the area is given by the simple expression $\dfrac{\pi}{4}\left(6^2+x^2\right)$.
$$dV=(dr)(rd\theta)(rcos\theta d\phi)=[(r^{2}cos\theta dr) d\theta] d\phi$$
$$\rightarrow V=\int \int \int [(r^{2}cos\theta dr) d\theta] d\phi$$
So I guess if you integrate w.r.t dr,d$\theta$ you get a disk.Now to 'rotate' this disk integrate the result w.r.t d$\phi$
Note: Using this method one can find that there are 6 ways to do this integration to find out the volume.(each way is unique and interesting)
Best Answer
The same holds, that in $\mathbb{R}^n$, $A = r \frac {V_b}{n}$, where $A$ is the volume of your simplex, and $V_b$ is the volume of your boundary.
To see why this works, simply show that a simplex with base of volume $B$ and height $r$ has volume $ r \frac {B}{n}$, and then add up over all faces.
Hint: The constant $\frac {1}{n}$ comes from $\int x^{n-1}\, dx = \frac {x^n}{n}$.