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 $\dfrac{\pi}{4}\left(6^2+x^2\right)$.
Instead of tetrahedron, let us work out a general formula for $n$-simplex first.
Given any non-degenerate $n$-simplex $S$ with vertices $v_1, \ldots, v_{n+1}$. Let
- $V$ be the volume of $S$.
- $\ell_{ij} = \| v_i - v_j \|$ be the edge lengths.
- $c$ and $R$ be the center and radius for the circumsphere.
- $\lambda_1, \ldots, \lambda_{n+1}$ be the barycentric coordinates of $c$ with respect to $S$.
i.e the list of $n+1$ real numbers such that
$$c = \sum_{i}\lambda_i v_i\quad\text{ with }\quad \sum_{i}\lambda_i = 1$$
Recall for any two points $u, v$ with barycentric coordinates $\mu_i, \nu_i$ with respect to $S$.
Their distance is given by the formula:
$$\| u - v \|^2 = -\sum_{i < j}\ell_{ij}^2(\mu_i - \nu_i)(\mu_j - \nu_j)$$
Apply this to circumcenter $c$ and vertex $v_k$, we obtain
$$R^2 = -\sum_{i<j}\ell_{ij}^2 (\lambda_i - \delta_{ki})(\lambda_j - \delta_{kj}) = \sum_{j}\ell_{kj}^2 \lambda_j - \frac12 \sum_{i,j} \ell_{ij}^2\lambda_i\lambda_j$$
Multiply by $\lambda_k$ and sum over $k$, we obtain
$$R^2 = \frac12\sum_{i,j} \ell_{ij}^2 \lambda_i\lambda_j
\quad\implies\quad
2R^2 = \sum_{j}\ell_{kj}^2 \lambda_j\;\;\text{ for all }k\tag{*1}$$
To proceed, rewrite everything in matrix form. Let
- ${\bf \theta}$ be the $(n+1)\times 1$ column vector with all entries $1$.
- ${\bf \lambda}$ be the $(n+1)\times 1$ column vector with entries $\lambda_i$.
- $\Lambda$ be the $(n+1)\times(n+1)$ matrix with entries $\ell_{ij}^2$.
- $\Delta$ be the $(n+2)\times(n+2)$ matrix
$\begin{bmatrix}0 & {\bf \theta}^T \\ {\bf \theta} & \Lambda\end{bmatrix}$
In terms of them, $(*1)$ becomes $2R^2{\bf \theta} = \Lambda {\bf \lambda}$.
Together with ${\bf \theta}^T{\bf \lambda} = \sum_{i} \lambda_i = 1$, we obtain
$$\Delta \begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix}
= \begin{bmatrix}0 & {\bf \theta}^T \\ {\bf \theta} & \Lambda\end{bmatrix}
\begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix}
= \begin{bmatrix} 1 \\ 0\end{bmatrix}
\quad\implies\quad
\begin{bmatrix} -2R^2 \\ {\bf \lambda}\end{bmatrix} = \Delta^{-1}\begin{bmatrix} 1 \\ 0\end{bmatrix}
$$
Extracting the first component of both sides and notice $\Delta^{-1} = (\det\Delta)^{-1} {\rm adj}\Delta$, we get
$$-2R^2 = \frac{\det\Lambda}{\det\Delta}$$
Notice $\det\Delta$ is simply the Cayley-Menger determiant for $S$ which satisfies
$$V^2 = \frac{(-1)^{n+1}}{2^n n!^2} \det\Delta$$
This means the volume $V$, the circumradius $R$ and $\det\Lambda$ of a $n$-simplex are related by following formula:
$$\bbox[border:1px solid blue;padding: 16px;]{(-1)^n 2^{n+1}(n!)^2 (VR)^2 = \det\Lambda}\tag{*2}$$
For example, when $n = 2$, the $n$-simplex becomes a triangle. If $a, b, c$ are
the side lengths of a triangle. $(*2)$ reduces to the familiar relation among its area, circumradius and side lengths.
$$32V^2R^2 = \det
\begin{bmatrix}
0 & a^2 & b^2\\
a^2 & 0 & c^2\\
b^2 & c^2 & 0
\end{bmatrix}
= 2a^2b^2c^2
\quad\iff\quad
4VR = abc
$$
Back to the original problem of tetrahedron which corresponds to $n = 3$.
Let $a, b, c$ be the lengths of the three edges attached to a vertex. Let $a_1, b_1, c_1$ be the lengths of corresponding opposite edges. $(*2)$ becomes
$$-576(VR)^2 =
\det\begin{bmatrix}
0 & a^2 & b^2 & c^2\\
a^2 & 0 & c_1^2 & b_1^2\\
b^2 & c_1^2 & 0 & a_1^2\\
c^2 & b_1^2 & a_1^2 & 0
\end{bmatrix}
$$
Reproducing the complicated formula appeared in the link in Edward Jiang's comment.
To simplify further, we first multiply
$2^{nd}/3^{rd}/4^{th}$ row/column of $\Lambda$ by $\frac{bc}{a}, \frac{ac}{b},
\frac{ab}{c}$ and then divide the $1^{st}$ row/column or resulting matrix by $abc$. At the end, we find
$$\det\Lambda = \det\begin{bmatrix}
0 & 1 & 1 & 1\\
1 & 0 & (cc_1)^2 & (bb_1)^2\\
1 & (cc_1)^2 & 0 & (aa_1)^2\\
1 & (bb_1)^2 & (aa_1)^2 & 0
\end{bmatrix}
$$
This has the form of a Cayley-Menger determinant for a "triangle" with sides $aa_1$, $bb_1$ and $cc_1$. Recall Heron's formula for computing area of triangle, $\det\Lambda$ can be factorized as follows:
$$\det\Lambda = -16 p(p-aa_1)(p-bb_1)(p - cc_1)\quad\text{ where }\quad p = \frac{aa_1 + bb_1 + cc_1}{2}$$
As a result, we obtain a much simpler relation for tetrahedron.
$$\bbox[border:1px solid blue;padding: 16px;]{6VR = \sqrt{p(p-aa_1)(p-bb_1)(p-cc_1)}}\tag{*3}$$
Update
I recently come across an elegant proof of formula $(*3)$ using sphere inversion (the 3d-counterpart of circle inversion in 2d). It explains why the RHS
is the area of a triangle with sides $aa_1$, $bb_1$, $cc_1$.
Let $OABC$ be a tetrahedron with sides $OA = a, OB = b, OC = c, BC = a_1, CA = b_1, AB = c_1$. Let $V$ and $R$ be its volume and circumradius.
Let $A', B', C'$ be the image of $A, B, C$ under a sphere inversion with respect to the sphere centered at $O$ with radius $\rho = \sqrt[3]{abc}$.
Notice tetrahedron $OA'B'C'$ is sharing a trihedral angle at $O$ with $OABC$.
Since
$$|OA'||OB'||OC'| = \frac{\rho^2}{a}\frac{\rho^2}{b}\frac{\rho^2}{c} = abc = |OA| |OB| |OC|$$
The volume of tetrahedron $OA'B'C'$ also equals to $V$.
Under the sphere inversion, the circumsphere of $OABC$ get mapped at a plane at a distance $\frac{\rho^2}{2R}$ from $R$. The points $A'$, $B'$, $C'$ belong to this plane and they are forming a triangle with sides $B'C' = \frac{a_1\rho^2}{bc}, C'A' = \frac{b_1\rho^2}{ac}, A'B' = \frac{c_1\rho^2}{ab}$. If $\Delta(u,v,w)$ is the area of a triangle of sides $u,v,w$, then
$$V = \frac{1}{3}\frac{\rho^2}{2R}\Delta\left(
\frac{a_1\rho^2}{bc},
\frac{b_1\rho^2}{ac},
\frac{c_1\rho^2}{ab}
\right)
= \frac{1}{6R}
\Delta\left(
\frac{a_1\rho^3}{bc},
\frac{b_1\rho^3}{ac},
\frac{c_1\rho^3}{ab}
\right)
= \frac{1}{6R}
\Delta(aa_1,bb_1,cc_1)
$$
Together with Heron's formula, $(*3)$ follows.
Best Answer
Are you aware of Eratosthenes' experiment?
A variation of this would involve taking a wedge formed by two planes and putting the sphere between the planes and measuring the distance from the corner of the wedge. Could you work the math for this example?