Jordan measure of a ball Currently I am trying to prove the following sentence:
Let $B(0,r)$ be an open ball. Show that $B(0,r)$ is Jordan measurable, with the Jordan measure
      $$m_{J} (B) = c_d \cdot r^d$$
      for some constant $c(d)$ depending only on the dimension $d$, such that 
$$\left(\frac{2}{\sqrt{d}}\right)^d \leq c_d \leq 2^d$$
For me a key observation was that the ball $B^{(d)}(0,r)$ is the same as the area under/above the graph of 
$$\pm f:B^{(d-1)}(0,r)\to\mathbb{R}, \quad x\mapsto \pm\sqrt{r^2-\|x\|^2}$$
I wrote a relatively lengthy proof that by induction shows that the ball/area under this graph is Jordan measurable (by building an outer cover and the inner cover that do not differ in measure by more than $\epsilon$). 
However, when it comes to establishing $c_d$ and its quantitative bounds, I am completely stuck. All of the sources I found in internet (such as this) use integrals to establish some kind of quantitative bounds, but I am not allowed to use them at all. Please help!
 A: Note that $B(0,1) \subset (-1,1)^d= B_\infty(0,1)$ and $m (-1,1)^d = 2^d$.
Note that if $x \in (-L,L)^d$ then $\|x\| \le \sqrt{d}L$ and so
$(-{1 \over \sqrt{d}}, {1 \over \sqrt{d}})^d \subset B(0,1)$ and 
$m (-{1 \over \sqrt{d}}, {1 \over \sqrt{d}})^d = ({ 2 \over \sqrt{d}})^d$.
Hence $({ 2 \over \sqrt{d}})^d \le c_d \le 2^d$.
Note:
If a set $A$ is Jordan measurable, then for $r>0$ the set $rA = \{ra | a \in A\}$ is also Jordan measurable. If $R$ is a rectangle, it is easy to see that $m (rR) = r^d \cdot m(R)$, so it follows that $m (rA) = r^d \cdot m(A)$.
In particular, $m(B(0,r)) = r^d \cdot m(B(0,1))$.
The measure of $B(0,1)$ only depends on the dimension $d$.
Addendum:
I struggled to come up with a short proof of Jordan measurability. My first attempt was circular. The following is reasonable:
It is sufficient to show that $S=\partial B(0,1)$ has content zero.
Clearly it is sufficient to show that $S_+ = \{ x \in S | x_d \ge 0 \}$ has content zero.
Define $f:[-2,2]^{d-1} \to \mathbb{R}$ by
$f(y) = \begin{cases} \sqrt{1-\|y\|^2}, & \|y\| \le 1 \\
0, & \text{otherwise} \end{cases}$. Since $f$ is continuous, it is Riemann integrable. Note that $S_+ \subset \operatorname{graph} f$, where
$\operatorname{graph} f = \{ (y,f(y)) | y \in [-2,2]^{d-1} \}$.
Since $f$ is Riemann integrable, for any $\epsilon>0$ there is some partition
$P$ such that $U(f,P)-L(f,P) < \epsilon$. Note that
$\operatorname{graph} f \subset \cup_{R \in P} R \times [\inf_R f, \sup_R f]$
and $m(\cup_{R \in P} R \times [\inf_R f, \sup_R f]) = U(f,P)-L(f,P) < \epsilon$.
Hence $S_+$ and consequently $S$ has Jordan content zero.
