Let me first state the statement which I want to prove (encountered while studying "Geometry of Number"):
Suppose $A$ is a convex, measurable, compact and centrally symmetric subset of $\mathbb{R}^n$ where $\mathbf{x}=(x_1, \ldots, x_r, x_{r+1}, \ldots, x_n)\in A$ iff $|x_i|\leq 1$ for $1\leq i \leq r$ and $x_{r+1}^2 + x_{r+2}^2 \leq 1, \ldots, x_{n-1}^2 + x_{n}^2 \leq 1$
Then $\text{vol}(A) = 2^r \pi^{\frac{n-r}{2}}$
The definition of various terms used above :
$A\subset \mathbb{R}^n$ is called a convex set if $\mathbf{x}$ and $\mathbf{y}$ are in $A$ then so is the entire line segment joining them.
Measurable refers to Lebesgue measure in $\mathbb{R}^n$; the Lebesgue measure $\text{vol}(A)$ coincides with any reasonable intuitive concept of n-dimensional volume, and Lebesgue measure is countably additive.
Centrally symmetric means symmetric around $\mathbf{0}$: if $\mathbf{x} \in A$ then so is $-\mathbf{x}$.
I have no idea about how to approach this problem.
Edit: It is also given that $n-r$ is an even number.
To keep things simple we will do some induction: Set $A_r = \{(x_1, \dots, x_r) \in \mathbb R^r : \forall i \in \{1,\dots, n\}: \vert x_i \vert \leq 1\}$. Consider the integral $\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r)$, we will prove by induction that $$\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) = 2^r.$$ For $n = 1$ we get $$\int_{A_1} 1 \ d x_1 = \int_{-1}^1 1\ dx_1 = 2.$$ Now let the statement hold for $r \in \mathbb N$. Then we get for $r + 1$ with Fubini that $$\int_{A_{r+1}} 1 \ d\lambda(x_1, \dots, x_{r+1}) = \int_{-1}^1 \bigg(\int_{A_r} 1 \ d\lambda(x_1, \dots, x_r) \bigg)dx_{r+1} = \int_{-1}^1 2^r\ dx_{r+1} = 2^{r +1}.$$ Now to the second part. Let $k := n - r$ and $B_{k} = \{(x_{r+1}, \dots, x_n) \in \mathbb R^{n-r} : x_{r+1}^2 + x_{r+2}^2 \leq 1, \ldots, x_{n-1}^2 + x_{n}^2 \leq 1 \}$ for even $k$. Now we prove by induction that $$\int_{B_k} 1 \ d\lambda(x_{r+1}, \dots, x_n) = \pi^{k/2}.$$ For $k = 2$ we get by using polar coordinates $$ \int_{B_2} 1\ d\lambda(x_{r+1}, x_{r+2}) = \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \int_0^1 2\pi r\ dr = \pi.$$ Now let the statement hold for $k \in \mathbb N$ even. Then we get for $k + 2$ with Fubini that \begin{align*}\int_{B_{k+2}} 1\ d\lambda(x_{r+1}, \dots, x_{n + 2}) &= \int_{B_2} \bigg( \int_{B_{k}} 1\ d\lambda(x_{r+1}, \dots, x_n)\bigg) d\lambda(x_{n+1}, x_{n + 2}) = \int_{B_2} \pi^{k/2}\ d\lambda(x_{n+1}, x_{n + 2}) \\ &= \pi^{k/2} \int_0^1 \int_0^{2 \pi} r\ d\varphi dr = \pi^{k/2} \pi = \pi^{(k + 1)/2}\end{align*} All together we can get know with Fubini \begin{align*}\operatorname{vol}(A) &= \int_A 1\ d\lambda(x_1, \dots, x_n) = \int_{A_r} \bigg(\int_{B_k} 1\ d\lambda(x_{r+1}, \dots, x_n) \bigg) d\lambda(x_1, \dots, x_r) \\ &= \int_{A_r} \pi^{k/2}\ d\lambda(x_1, \dots, x_r) = \pi^{k/2} \int_{A_r}1 \ d\lambda(x_1, \dots, x_r) = \pi^{k/2} 2^r.\end{align*} And that was the result we were aiming for.
The "physics" version: Denote the unit disk by $D = \{x \in R^2 : x_1^2 + x_2^2 \leq 1\}$. Then $A = [-1,1]^r \times D^{(n - r)/2}$. Show like above that $\operatorname{vol}([-1,1]) = 2$ and $\operatorname{vol}(D) = \pi$. Then we got $$\operatorname{vol}(A) = \operatorname{vol}([-1,1]^r \times D^{(n - r)/2}) = \operatorname{vol}([-1,1])^r \operatorname{vol}(D)^{(n - r)/2} = 2^r \pi^{(n - r)/2}.$$ This prove makes use of the fact you can write $A$ as $A = [-1,1]^r \times D^{(n - r)/2}$ and the relation between the one-dimensional Lebesgue measure and the multi-dimensional Lebesgue measure.
Hope it helps :)