Zero-volume sets We are going to define integration on manifolds, I guess, and we first gave a definition for zero-volume sets. 
Defn:$X \in \mathbf{R^n}$ is called a $\textit{zero-volume set}$ if for every $\epsilon > 0$, there is a set of finite number of rectangles $R_1 \ldots R_s$ so that $X\subseteq \cup_{i=1}^{s}R_i$ and
$$\sum_{i=1}^{s} volume(R_i) < \epsilon$$
This definition bugs me a little. I think of it like submanifolds. For example, a circle in $\mathbf{R}^2$ or $\mathbf{R}^3$ is a zero-volume set and a sphere in $\mathbf{R}^3$ is not a zero-volume set.
In general higher dimensions, $\mathbf{R}^n$, this definition confuses me. Is a 3D sphere in, let's say $\mathbf{R}^8$, a zero-volume set? It has a volume after all, I can compute it using its radius. But It might have $0$ hypervolume in $\mathbf{R}^8$, I am not sure about this.  
So, should this definition be read as hyperrectangles and hypervolumes in a general $\mathbf{R}^n$ setting? This definition seems so geometric, or let's say qualitative, that I find it hard to construct such hyperrectangles in higher dimensions and bound the total hypervolume at the same time.
NOTE: The suggested similar questions are all about "measure zero" and I don't yet know what it is.
 A: Both your notion of rectangle and volume change in dimension. 
In $\mathbb{R}^d$ the correction notion of a rectangle is that of a $d-$dimensional box $[a_1,b_1]\times ... \times[a_d,b_d]$. In dimension $1$ this is an interval, in dimension $2$ it coincides with an actual rectangle and in dimension $3$ this is a box.
In $\mathbb{R}^d$ you have to use the volume $\text{vol}_d$ defined by $\text{vol}_d([a_1,b_1]\times ... \times[a_d,b_d])=\Pi_{i=1}^d(b_i-a_i)$, this is the so called $d-$dimensional Lebesgue-measure. In dimension $1$ we also call it length, in dimension $2$ we call it area and in dimension $3$ we usually just call it volume.
Thus a subsest $S\subset \mathbb{R}^d$ has zero $d-$dimensional volume, if and only if for every $\epsilon>0$ it is covered by $d-$boxes $R_1,...,R_k$ with $\text{vol}_d(R_1) + ... \text{vol}_d(R_k) \le \epsilon$. (Edit: If $S$ is unbounded you need to allow for infinitely many $d-$boxes. Consider for example the line $S=\{(x,0): x\in \mathbb{R}\} \subset \mathbb{R}^2$, then there is simply no finite collection of rectangles that covers $S$. On the other hand you can choose the rectangles $\{R_i\}_{i=0}^\infty$ with $R_j=[-2^{j-1},2^{j-1}]\times[-2^{-2j-1}\epsilon,2^{-2j-1}\epsilon]$, then $\sum_i\text{vol}_2(R_i) = \epsilon$, thus $S$ indeed has volume zero.)
That means a circle in $\mathbb{R}^2$ has $2-$dimensional volume $0$ and also a $2-$sphere in $\mathbb{R}^3$ has $3$-dim volume $0$. In general you can show that if $S \subset \mathbb{R}^d$ is a submanifold of dimension $m < d$, then it's $d-$dimensional volume is $0$. 
Your intuition of course tells you that for example the $2-$sphere in $\mathbb{R}^3$ has a positive $2-$dimensional volume, but defining this is a little tricky. One way to define the $k-$dimensional volume of a subset $S\subset \mathbb{R}^d$ is using the Hausdorff-measure of dimension $k$, this means that you cover $S$ by smaller and smaller open $d-$balls, and measure how their diameters to the $k-$th power behave in the limit. 
If $S$ happens to be a submanifold, one can also define the volume in terms of differential forms. This breaks down to cutting it up into pieces $S_1, ... , S_l$, each of which is isometric to an open set $T_j \subset\mathbb{R}^k$, then you put $\text{vol}_k(S) = \text{vol}_k(T_1) + ... +  \text{vol}_k(T_l)$. This gives the same result as the Hausdorff-measure.
