Is there something missing in the usual calculation by integral for hyper-surface volume of a $3$-D ball? Where is my $4$-D intuition going wrong about hypersurface volume of a $3$-D ball?
There are plenty of examples of how to calculate the hypersurface volume and hypervolume of a $3$-D ball (eg. wikipedia) which clearly infer that for a unit $3$-D ball ($r = 1$) the hyper-volume is
$$\frac{1}{2} \pi^2 = 4.935$$
Also the volume of its $3$-D hyper-surface is
$$2 \pi^2 = 19.739 \tag{1}$$
And the volume of a $2$-D ball is
$$\frac{4}{3} \pi = 4.189 \tag{2}$$
Also in higher dimensions volume and surface are are said to decrease with increasing dimension, which I do get. But I have for a long time thought that something was missing in the integration process, because it appears to me (who gave up on maths $40$ years ago) that we go from one $2$-D ball to one 3 ball, ignoring the other $2$-D balls needed, which would double in number for each dimension added after $4$.
My intuitional imagining of the $3$-D ball tells me it has [is composed of, in part] four $2$-D balls, each with unit radius, and each with one coordinate value of 0. Indeed we would see $x=y=z=1$ with $w=0$ as a simple $2$-D ball in $x,y,z$, and I assume there must also be $w=x=y=1$ with $z=0$ and so on for $x=0$ and $y=0$. 
These would have a combined volume of
$$4 \cdot 4.189 = 16.756 \tag{3}$$ 
Subtracting from $(1)$ the total of $19.739$, gives a difference of
$$2.983 \tag{4}$$
Also, my intuitional imagining says there is a minimum radius 2 ball that is a surface at the “center” of the $3$-D ball, where $w,x,y$ and $z$ are equal to $\sqrt{0.25}r = 0.5r$. 
Actually that there are $4$ of these superimposed, each has a volume of $\frac{4}{3} \pi r^3 = 0.523$ so the total is $2.094$.
Subtracting from $(3)$ leaves only $0.888$  for the rest of the hyper-surface, which doesn’t seem like very much.
So where am I going wrong, or is $0.888$ enough of a figleaf to cover it?
Or is it the case that $(2)$ is not included in $(1)$? 
 A: Using wikipedia notation, I guess that:


*

*$\frac 12\pi^2$ refers to $V_4(1)$ which is the $4-$dimensional volume of the (unit) $4-$ball,

*$2\pi^2$ refers to $S_3(1)$, which is the $3-$dimensional volume of the (unit) $3-$sphere,

*$\frac 43\pi$ refers to $V_3(1)$, which is the $3-$dimensional volume of the (unit) $3-$ball.


Now if I've understood your question/reasoning, you're listing some $3-$balls of varying radii that are contained within the (unit) $4-$ball. From there, you sum up the $3-$volumes of these $3-$balls and substract it from the $3-$volume of the $3-$sphere. Then, you observe that the resulting value is small and that something is wrong.

The short answer is that as you've surmised, $(2)$ is not included in $(1)$, at least not in the way you're manipulating those values. If we go down one dimension, back in good old regular $3$D, it should be simpler to vizualize what you tried to compute:


*

*$S_2(1) = 4\pi$ is the surface of a unit sphere,

*$V_2(1) = \pi$ is the surface of a unit disk.


Inside of a unit ball, you can easily fit three unit disks, one for $x=0$, another for $y=0$, and $z=0$. That's already $3\pi$'s worth of surface, so you're left with only one $\pi$ worth of surface to fill the whole unit ball... except that you don't fill a ball with "surfaces", you're supposed to use "volumes". Also, basing your computation on the surface of the sphere will not help you here. Although the surface of a sphere, and the volume of the ball that it delimits, share somewhat of a relationship, these two values represent two different things of very different nature. In maths, you'll never obtain a volume by just adding together surfaces.
In the real world, if you stack enough sheets of paper, you'll get thick books. That's because no matter how thin your sheets are, they will still have some width/thickness to them. In maths, an "ideal" $2$D disk has zero width. So no matter how many disks you stack, you'll be stuck at zero width.
Back to your computation, I'll try to give you some understanding of what you did. Let's stay in $3$D. Say you bought a spherical chocolate Easter egg for your kids, but you forgot it in your car, and it melted into a (weirdly-shaped, but) perfectly flat tablet. Since you don't want to waste the chocolate, you decide to repackage it. To do that, you take a cookie-cutter, and cut circular disk shapes into your (weird) chocolate tablet. Since the shape is irregular, you get a couple of nice, circular disks, but also a bunch of leftover chocolate. That is what you computed.
Figuring out how to split a sphere into a bunch of non-overlapping disks (or other shapes) can be an interesting topic, but it has, a priori, little to do with how to split a ball into disks. Even more so with mathematical disks with zero width/height.
