Volumes and surface areas for the $d-$sphere $S^d$ rather than the $d-$ball. I've been struggling with the concept of volume and surface area for the $d-$sphere $S^d$. For concreteness, I can give the $d-$sphere of radius $R$ an extrinsic definition by embedding it in $(d+1)-$dimensional Euclidean space:
$$S^d = \{x\in \mathbb{R}^{d+1}: |x|=R\}$$
(although I can't see how one defines this intrinsically, any suggestions?)
$S^d$ is a d-dimensional manifold without boundary.
My confusion begins from here:


*

*Does it make sense to talk about the volume of $S^d$ given that it's merely a spherical shell and isn't "solid" in the conventional sense? Presumably, it's volume should be zero when viewed as an embedding in $R^{d+1}$? Note: I stress, I'm not asking for the volume of the region in $\mathbb{R}^{d+1}$ enclosed by this spherical surface, but rather the volume of the shell $S^d$ itself. What is the precise definition for the volume of a manifold without boundary?

*Does it make sense to speak of a surface area of $S^d$ given that it has no boundary itself? What is the precise definition for the surface area of a manifold without boundary?
Note, I can see both concepts being totally clear for the $(d+1)-$ball because:


*

*The $(d+1)-$ball is "solid" in $(d+1)-$dimensional space and has a boundary $S^d$, so the concept of volume is intuitively clear to me here. 

*The surface area of the $(d+1)-$ball is equally easy to visualise given that it possesses a boundary, and one would get the usual "surface area of a sphere". But, surely, the surface area of $B^{d+1}$ and the surface area of $S^{d}$ cannot be defined in the same way, whilst also giving the same answer?


I just can't make sense of this question when I consider $S^d$ as a manifold in it's own right, rather than through embedding. For instance, the volume of the closed unit disk $B^2$ in two dimensions is clearly $V(B^2)=\pi R^2$, and the surface area is corresponding the circumference of the boundary $S^1$: $A(B^2) = 2\pi R$.
My question is, what are $V(S^1)$ and $A(S^1)$ exactly? Is $V(S^1) = A(B^2)$? If there are issues with the small dimensionality here, one can ask the same question for $B^3$ and $S^2$.
I appreciate that this question is very muddled, so please fire away for any clarifications. It has very been painful to search the internet for answers to this.
 A: I think the issue comes from the meaning of the words 'volume' and 'area'. In every day life volume is 3-dimensional and area is 2-dimensional and you never need to go higher than that so there is no confusion.
If you have a d-dimensional manifold, the natural thing to consider is its d-dimensional measure, this is usually called the volume for any d but if there is more than one manifold around you have to be careful which dimension of measure you mean. The d-dimensional sphere can be defined through an embedding in $\mathbb{R}^{d+1}$ but you could also define it abstractly through charts which are all maps to $\mathbb{R}^d$ without ever mentioning $\mathbb{R}^{d+1}$. The d+1-volume of the d-sphere is zero but the d-volume is not. 
Similarly the word surface or area is often used for the d-1-dimensional measure of the boundary of a d-dimensional manifold. So you can consider the d-dimensional unit ball, its boundary is the d-1-dimensional sphere and the area of the boundary of the ball is the volume of the d-1-dimensional sphere.
Edit: I would say talking of the surface of a d-dimensional manifold without boundary is missleading and ideally shouldn't be used. The d-sphere has a (d-dimensional) volume but it has no boundary so at most you could say the area of its boundary is zero.
