# 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.

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.

• This clarifies things a bit! So $V(S^{d-1}) = A(B^d)$? Where $V$ is understood to relate to the $(d-1)$-dimensional volume measure for the $(d-1)-$sphere, and similarly $A$ is related to the $(d-1)$-dimensional volume measure for the $d-$ball? Commented Mar 3, 2020 at 13:06
• Is it correct to say that the 3-volume of the 3-sphere is $4\pi R^2$? Commented Mar 3, 2020 at 13:07
• @TheoreticalConfusion The 3-sphere is 3-dimensional, so the formula should have an $R^3$ at the end. Looking at wikipedia en.wikipedia.org/wiki/N-sphere the constant term is $2\pi^2$ so the volume is $2\pi^2R^3$. Commented Mar 3, 2020 at 13:13
• You are, of course, absolutely right. The 2-volume of the 2-sphere is $4\pi R^2$. Many thanks! Commented Mar 3, 2020 at 13:15
• @quarague Quick clarification re. final paragraph: you say that surface area refers to the $(d-1)-$dimensional measure of the boundary of a $d-$manifold. $S^d$ has no boundary, so does this mean we can't ascribe a genuine surface area to $S^d$? Your post appears to define the surface area of the $d-$ball (and therefore the volume of $S^{d-1}$) rather than the surface area of $S^d$.
– FH93
Commented Mar 3, 2020 at 17:19