I am having trouble in confirming that the volume of a sphere $V=(4\pi/3) R^3$ is equal to the sum of the volumes of a set of contiguous, nested spherical shells.

I start by obtaining the following expression for the volume $W_x$ of a single spherical shell with outer radius $x$ and radial thickness $t$ based upon the difference in volumes of the outer sphere and inner sphere:-

$$W_{x(t)} = (4\pi/3) x^3 - (4/3)\pi (x-t)^3 $$ $$W_{x(t)} = (4\pi/3) ( x^3 - (x-t)^3) $$ $$W_{x(t)} = (4\pi/3) ( x^3 - (x^3-3tx^2+3t^2x-t^3)) $$ $$W_{x(t)} = (4\pi/3) ( 3tx^2 -3t^2x +t^3) $$.

This formula can be checked by setting $t=x=R$ giving:-
$$V = W_{R(R)} = (4\pi/3) ( 3R^3 -3R^3 +R^3) = (4\pi/3) R^3 $$ which is the volume $V$ of a sphere with radius $R$, as expected.

Next I wish to obtain the volume $V$ of the sphere (radius $R$) by summing the volumes of a series of contiguous, nested spherical shells of common thickness $t$. $$V_R = \Sigma_t^R W_{x(t)} = (4\pi/3) \Sigma_t^R ( 3tx^2 -3t^2x +t^3) $$ $$V_R = 4\pi \Sigma_t^R \left[ tx^2 -t^2x +(1/3)t^3 \right]. $$ This formula has been checked by numerical testing. Now I wish to go from the above "discrete shell formula" to an integral expression of the form:- $$ V = \int _0^R G(x) dx = (4 \pi /3)R^3 $$ I realize that setting the function $G(x) = 4 \pi x^2$ will give an anti-derivative function $F(x) = (4 \pi/3)x^3$.

But, here is my question: how do I get from the "discrete shell formula" to a suitable form of the integral formula involving $G(x)$?

In particular I am unsure how to handle the "trailing terms" containing $t^2$ and $t^3$.


I have learned that it is not essential to obtain an integral formula. The exact formula $V=(4/3)\pi R^3$ can be obtained using power sums (see Dr.MV's answer (Method 1) and my own answer).

A.G. referred to the multiple thin shell approximation method for deriving sphere volume using the formula for sphere surface area. I have applied the power sum approach to that method in this answer.

  • 3
    $\begingroup$ Note that $4\pi x^2$ is exactly the area of a sphere of radius $x$. If you know that, so you can approximate the volume of a thin spherical shell of radial thickness $dx$ by $4\pi x^2\,dx$ and adding up those volumes for "all $x$" gives you a Riemann sum that converges to $\int_0^R 4\pi x^2\,dx$. $\endgroup$
    – A.Γ.
    Jul 19 '15 at 18:33
  • $\begingroup$ @A.G. Thanks, I am familiar with that approach. The approximation is equivalent to taking the first term of my discrete shell formula. But I want justification that the trailing terms converge to zero. $\endgroup$
    – steveOw
    Jul 19 '15 at 19:08
  • $\begingroup$ Do you not mean the summing of surfaces? $\endgroup$ Jul 19 '15 at 20:37
  • $\begingroup$ The whole idea of splitting an apriori known volume into a finite number of known volumes and then sum them up is a bit strange. It does not help to understand integration unless some kind of approximation is involved. An interesting step would be if you did not assume knowing the ball volume and tried to figure out approximations to calculate it using, for example, Riemann sums. $\endgroup$
    – A.Γ.
    Jul 19 '15 at 21:02
  • $\begingroup$ @A.G. Yes I see your point. But I am not trying to prove the formula for a sphere...just trying to prove/confirm that the sum of the parts is equal to the whole when I using a particular formula for the volume of a part. As I have now learned, there is no need for any approximation/calculus/integration. $\endgroup$
    – steveOw
    Jul 19 '15 at 21:17


We will use the following Power Sums

$$\begin{align} \sum_{k=1}^{n}\,1&=n \tag 1\\\\ \sum_{k=1}^{n}\,k&=\frac12\,n(n+1) \tag 2\\\\ \sum_{k=1}^{n}\,k^2&=\frac16\,n(n+1)(2n+1) \tag 3 \end{align}$$

The approximated volume $W$ by summing the volume of spherical shells is given by the sum

$$W(n)=\frac{4\pi}{3}\sum_{k=1}^n\left(3tx_k^2-3t^2x_k+t^3\right) \tag 4$$

where $t=R/n$ and $x_k=kR/n$. We can rewrite $(4)$ as

$$W(n)=\frac{4\pi}{3}\frac{R^3}{n^3}\sum_{k=1}^n\left(3k^2-3k+1\right) \tag 5$$

whereupon using $(1)-(3)$ in $(5)$ reveals that

$$\begin{align} W(n)&=\frac{4\pi}{3}\frac{R^3}{n^3}\left(\left(n^3+\frac32 n^2+\frac12 n\right)- \left(\frac32 n^2+\frac32 n\right) +(n) \right)\\\\ &=\frac{4\pi R^3}{3} \end{align}$$

Finally, we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}W(n)=\frac{4\pi R^3}{3}}$$

as expected!


As an alternative, we may rewrite $(4)$ in terms of Riemann sums as

$$W(n)=\frac{4\pi}{3}\left(\sum_{k=1}^n\, 3x_k^2\times (t)-(t)\,\sum_{k=1}^n 3x_k \times (t)+(t)^2\,\sum_{k=1}^n\,1\,\times (t)\right)$$

Note that the first term is the Riemann sum for $3x^2$, the second term it $t$ times the Riemann sum for $3x$, and the third term is $t^2$ times the Riemann for $1$. Inasmuch as the $t\to 0$ as $n\to \infty$, the second and third terms vanish in the limit. Thus,

$$\lim_{n\to \infty}W(n)=\frac{4\pi }{3}\int_0^R 3x^2\,dx=\frac{4\pi R^3}{3}$$

and therefore we have

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}W(n)=\frac{4\pi R^3}{3}}$$

as in the previous development.

  • $\begingroup$ Ah yes I have just been looking at power sums myself but you have beaten me to it! $\endgroup$
    – steveOw
    Jul 19 '15 at 20:30
  • $\begingroup$ @steveOw If you have a look at METHOD 2, you will see that we can forgo the need for the power sums and use Riemann sums instead. Kind of cool I hope. $\endgroup$
    – Mark Viola
    Jul 19 '15 at 20:35
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Jul 19 '15 at 20:36
  • 1
    $\begingroup$ @johannesvalks Thanks!!! And I really appreciate the nice comment! $\endgroup$
    – Mark Viola
    Jul 19 '15 at 20:43
  • 1
    $\begingroup$ @Dr.MV In the next formula after (5) a minus sign is missing in the middle term. The formula for $W(n)$ should not depend on $n$ since it is an exact sum of the exact shell volumes. $\endgroup$
    – A.Γ.
    Jul 19 '15 at 20:54

(Note: this answer is essentially the same as Dr.MV's corrected Method 1. If you wish to upvote, please upvote Dr. MV's answer, not mine).

We have $V_r = 4\pi \sum_{x=t}^{x=R}(tx^2-t^2x+(1/3)t^3)$ which can be expressed as:-

$$ V_r = 4\pi \sum_{x=t}^{x=R} tx^2 - 4\pi \sum_{x=t}^{x=R} t^2x + 4\pi \sum_{x=t}^{x=R} (1/3)t^3 $$

$$ V_r = 4t\pi \sum_{x=t}^{x=R} x^2 - 4t^2\pi \sum_{x=t}^{x=R} x + (4\pi/3)t^3 \sum_{x=t}^{x=R} 1 $$

We can define $t=R/N$ and thus $Nt=R$. Also we can replace $x$ by $it$. So we can modify the expression to:- $$ V_r = 4t\pi \sum_{i=1}^{i=N} i^2t^2 - 4t^2\pi \sum_{i=1}^{i=N} it + 4t^3\pi/3 \sum_{i=1}^{i=N} 1 $$ and moving the $t$ terms in front of the sums $$ V_r = 4t^3\pi \sum_{i=1}^{i=N} i^2 - 4t^3\pi \sum_{i=1}^{i=N} i + 4t^3\pi/3 \sum_{i=1}^{i=N} 1 $$

Using the known power sums (as per Dr. MV's answer) this becomes:- $$ V_r = 4t^3\pi (1/6)(2N^3+3N^2+N) - 4t^3\pi (N^2+N)/2 + 4Nt^3\pi/3 $$ So $$ V_r = 4t^3\pi \left[ (1/6)(2N^3+3N^2+N) - (N^2+N)/2 + N/3 \right] $$ then $$ V_r = 4t^3\pi \left[ \frac{N^3}{3}+\frac{N^2}{2}+\frac{N}{6} - \frac{N^2}{2} -\frac{N}{2} + \frac{N}{3} \right] = 4t^3\pi \left[ \frac{N^3}{3} \right]$$ and, using $Nt=R$, we obtain:- $$ V_r = \frac{4}{3}\pi R^3$$ as expected.

  • $\begingroup$ It should not be surprising given that the method presumes a volume for the shells and then proves that summing volumes gives the total volume. A different way to go here would be to assume a known value for the surface area $4\pi R^2$ and then sum $4\pi R^2 t$ to see what volume results. It has more than the one-term answer. $\endgroup$
    – Mark Viola
    Jul 19 '15 at 21:17
  • $\begingroup$ @Dr.MV Quite so. My original problem was that I could not work out how to make the sum of the parts equal the known value of the whole. $\endgroup$
    – steveOw
    Jul 19 '15 at 21:33
  • $\begingroup$ @Dr. MV, regarding "summing $4\pi R^2t$ to see what volume results" I have now done this in my answer to another question. $\endgroup$
    – steveOw
    Jul 20 '15 at 22:21

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