Confirm sphere volume is equalled by the sum of volumes of contiguous, nested spherical shells.

Question

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.

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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 $t^3$.

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Update

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.

Answer 1

METHOD 1:

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!

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METHOD 2:

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.

Answer 2

(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.