Consider the function f ( x ) {\displaystyle f(x)} which describes our cross-section of the solid, now the integral of the function can be described as a Riemann integral: ∫ a b f ( x ) d x = lim n → ∞ ∑ i = 1 n f ( a + i Δ x ) Δ x {\displaystyle \int \limits _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{i=1}^{n}f(a+i\Delta x)\Delta x}

Where Δ x = b − a n {\displaystyle \Delta x={\frac {b-a}{n}}} is a small difference in x {\displaystyle x}

The Riemann sum can be thought up as a sum of a number n of rectangles with ever shrinking bases, we might focus on one of them:

f ( a + k Δ x ) Δ x {\displaystyle f(a+k\Delta x)\Delta x}

Now, when we rotate the function around the axis of revolution, it is equivalent to rotating all of these rectangles around said axis, these rectangles end up becoming a hollow cylinder, composed by the difference of two normal cylinders. For our chosen rectangle, its made by obtaining a cylinder of radius a + ( k + 1 ) Δ x {\displaystyle a+(k+1)\Delta x} with height f ( a + k Δ x ) {\displaystyle f(a+k\Delta x)} , and substracting it another smaller cylinder of radius a + k Δ x {\displaystyle a+k\Delta x} , with the same height of f ( a + k Δ x ) {\displaystyle f(a+k\Delta x)} , this difference of cylinder volumes is:

π ( a + ( k + 1 ) Δ x ) 2 f ( a + k Δ x ) − π ( a + k Δ x ) 2 f ( a + k Δ x ) {\displaystyle \pi (a+(k+1)\Delta x)^{2}f(a+k\Delta x)-\pi (a+k\Delta x)^{2}f(a+k\Delta x)}

= π f ( a + k Δ x ) ( ( a + ( k + 1 ) Δ x ) 2 − ( a + k Δ x ) 2 ) {\displaystyle =\pi f(a+k\Delta x)((a+(k+1)\Delta x)^{2}-(a+k\Delta x)^{2})}

By difference of squares , the last factor can be reduced as:

π f ( a + k Δ x ) ( 2 a + 2 k Δ x + Δ x ) Δ x {\displaystyle \pi f(a+k\Delta x)(2a+2k\Delta x+\Delta x)\Delta x}

The third factor can be factored out by two, ending up as:

2 π f ( a + k Δ x ) ( a + k Δ x + Δ x 2 ) Δ x {\displaystyle 2\pi f(a+k\Delta x)(a+k\Delta x+{\frac {\Delta x}{2}})\Delta x}

This same thing happens with all terms, so our total sum becomes:

lim n → ∞ 2 π ∑ i = 1 n f ( a + i Δ x ) ( a + i Δ x + Δ x 2 ) Δ x {\displaystyle \lim _{n\to \infty }2\pi \sum _{i=1}^{n}f(a+i\Delta x)(a+i\Delta x+{\frac {\Delta x}{2}})\Delta x}

In the limit of n → ∞ {\displaystyle n\rightarrow \infty } , we can clearly identify that:

• f ( a + i Δ x ) {\displaystyle f(a+i\Delta x)} as Δ x {\displaystyle \Delta x} tends to 0 ends up becoming f ( x ) {\displaystyle f(x)}
• ( a + i Δ x + Δ x 2 ) {\displaystyle (a+i\Delta x+{\frac {\Delta x}{2}})} becomes x {\displaystyle x} itself, going from a to b (ignoring the last term which vanishes)
• Δ x {\displaystyle \Delta x} becomes the infinitesimal d x {\displaystyle dx}

Thus, at the limit of infinity, the sum becomes the integral:

2 π ∫ a b x f ( x ) d x {\displaystyle 2\pi \int \limits _{a}^{b}xf(x)dx}

QED ◻ {\displaystyle \square } .