Shell integration

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

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

${\displaystyle f(a+k\Delta x)\Delta x}${\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 ${\displaystyle a+(k+1)\Delta x}${\displaystyle a+(k+1)\Delta x} with height ${\displaystyle f(a+k\Delta x)}${\displaystyle f(a+k\Delta x)} , and substracting it another smaller cylinder of radius ${\displaystyle a+k\Delta x}${\displaystyle a+k\Delta x}, with the same height of ${\displaystyle f(a+k\Delta x)}${\displaystyle f(a+k\Delta x)} , this difference of cylinder volumes is:

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

${\displaystyle =\pi f(a+k\Delta x)((a+(k+1)\Delta x)^{2}-(a+k\Delta x)^{2})}${\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:

${\displaystyle \pi f(a+k\Delta x)(2a+2k\Delta x+\Delta x)\Delta x}${\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:

${\displaystyle 2\pi f(a+k\Delta x)(a+k\Delta x+{\frac {\Delta x}{2}})\Delta x}${\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:

${\displaystyle \lim _{n\to \infty }2\pi \sum _{i=1}^{n}f(a+i\Delta x)(a+i\Delta x+{\frac {\Delta x}{2}})\Delta x}${\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 ${\displaystyle n\rightarrow \infty }${\displaystyle n\rightarrow \infty }, we can clearly identify that:

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

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

${\displaystyle 2\pi \int \limits _{a}^{b}xf(x)dx}${\displaystyle 2\pi \int \limits _{a}^{b}xf(x)dx}

QED ${\displaystyle \square }${\displaystyle \square }.