volume of solid of revolution

Let us consider a solid of revolution, which is generated when a planar domain $D$ rotates about a line of the same plane. We chose this line for the $x$-axis, and for simplicity we assume that the boundaries of $D$ are the mentioned axis, two ordinates   $x=a$,  $x=b(>a)$, and a continuous curve   $y=f(x)$.

Between the bounds $a$ anb $b$ we fit a sequence of points  $x_1,x_2,\mathrm{\dots },x_{n-1}$  and draw through these the ordinates which divide the domain $D$ in $n$ parts. Moreover we form for every part the (maximal) inscribed and the (minimal) circumscribed rectangle. In the revolution of $D$, each rectangle generates a circular cylinder. The considered solid of revolution is part of the volume $V_{>}$ of the union of the cyliders generated by the circumscribed rectangles and at the same time contains the volume of the union of the cylinders generated by the inscribed rectangles.

Now it is apparent that

$$V_{>}=\pi [M_1^2(x_1-a)+M_2^2(x_2-x_1)+\mathrm{\dots }+M_n^2(b-x_{n-1})],$$

where  $M_1,M_2,\mathrm{\dots },M_n$  are the greatest and  $m_1,m_2,\mathrm{\dots },m_n$  the least values of the continuous function $f$ on the intervals (http://planetmath.org/Interval)   $[a,x_1]$,  $[x_1,x_2]$, …, $[x_{n-1},b]$. The volume $V$ of the solid of revolution thus satisfies

and this is true for any     of the interval  $[a,b]$.The theory of the Riemann integral guarantees that there exists only one real number $V$ having this property and that it is also the definition of the integral ${\displaystyle {\int }_a^b}\pi {[f(x)]}^2𝑑x.$ Therefore the volume of the given solid of revolution can be obtained from

$$V=\pi {\int }_a^b{[f(x)]}^2𝑑x.$$