# Volume of a solid of revolution

Let $[a,b]$ be an interval, $a\geq 0$ and $f:[a,b]\to \mathbb{R}_+$ continuous.

I want to calculate the volume of the solid of revolution obtained by rotating the area below the graph of $f$ around the $y$-axis. The result should be $$2\pi\int_{a}^bxf(x)~dx.$$

For $h,r,t\geq 0$ the volume of a cylinder of radius $r+t$ in which a centred cylinder of radius $r$ is removed, both of height $h$, is $$\pi(r+t)^2h-\pi r^2h=\pi h(2rt+t^2).$$

This formula in mind, it seems reasonable to me that the volume of the solid is $$\begin{array}{rl} \lim_{k\to\infty} ~\sum_{i=1}^k\pi f(a+i{\Tiny \frac{b-a}{k}})(2(a+i{\Tiny \frac{b-a}{k}}){\Tiny\frac{1}{k}}+{\Tiny\frac{1}{k^2}})&=\\ \pi\lim_{k\to\infty} ~\sum_{i=1}^k\left(\left( f(a+i{\Tiny \frac{b-a}{k}})2(a+i{\Tiny \frac{b-a}{k}}){\Tiny\frac{1}{k}}\right)+\left(f(a+i{\Tiny \frac{b-a}{k}}){\Tiny\frac{1}{k^2}}\right)\right)& \end{array}$$ With the 'definition' $\int_{a}^b g(x)dx=\lim_{k\to\infty} ~\sum_{i=1}^k f(a+i{\Tiny \frac{b-a}{k}}){\Tiny\frac{1}{k}}$, the first 'summand' of the infinite sum looks exactly like the solution integral. Why does the second summand disappear?

-
@Ana Lytics: It is because in the second term there is a $k^2$ in the denominator. – André Nicolas Jun 29 '11 at 19:16

If $f$ is continuous on $[a,b]$, then $|f(x)| \leq M$ for all $x \in [a,b]$, for some $M > 0$ fixed. Hence, $$\Bigg|\frac{{\sum\nolimits_{i = 1}^k {f(a + i\frac{{b - a}}{k})} }}{{k^2 }}\Bigg| \le \frac{{Mk}}{{k^2 }} = \frac{M}{k},$$ which tends to $0$ as $k \to \infty$.
The second summand dissapears because you have the factor $1/k^2$. Group in the following way: $$\lim_{k \to \infty} \frac{1}{k} \sum_{i=1}^k\frac{1}{k}f(a+i\frac{b-a}{k})$$
The sum converges to an integral, something finite, and divided by $k$ it converges to $0$.