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?