Area and Volume Formulas — Question 4

Question 4 - Solution

Take a partition $a=x_0<\cdots<x_n=b$ and let $m_i,M_i$ be the minimum and maximum of $f$ on $[x_{i-1},x_i]$.

Since $a\geq0$, the annular cylinders occupy disjoint radial intervals. Their volumes bound the desired volume:

$$\sum_i\pi(x_i^2-x_{i-1}^2)m_i\ \leq V\ \leq\sum_i\pi(x_i^2-x_{i-1}^2)M_i.$$

Let $c_i=(x_i+x_{i-1})/2$. Then $\pi(x_i^2-x_{i-1}^2)=2\pi c_i\Delta x_i$.

Uniform continuity of $f$ makes $\max_i(M_i-m_i)\to0$ as the mesh tends to zero, so the difference between the upper and lower bounds tends to zero.

Both bounds approach the Riemann integral of $2\pi x f(x)$. Hence

$$\boxed{V=2\pi\int_a^b x f(x)\,dx.}$$