Area and Volume Formulas — Question 5

Question 5 - Solution

We derive the formula using the idea of approximating the solid with thin disks.

Partition the interval $[a,b]$ into subintervals $$a=x_0<x_1<\cdots<x_n=b,$$ and let $$\Delta x_i=x_i-x_{i-1}.$$

Choose a sample point $x_i^*$ in each subinterval $[x_{i-1},x_i]$.

Over a small interval $[x_{i-1},x_i]$, the graph of $y=f(x)$ is approximately constant at height $f(x_i^*)$. When this vertical strip is revolved about the $x$-axis, it forms a thin circular disk.

The radius of the disk is $$r_i=f(x_i^*).$$

The area of the circular face of the disk is $$A_i=\pi r_i^2=\pi [f(x_i^*)]^2.$$

Since the thickness of the disk is $\Delta x_i$, its volume is approximately $$\Delta V_i=\pi [f(x_i^*)]^2\,\Delta x_i.$$

Adding the volumes of all disks gives an approximation to the total volume: $$\sum_{i=1}^n \pi [f(x_i^*)]^2\,\Delta x_i.$$

As the partition is refined and the maximum subinterval length approaches zero, these disks more accurately fill the solid. Because $f$ is continuous on $[a,b]$, the limit of the sum exists.

Taking the limit yields $$V=\lim_{\|P\|\to 0}\sum_{i=1}^n \pi [f(x_i^*)]^2\,\Delta x_i =\pi\int_a^b [f(x)]^2\,dx.$$

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