Area and Volume Formulas — Question 6

Question 6 - Solution

We derive the formula using washers (disks with holes).

Partition the interval $[a,b]$ as $$a=x_0<x_1<\cdots<x_n=b,$$ 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 near $x_i^*$, the region between the curves has an outer radius $$R_i=f(x_i^*)$$ and an inner radius $$r_i=g(x_i^*).$$

When this vertical strip is revolved about the $x$-axis, it forms a thin washer.

The area of the outer disk is $$\pi R_i^2=\pi [f(x_i^*)]^2,$$ and the area of the inner disk (the hole) is $$\pi r_i^2=\pi [g(x_i^*)]^2.$$

Thus, the area of the washer is $$\pi\bigl([f(x_i^*)]^2-[g(x_i^*)]^2\bigr).$$

Multiplying by the thickness $\Delta x_i$, the volume of the washer is $$\Delta V_i = \pi\bigl([f(x_i^*)]^2-[g(x_i^*)]^2\bigr)\Delta x_i.$$

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

As the partition is refined and $\|P\|\to 0$, the washers fill the solid more accurately. Since $f$ and $g$ are continuous on $[a,b]$, the limit of the sum exists.

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

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