Volumes of Revolution

Take the graph of y = f (x) on the interval [a, b] and spin it round the x-axis so as to produce what is known as a solid of revolution as shown in Fig. 2.9. We want to get a formula for the volume of this solid.

Figure 2.9: A volume of revolution

The method is almost exactly the same as in the previous examples. Think of the interval [a, b] being subdivided into lots of little bits. Now look at one of the bits and try to get an approximation for the volume of the `thin slice' of the solid obtained by rotating the piece of the graph on this interval.

Figure 2.10: The volume of a slice.

25cm

In the notation of the diagram, the thin slice of the solid is virtually a cylinder of radius y and thickness $\delta$x. The volume of a cylinder is the product of its height and the area of its base. So we get the approximation

$$\delta$$V = $$\pi$$y² $$\delta$$x

for the volume of the slice.

The approximation to the total volume can then be written as

V $$\approx$$ $$\sum$$$$\pi$$y² $$\delta$$x

Now take the limit as n$\to$$\infty$ and get

$$\boxed{ \text{Volume} = \int_a^b \pi y^2 dx }$$

Example 2.9   Cone

Take the line y = mx on [0, h] and spin it round the x-axis so as to produce a cone of height h and `semi-angle' $\alpha$, where tan$\alpha$ = m.

The volume of this cone is, by our formula,

V = $$\int_{0}^{h}$$$$\pi$$m²x² dx = $$\left[\vphantom{\pi m^2\frac{1}{3}x^3}\right.$$$$\pi$$m²$${\textstyle\frac{1}{3}}$$x³$$\left.\vphantom{\pi m^2\frac{1}{3}x^3}\right]_{0}^{h}$$ = $${\textstyle\frac{1}{3}}$$$$\pi$$m²h³

If R is the radius of the base of the cone then m = tan$\alpha$ = R/h so, with a bit of rearranging, we get

V = $${\textstyle\frac{1}{3}}$$$$\pi$$R²h = $${\textstyle\frac{1}{3}}$$ base × height

Example 2.10   Sphere

Take the semicircle y = $\sqrt{r^2-x^2}$ on [- r, r] and spin it round the x-axis. We get, as our solid of revolution, a Sphere of radius r.

Our formula tells us that the volume of this sphere is

V = $$\int_{-r}^{r}$$$$\pi$$(r² - x²) dx = $$\left[\vphantom{ \pi r^2x - \frac{1}{3}\pi x^3}\right.$$$$\pi$$r²x - $${\textstyle\frac{1}{3}}$$$$\pi$$x³$$\left.\vphantom{ \pi r^2x - \frac{1}{3}\pi x^3}\right]_{-r}^{r}$$ = $${\textstyle\frac{4}{3}}$$$$\pi$$r³

So the volume of a sphere of radius r is

V = $${\textstyle\frac{4}{3}}$$$$\pi$$r³

Example 2.11   Consider the funnel formed by taking the curve y = 1/x and rotating it round the x-axis on the interval [1, a], as shown in Fig. 2.11

Figure 2.11: Volume of rotation.

The volume of this funnel is

V = $$\int_{1}^{a}$$$$\pi$$$${\frac{1}{x^2}}$$ dx = $$\left[\vphantom{-\frac{\pi}{x}}\right.$$ - $${\frac{\pi}{x}}$$$$\left.\vphantom{-\frac{\pi}{x}}\right]_{1}^{a}$$ = $$\pi$$(1 - $${\frac{1}{a}}$$)

Now notice that, as a$\to$$\infty$, this volume tends to the finite value $\pi$ (because $\pi$/a$\to$ 0).

We write

$$\int_{1}^{\infty}$$$${\frac{\pi}{x^2}}$$ dx = $$\lim_{a\to\infty}^{}$$$$\int_{1}^{a}$$$${\frac{\pi}{x^2}}$$ dx = $$\lim_{a\to\infty}^{}$$$$\left(\vphantom{ \pi-\frac{\pi}{a}}\right.$$$$\pi$$ - $${\frac{\pi}{a}}$$$$\left.\vphantom{ \pi-\frac{\pi}{a}}\right)$$ = $$\pi$$