Volumes of revolution.

Proposition 8.7.1   Let $f$ be a continuous function on the interval $[a,b]$ , with $a<b$ . Suppose that for any $x \in [a,b]$ , $f(x) \geq 0$ . The volume of the solid of revolution obtained by revolving about the $x-$ axis the region between the $x-$ axis and the graph of $f$ id equal to:

$$V= \pi \int_a^b [f(x)]^2 \; dx$$

Figure 5: A solid of revolution.

Example 8.7.2   Revolve about the $x-$ axis the arc of parabola defined by $y=x^2, \; 0 \leq x \leq 2$ (v.s. Fig. 6.4). The volume of the ``horn'' is given by:

$$V= \pi \int_0^2 (x^2)^2 \; dx = \pi \int_0^2 x^4 \; dx = \pi \left[ \frac 15 x^5 \right]_0^2 = \frac {32 \pi}{5}.$$

If the region $R$ to be revolved is not bordered by the $x-$ axis, the section of the solid is not a disk, but an annulus.

Figure 6: A washer.

Denote by $f$ the function whose graph is the outer boundary of the region $R$ , and by $g$ the function whose graph is the inner boundary of $R$ ; suppose that both functions are continuous on $[a,b]$ , where $a<b$ .

Proposition 8.7.3   The volume of the solid obtained by revolving the region $R$ about the $x-$ axis is given by:

$$V=\pi \int_a^b [ f(x)^2 - g(x)^2 ] \; dx.$$

Example 8.7.4  

We revolve about the $x-$ axis the region bounded by the parabolas whose respective equations are $y=x^2+1$ and $y=\frac 12 x^2 + \frac 14$ , by the $y-$ axis and the line whose equation is $x=2$ , i.e. the region $ABCD$ in Figure  7.

Figure 7: A washer between two parabolas.

The volume of the solid is given by:

$$V = \pi \int_0^2 \left[ (x^2+1)^2-\left( \frac 12 x^2 + \frac 14 \right)^2 \right] \; dx$$

$$\quad =\pi \int_0^2 \left[ \frac 54 x^4 +\frac 94 x^2 + \frac {17}{16} \right] \; dx$$

$$\quad =\pi \left[ \frac 14 x^5 + \frac 34 x^3 + \frac {17}{16} x \right]_0^2 =\frac {129 \pi}{8}.$$