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:

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

Figure 5: A solid of revolution.
\begin{figure}\mbox{\epsfig{file=VolumeRevolution.eps,height=4cm}}\end{figure}

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:

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

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.
\begin{figure}\mbox{\epsfig{file=AnnulusRevolution.eps,height=4cm}}\end{figure}

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:

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

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.
\begin{figure}\begin{center}
\mbox{\epsfig{file=washer.eps,height=4cm}}\end{center}\end{figure}

The volume of the solid is given by:

$\displaystyle V$ $\displaystyle = \pi \int_0^2 \left[ (x^2+1)^2-\left( \frac 12 x^2 + \frac 14 \right)^2 \right] \; dx$    
$\displaystyle \quad$ $\displaystyle =\pi \int_0^2 \left[ \frac 54 x^4 +\frac 94 x^2 + \frac {17}{16} \right] \; dx$    
$\displaystyle \quad$ $\displaystyle =\pi \left[ \frac 14 x^5 + \frac 34 x^3 + \frac {17}{16} x \right]_0^2 =\frac {129 \pi}{8}.$    

Noah Dana-Picard 2007-12-28