6.2 Volumes of Revolution

Figure 6.4: A surface of revolution.

Figure 6.5: The volume of a small disc.

Again divide the area into strips of width δx. Since the height is f(x), when we rotate the strip we get a disc of radius r = f(x), see Fig. 6.5. The area of this disc is π{r}^{2} = πf{(x)}^{2}, and the volume of the disc is δV = π{r}^{2}δx. The total volume is again a sum,

Now take limit where δx becomes infinitesimal, and thus

This is the formula for the volume of a solid of revolution.

Example 6.3: 

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Figure 6.6: The surface of revolution for y = (1∕x), 1 < x < 2.

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Find the volume formed when the curve y = 1∕x, between x = 1 and x = 2 is rotated around the x axis, see Fig. 6.6

Solution: 

\begin{eqnarray*} V & =& π{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{1}^{2}{(1∕x)}^{2}\kern 1.66702pt dx %& \\ & =& π{\left (−1∕x\right )}_{1}^{2} %& \\ & =& π(−(1∕2) − (−1))%& \\ & =& π∕2\quad . %& \\ \end{eqnarray*}

Example 6.4: 

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Figure 6.7:

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Find the volume formed when equilateral triangle with corners at O = (0,0), A = (1,\sqrt{3}), B = (2,0) is rotated around the x axis, see Fig. 6.7.

Solution: 

Along OA the curve is y = \sqrt{3}x, along AB the curve is y = 2\sqrt{3} −\sqrt{3}x. Thus

\begin{eqnarray*} V & =& π{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{1}{(\sqrt{3}x)}^{2}\kern 1.66702pt dx + π{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{1}^{2}{(2\sqrt{3} −\sqrt{3}x)}^{2}\kern 1.66702pt dx%& \\ & =& 3π{\left ({x}^{3}∕3\right )}_{ 0}^{1} + 3π{\left (−{(2 − x)}^{3}∕3\right )}_{ 1}^{2} %& \\ & =& π + π(0 + 1) %& \\ & =& 2π. %& \\ \end{eqnarray*}