What is the volume of the solid produced by revolving f(x)=1/x-1/x^2, x in [2,6] around the x-axis?

Dec 18, 2016

$\frac{49}{324}$

Explanation:

The portion of the graph of f(x) in the interval [2,6] looks like in the figure below:

If this portion is revolved around x-axis, the volume of the solid so generated can be worked out as follows:

Consider an element of the area enclosed by the x-axis and the curve, at a distance x from the origin and length y= f(x). If this is rotated about x-axis a circular disc of area $\pi {y}^{2}$ If width of this elementary disc is dx, its volume would be $\pi {y}^{2} \mathrm{dx}$. The volume of the entire solid would be

${\int}_{2}^{6} \pi {y}^{2} \mathrm{dx} = {\int}_{2}^{6} \pi {\left(\frac{1}{x} - \frac{1}{x} ^ 2\right)}^{2} \mathrm{dx}$

=$\pi {\int}_{2}^{6} \left(\frac{1}{x} ^ 2 - \frac{2}{x} ^ 3 + \frac{1}{x} ^ 4\right) \mathrm{dx}$

=$\pi {\left[- \frac{1}{x} + \frac{1}{x} ^ 2 - \frac{1}{3 {x}^{3}}\right]}_{2}^{6}$

=$\pi \left[- \frac{1}{6} + \frac{1}{36} - \frac{1}{648} + \frac{1}{2} - \frac{1}{4} + \frac{1}{24}\right] = \frac{49}{324}$