Question

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?

Answer 1

`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 dx`. The volume of the entire solid would be

`int_2^6 pi y^2 dx= int_2^6 pi (1/x- 1/x^2)^2 dx`

=`pi int_2^6 (1/x^2 -2/x^3 +1/x^4)dx`

=`pi[-1/x +1/x^2 -1/(3x^3)]_2^6`

=`pi[-1/6 +1/36 -1/648 +1/2 -1/4 +1/24]=49/324`