Question

The region under the curve `y=lnx/x^2, 1<=x<=2` is rotated about the x axis. How do you find the volume of the solid of revolution?

Answer 1

See below.

Explanation:

To find the volume of revolution, we use the idea of a series of discs. We find the volume of each disc and these are summed together. From the diagram, we can see that the rectangle has a height of `ln(x)/x^2` and a width of `Deltax`. This rectangle is rotated around the x axis through an angle `pi` and results in a disc. The volume of this disc will then be:

`V=pi(ln(x)/x^2)^2*Deltaxcolor(white)(88)` ( `V=pir^2h`)

We then sum all the disc in the interval together to find the total volume of the solid. This is the same idea as when we find the area between a curve and the axes.

`:.`

`int_(1)^(2)(ln(x)/x^2)^2dx`

`(ln(x)/x^2)^2=(ln(x)^2)/x^4`

`V=piint_(1)^(2)(ln(x)/x^2)^2dx=-(9(ln(x))^2+6ln(x)+2)/(27x^3)`

`V=pi[-(9(ln(x))^2+6ln(x)+2)/(27x^3)]^(2)-[-(9(ln(x))^2+6ln(x)+2)/(27x^3)]_(1)`

Plugging in the upper and lower bounds:

`V=pi[-(9(ln(2))^2+6ln(2)+2)/(27(2)^3)]^(2)-[-(9(ln(1))^2+6ln(1)+2)/(27(1)^3)]_(1)=0.025542pi`

Volume of revolution: