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?

1 Answer
Somebody N.
Dec 24, 2017

See below.

Explanation:

enter image source here

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:

enter image source here

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