Calculating the volume of a function that rotates around the x-axis

Why are you multiplying in your last line:

V=∫²₁ (pi)...

Once you have the antiderivative , you evaluate the antiderivative at the endpoints

2,1 respectively, and that is the value of the integral.

 

Yes, but a nicer answer would have been [itex] \frac{\pi}{2} [/itex] units³.

 

Easily stated you want :

[itex]2π\int_{1}^{2}\frac{1}{x^2}dx[/itex]

You could have also used :

[itex]π\int_{1}^{2}1dx[/itex]

 

The OP correctly identified the volume of revolution to be [tex] V = \pi\int_{1}^{2} \frac{1}{x^2} dx. [/tex]

What mathematical reasoning gives this?

 

There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.

[tex]\pi\int_{a}^{b}xf(x)dx[/tex]

[tex]2\pi\int_{a}^{b}f^2(x)dx[/tex]

 

I have come across the method of cylindrical shells, where for some rotation about y axis, [tex] V = 2\pi \int_{a}^{b} x f(x) dx [/tex]
I was wondering if someone could attempt the problem in the OP using cylindrical shells, because I am always getting either π or π/4, not the required π/2.

 

You've remembered incorrectly.

 

Show us what you tried. You should have had to split it up into two integrals.

 

I split it up into 2 regions: one integral using y limits from 1/2 to 1 and another from 0 to 1/2.
For the first region: [tex] V = 2\pi \int_{\frac{1}{2}}^{1} (\frac{1}{2} + y)(\frac{1}{y} - 1) dy ,[/tex] where the shell radius is 1/2 + y and it's height is 1/y - 1.

Then the region using y limits 0 to 1/2 was a rectangular region:
[tex] V = 2\pi\int_{0}^{\frac{1}{2}} y dy, [/tex] where the height is a constant 1 unit.

Doing the integral does not give me the correct answer. Where am I going wrong?
Many thanks.

 

The shell radius in the first integral should simply be y.

 

Perfect, I must have thought that I was rotating about the line y = 1/2 at my first attempt, but I see why it is simply y: y is the distance from the axis of rotation, which is the x axis. I now have the correct answer.
Thanks again.