# Homework Help: Calculating the volume of a function that rotates around the x-axis

1. Sep 2, 2012

### jakobs

1. The problem statement, all variables and given/known data

y=(1/x) is rotating around the x-axis.
find the volume for the figure between x=1 and x=2

2. Relevant equations

3. The attempt at a solution
V=pi y2 = pi*(1/x)2

V= 1∫2 pi(1/x)2dx wich gives V= 1∫2 pi(1/x2)dx

A primitive function for (1/x2) is -(1/x)

V=1∫2 pi((-(1/2))-(-(1/1)))
How do I go from here?

2. Sep 2, 2012

### Bacle2

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.

3. Sep 2, 2012

### jakobs

Sorry, I posted wrong, don't know what happened.

but pi*(-(1/2))=-1,57
pi*(-(1/1))=-3,14

-1,57-(-3,14)= 1,57

4. Sep 2, 2012

### CAF123

Yes, but a nicer answer would have been $\frac{\pi}{2}$ units3.

5. Sep 2, 2012

### Zondrina

Easily stated you want :

$2π\int_{1}^{2}\frac{1}{x^2}dx$

You could have also used :

$π\int_{1}^{2}1dx$

6. Sep 3, 2012

### CAF123

The OP correctly identified the volume of revolution to be $$V = \pi\int_{1}^{2} \frac{1}{x^2} dx.$$
What mathematical reasoning gives this?

7. Sep 3, 2012

### Zondrina

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

$$\pi\int_{a}^{b}xf(x)dx$$

$$2\pi\int_{a}^{b}f^2(x)dx$$

Last edited: Sep 3, 2012
8. Sep 5, 2012

### CAF123

I have come across the method of cylindrical shells, where for some rotation about y axis, $$V = 2\pi \int_{a}^{b} x f(x) dx$$
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.

9. Sep 5, 2012

### vela

Staff Emeritus
You've remembered incorrectly.

10. Sep 5, 2012

### vela

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

11. Sep 5, 2012

### CAF123

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: $$V = 2\pi \int_{\frac{1}{2}}^{1} (\frac{1}{2} + y)(\frac{1}{y} - 1) dy ,$$ 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:
$$V = 2\pi\int_{0}^{\frac{1}{2}} y dy,$$ 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.

12. Sep 5, 2012

### vela

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

13. Sep 5, 2012

### CAF123

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.

14. Sep 5, 2012

### Staff: Mentor

It would be a good idea for you to review formulas that you "remember" to see if they are actually valid. The one above has nothing to do with this problem.