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

## Homework Statement

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

## 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?

## Answers and Replies

Bacle2
Science Advisor
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.

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

is this the right answer?

CAF123
Gold Member
Yes, but a nicer answer would have been $\frac{\pi}{2}$ units3.

STEMucator
Homework Helper

## Homework Statement

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

## 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?

Easily stated you want :

$2π\int_{1}^{2}\frac{1}{x^2}dx$

You could have also used :

$π\int_{1}^{2}1dx$

CAF123
Gold Member
Easily stated you want :

$2π\int_{1}^{2}\frac{1}{x^2}dx$
The OP correctly identified the volume of revolution to be $$V = \pi\int_{1}^{2} \frac{1}{x^2} dx.$$
You could have also used :

$π\int_{1}^{2}1dx$

What mathematical reasoning gives this?

STEMucator
Homework Helper
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?

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:
CAF123
Gold Member
There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.

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

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

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
Easily stated you want :

$2π\int_{1}^{2}\frac{1}{x^2}dx$

You could have also used :

$π\int_{1}^{2}1dx$

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

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

$$2\pi\int_{a}^{b}f^2(x)dx$$
You've remembered incorrectly.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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.
Show us what you tried. You should have had to split it up into two integrals.

CAF123
Gold Member
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.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
The shell radius in the first integral should simply be y.

CAF123
Gold Member
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.

Mark44
Mentor
You could have also used :

$π\int_{1}^{2}1dx$

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.