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

jakobs

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 which 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?

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.

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

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

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 which 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 :

$π\int_{1}^{2}1dx$

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?

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:
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.

Staff Emeritus
Homework Helper
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.

Staff Emeritus
Homework Helper
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.

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.

Staff Emeritus
Homework Helper
$π\int_{1}^{2}1dx$