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

Homework Equations

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?

 

[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.

 

[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

is this the right answer?

 

[CAF123]

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

 

[STEMucator]

Homework Statement

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

Homework Equations

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

 

[CAF123]

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]

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

 

[CAF123]

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]

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]

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]

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]

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

 

[CAF123]

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]

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.