Solids of Revolution by Disks

We can have a function, like this one:

Solids of Revolution y=f(x)

And revolve it around the x-axis like this:

Solids of Revolution y=f(x)

To find its volume we can add up a series of disks:

Solids of Revolution y=f(x)

Each disk's face is a circle:

Solids of Revolution y=f(x)

The area of a circle is π times radius squared:

A = π r2

And the radius r is the value of the function at that point f(x), so:

A = π f(x)2

And the volume is found by summing all those disks using Integration:

Volume =
      b
a
π f(x)2 dx

And that is our formula for Solids of Revolution by Disks

In other words, to find the volume of revolution of a function f(x): integrate pi times the square of the function.

Example: A Cone

Take the very simple function y=x between 0 and b

Solids of Revolution y=f(x)

Rotate it around the x-axis ... and we have a cone!

Solids of Revolution y=f(x)

The radius of any disk is the function f(x), which in our case is simply x

Solids of Revolution y=f(x)

What is its volume? Integrate pi times the square of the function x :

Volume =
      b
0
π (x)2 dx

pie outside

First, let's have our pi outside (yum).

Seriously, it is OK to bring a constant outside the integral:

Volume = π
      b
0
x2 dx

Using Integration Rules we find the integral of x2 is x3/3 + C

To calculate this definite integral, we calculate the value of that function for b and for 0 and subtract, like this:

Volume = π (b3/3 − 03/3)

= π b3/3

Compare that result with the more general volume of a cone:

Volume = 1 3 π r2 h

When both r=b and h=b we get:

Volume = 1 3 π b3

As an interesting exercise, why not try to work out the more general case of any value of r and h yourself?

 

We can also rotate about other lines, such as x = −1

Example: Our Cone, But About x = −1

So we have this:

Solids of Revolution y=f(x)

Rotated about x = −1 it looks like this:

Solids of Revolution y=f(x)
The cone is now bigger, with its sharp end cut off (a truncated cone)

Let's draw in a sample disk so we can work out what to do:

Solids of Revolution y=f(x)

OK. Now what is the radius? It is our function y=x plus an extra 1:

y = x + 1

Then integrate pi times the square of that function:

Volume =
      b
0
π (x+1)2 dx

Pi outside, and expand (x+1)2 to x2+2x+1 :

Volume = π
      b
0
(x2+2x+1) dx

Using Integration Rules we find the integral of x2+2x+1 is x3/3 + x2 + x + C

And going between 0 and b we get:

Volume = π (b3/3+b2+b − (03/3+02+0))

= π (b3/3+b2+b)

Now for another type of function:

Example: The Square Function

Take y = x2 between x=0.6 and x=1.6

Solids of Revolution y=x^2

Rotate it around the x-axis:

Solids of Revolution y=x^2

What is its volume? Integrate pi times the square of x2:

Volume =
      1.6
0.6
π (x2)2 dx

Simplify by having pi outside, and also (x2)2 = x4 :

Volume = π
      1.6
0.6
x4 dx

The integral of x4 is x5/5 + C

And going between 0.6 and 1.6 we get:

Volume = π ( 1.65/5 − 0.65/5 )

6.54

Can you rotate y = x2 about x = −1 ?

In summary:

pie outside

 

About The Y Axis

We can also rotate about the Y axis:

Example: The Square Function

Take y=x2, but this time using the y-axis between y=0.4 and y=1.4

Solids of Revolution about Y

Rotate it around the y-axis:

Solids of Revolution about Y

And now we want to integrate in the y direction!

So we want something like x = g(y) instead of y = f(x). In this case it is:

x = √(y)

Now integrate pi times the square of √(y)2 (and dx is now dy):

Volume =
      1.4
0.4
π √(y)2 dy

Simplify with pi outside, and √(y)2 = y :

Volume = π
      1.4
0.4
y dy

The integral of y is y2/2

And lastly, going between 0.4 and 1.4 we get:

Volume = π ( 1.42/2 − 0.42/2 )

≈ 2.83...

 

Washer Method

Washers (various)
Washers: Disks with Holes

What if we want the volume between two functions?

Example: Volume between the functions y=x and y=x3 from x=0 to 1

These are the functions:

Solids of Revolution between y=x and y=x^3

Rotated around the x-axis:

Solids of Revolution between y=x and y=x^3

The disks are now "washers":

Solids of Revolution between y=x and y=x^3

And they have the area of an annulus:

annulus r and R
In our case R = x and r = x3

In effect this is the same as the disk method, except we subtract one disk from another.

And so our integration looks like:

Volume =
      1
0
π (x)2 π (x3)2 dx

Have pi outside (on both functions):

Volume = π
      1
0
(x)2 (x3)2 dx

Simplify:

Volume = π
      1
0
x2 − x6 dx

The integral of x2 is x3/3 and the integral of x6 is x7/7

And so, going between 0 and 1 we get:

Volume = π [ (13/3 − 17/7 ) − (0−0) ]

≈ 0.598...

So the Washer method is like the Disk method, but with the inner disk subtracted from the outer disk.