We can have a function, like this one:

And revolve it around the x-axis like this:

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

Each disk's face is a circle:

The area of a circle is π times radius squared:

A = π r^{2}

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

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

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

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

Volume =

b

π x^{2} dx

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

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

Volume = π

b

x^{2} dx

Using Integration Rules we find the integral of x^{2} is: *x ^{3}*

**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** (*b ^{3}*

**3**−

*0*

^{3}**3**)

= π *b ^{3}*

**3**

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

Volume = *1* **3** π r^{2} h

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

Volume = *1* **3** π b^{3}

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 y = −1

### Example: Our Cone, But About y = −1

So we have this:

Rotated about y = −1 it looks like this:

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:

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

π (x+1)^{2} dx

**Pi outside**, and expand (x+1)^{2} to x^{2}+2x+1 :

Volume = π

b

(x^{2} + 2x + 1) dx

Using Integration Rules we find the integral of x^{2}+2x+1 is **x ^{3}/3 + x^{2} + x + C**

And going between **0** and **b** we get:

Volume = π (b^{3}/3+b^{2}+b − (0^{3}/3+0^{2}+0))

= π (b^{3}/3+b^{2}+b)

Now for another type of function:

### Example: The Square Function

Take **y = x ^{2}** between x=0.6 and x=1.6

Rotate it around the x-axis:

What is its volume? **Integrate pi times the square of x ^{2}**:

Volume =

1.6

0.6

π (x^{2})^{2} dx

Simplify by having pi outside, and also (x^{2})^{2} = x^{4} :

Volume = π

1.6

0.6

x^{4} dx

The integral of x^{4} is **x ^{5}/5 + C**

And going between 0.6 and 1.6 we get:

Volume = π ( 1.6^{5}/5 − 0.6^{5}/5 )

≈ 6.54

Can you rotate **y = x ^{2}** about y = −1 ?

## In summary:

- Have pi outside
- Integrate the
**function squared** - Subtract the lower end from the higher end

## About The Y Axis

We can also rotate about the Y axis:

### Example: The Square Function

Take y=x^{2}, but this time using the **y-axis** between y=0.4 and y=1.4

Rotate it around the **y-axis**:

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 y^{2}/2

And lastly, going between 0.4 and 1.4 we get:

Volume = π ( 1.4^{2}/2 − 0.4^{2}/2 )

≈ **2.83...**

## Washer Method

Washers: Disks with Holes

What if we want the volume **between two functions**?

### Example: Volume between the functions **y=x** and **y=x**^{3} from x=0 to 1

^{3}

These are the functions:

Rotated around the x-axis:

The disks are now "washers":

And they have the area of an annulus:

In our case **R = x** and **r = x ^{3}**

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

π (x)^{2} − π (x^{3})^{2} dx

Have pi outside (on both functions) and simplify (x^{3})^{2} = x^{6}:

Volume = π

1

x^{2} − x^{6} dx

The integral of x^{2} is x^{3}/3 and the integral of x^{6} is x^{7}/7

And so, going between 0 and 1 we get:

Volume = π [ (1^{3}/3 − 1^{7}/7 ) − (0−0) ]

≈ 0.598...

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

Solids of Revolution by Shells Calculus Index