# Solids of Revolution by Shells

We can have a function, like this one:

And revolve it around the y-axis to get a solid like this:

Now, to find its **volume** we can **add up "shells"**:

Each shell has the curved surface area of a cylinder, so its area is **2πr** times its height:

A = 2π(radius)(height)

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

That is our formula for **Solids of Revolution by Shells**

These are the steps:

- sketch the volume and how the shell fits inside it
- integrate
**2π**times the**shell's radius**times the**shell's height**, - put in the values for b and a, subtract, and you are done.

As in this example:

### Example: A Cone!

Take the simple function **y = b − x** between x=0 and x=b

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

Now let us imagine a shell inside:

What is the shell's radius? It is simply **x**

What is the shell's height? It is **b−x**

What is the volume? **Integrate 2π times x times (b−x)** :

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

Seriously, we can bring a constant like 2π outside the integral:

Expand x(b−x) to bx − x^{2}:

^{2}) dx

Using Integration Rules we find the integral of bx − x^{2} is:

\frac{bx^{2}}{2} − \frac{x^{3}}{3} + C

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

Volume = \frac{1}{3} π r^{2} h

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

Volume = \frac{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 values, such as x = 4

### Example: y=x, but rotated around x = 4, and only from x=0 to x=3

So we have this:

Rotated about x = 4 it looks like this:

It is a cone, but with a hole down the center

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

What is the shell's radius? It is **4−x** *(not just x, as we are rotating around x=4)*

What is the shell's height? It is **x**

What is the volume? **Integrate 2π times (4−x) times x** :

**2π outside**, and expand **(4−x)x** to **4x − x ^{2}** :

^{2}) dx

Using Integration Rules we find the integral of 4x − x^{2} is:

\frac{4x^{2}}{2} − \frac{x^{3}}{3} + C

And going between **0** and **3** we get:

Volume = 2π(\frac{4×3^{2}}{2} − \frac{3^{3}}{3}) − 2π(\frac{4×0^{2}}{2} − \frac{0^{3}}{3})

= 2π(18−9)

= 18π

We can have more complex situations:

### Example: From y=x down to y=x^{2}

Rotate around the y-axis:

Let's draw in a sample shell:

What is the shell's radius? It is simply **x**

What is the shell's height? It is **x − x ^{2}**

Now **integrate 2π times x times x − x ^{2}**:

^{2}) dx

Put 2π outside, and expand x(x−x^{2}) into x^{2}−x^{3} :

^{2}− x

^{3}) dx

The integral of x^{2} − x^{3} is (\frac{x^{3}}{3} − \frac{x^{4}}{4})

Now calculate the volume between a and b ... but what* is* a and b? a is 0, and b is where x crosses x^{2}, which is 1

## In summary:

- Draw the shell so you know what is going on
**2π**outside the integral- Integrate the
**shell's radius**times the**shell's height**, - Subtract the lower end from the higher end