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 − x²:

Using Integration Rules we find the integral of bx − x² is:

bx²2 − x³3 + C

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

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 − x² :

Using Integration Rules we find the integral of 4x − x² is:

4x²2 − x³3 + C

And going between 0 and 3 we get:

Volume = 2π(4×3²2 − 3³3) − 2π(4×0²2 − 0³3)

= 2π(18−9)

= 18π

Example: From y=x down to y=x²

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 − x²

Now integrate 2π times x times x − x²:

Put 2π outside, and expand x(x−x²) into x²−x³ :

The integral of x² − x³ is x³3 − x⁴4

Now calculate the volume between a and b … but what is a and b? a is 0, and b is where x crosses x², which is 1

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 :

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

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

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

Example: Our Cone, But About x = −1

So we have this:

Rotated about x = −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:

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

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)

Example: The Square Function

Take y = x² between x=0.6 and x=1.6

Rotate it around the x-axis:

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

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

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

Example: The Square Function

Take y=x², 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):

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

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…

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

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 = x³

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

And so our integration looks like:

Have pi outside (on both functions) and simplify (x³)² = x⁶:

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…