# Disc Integration

How to use the disc integration to find the volume of a rotated figure about the axis (x-axis and y-axis): formula, 3 examples, and their solutions.

## Formula

### Formula: Rotated about the x-axis

To find the volume of a rotated figure

about the x-axis,

add up the cross sectional slices,

which are discs.

Volume from Its Slices

The cross section is a circle

whose radius is y.

So the cross sectional area is πy^{2}.

Area of a Circle

Then volume of each slice is

πy^{2}⋅dx = πy^{2} dx.

So the volume of the rotated figure

about the x-axis is

V = ∫_{a}^{b} πy^{2} dx.

### Formula: Rotated about the y-axis

By the same way,

the volume of a rotated figure

about the y-axis is

V = ∫_{a}^{b} πx^{2} dy.

## Example 1

### Example

### Solution

Draw y = x^{4}.

Draw x = 1.

Then color the region

that is bounded by these two graphs and the x-axis.

Draw the rotated figure.

Draw a cross sectional disc.

The radius of the disc is y.

So write the radius y = x^{4}.

x is from 0 to 1.

The radius of the disc is y = x^{4}.

And the region is rotated about the x-axis.

Then V = ∫_{0}^{1} π⋅(x^{4})^{2} dx.

∫_{0}^{1} π⋅(x^{4})^{2} dx

= π∫_{0}^{1} x^{8} dx

Power of a Power

Solve the integral.

Definite Integral: How to Solve

The integral of x^{8} is

[1/9]x^{9}.

Integral of a Polynomial

Put 1 and 0

into [1/9]x^{9}.

[1/9]⋅1^{9} = 1/9

-[1/9]⋅0^{9} = -0

π(1/9 - 0) = π/9

So π/9 is the answer.

## Example 2

### Example

### Solution

Draw y = x^{4}.

Draw y = 1.

Then color the region

that is bounded by these two graphs and the y-axis.

Draw the rotated figure.

Draw a cross sectional disc.

The radius of the disc is x.

So write the radius x = y^{1/4}.

y is from 0 to 1.

The radius of the disc is x = y^{1/4}.

And the region is rotated about the y-axis.

Then V = ∫_{0}^{1} π⋅(y^{1/4})^{2} dy.

∫_{0}^{1} π⋅(y^{1/4})^{2} dy

= π∫_{0}^{1} y^{1/2} dy

Put 1 and 0

into [2/3]y^{3/2}.

[2/3]⋅1^{3/2} = 2/3

-[2/3]⋅0^{3/2} = -0

π(2/3 - 0) = 2π/3

So 2π/3 is the answer.

## Example 3: Proof of the Volume of a Sphere Formula

### Example

Recall that

if the radius of a sphere is r,

then the volume of the sphere is

[4/3]πr^{3}.

Let's see the proof of this formula.

### Solution

To make a sphere whose radius is r,

first draw a circle

whose radius is r.

The equation of the circle is

x^{2} + y^{2} = r^{2}.

Rotate the circle about the x-axis.

Then the rotated figure is a sphere

whose radius is r.

Draw a cross sectional disc.

The radius of the disc is y.

So write y^{2} = r^{2} - x^{2}.

x is from -r to r.

The radius of the disc is y

that satisfies y^{2} = r^{2} - x^{2}.

And the region is rotated about the x-axis.

Then V = ∫_{-r}^{r} π⋅(r^{2} - x^{2}) dx.

Take the constant π out from the integral.

The upper limit (r) and the lower limit (-r)

are the opposites.

r^{2} and -x^{2} are even functions.

So π∫_{-r}^{r} (r^{2} - x^{2}) dx

= π⋅2∫_{0}^{r} (r^{2} - x^{2}) dx.

Definite Integral: Odd and Even Functions

π⋅2 = 2π

Solve the integral.

The integral of r^{2} is

r^{2}x.

The integral of -x^{2} is

-[1/3]x^{3}.

Put r and 0

into r^{2}x - [1/3]x^{3}.

r^{2}⋅r = r^{3}

-(r^{2}⋅0 - [1/3]⋅0^{3}) = -(0)

[r^{3} - [1/3]r^{3} - 0] = r^{3}(1 - 1/3)

1 - 1/3 = 2/3

2π⋅r^{3}⋅[2/3] = [4/3]πr^{3}

So the volume of the sphere is [4/3]πr^{3}.

This is the proof of the volume of a sphere formula.