Disc Integration

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 πy².

Area of a Circle

Then volume of each slice is
πy²⋅dx = πy² dx.

So the volume of the rotated figure
about the x-axis is
V = ∫_a^b πy² dx.

By the same way,
the volume of a rotated figure
about the y-axis is
V = ∫_a^b πx² dy.

Draw y = x⁴.
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 = x⁴.

x is from 0 to 1.
The radius of the disc is y = x⁴.
And the region is rotated about the x-axis.

Then V = ∫₀¹ π⋅(x⁴)² dx.

∫₀¹ π⋅(x⁴)² dx
= π∫₀¹ x⁸ dx

Power of a Power

Solve the integral.

Definite Integral: How to Solve

The integral of x⁸ is
[1/9]x⁹.

Integral of a Polynomial

Put 1 and 0
into [1/9]x⁹.

[1/9]⋅1⁹ = 1/9
-[1/9]⋅0⁹ = -0

π(1/9 - 0) = π/9

So π/9 is the answer.

Draw y = x⁴.
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 = ∫₀¹ π⋅(y^1/4)² dy.

∫₀¹ π⋅(y^1/4)² dy
= π∫₀¹ y^1/2 dy

Solve the integral.

The integral of y^1/2 is
[2/3]y^3/2.

Integral of a Radical

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.

Recall that
if the radius of a sphere is r,
then the volume of the sphere is
[4/3]πr³.

Let's see the proof of this formula.

To make a sphere whose radius is r,
first draw a circle
whose radius is r.

The equation of the circle is
x² + y² = r².

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 y² = r² - x².

x is from -r to r.
The radius of the disc is y
that satisfies y² = r² - x².
And the region is rotated about the x-axis.

Then V = ∫_-r^r π⋅(r² - x²) dx.

Take the constant π out from the integral.

The upper limit (r) and the lower limit (-r)
are the opposites.

r² and -x² are even functions.

So π∫_-r^r (r² - x²) dx
= π⋅2∫₀^r (r² - x²) dx.

Definite Integral: Odd and Even Functions

π⋅2 = 2π

Solve the integral.

The integral of r² is
r²x.

The integral of -x² is
-[1/3]x³.

Put r and 0
into r²x - [1/3]x³.

r²⋅r = r³

-(r²⋅0 - [1/3]⋅0³) = -(0)

[r³ - [1/3]r³ - 0] = r³(1 - 1/3)

1 - 1/3 = 2/3

2π⋅r³⋅[2/3] = [4/3]πr³

So the volume of the sphere is [4/3]πr³.

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