# Washer Method

How to use the washer method to find the volume of a rotated figure about the x-axis: formula, 2 examples, and their solutions.

## Formula

This rotated 3D figure has a hole inside.

To find the volume of this figure,

first find the volume of the outer rotated figure,

then subtract the volume of the inner rotated figure.

So the volume of the rotated figure

about the x-axis is,

the volume of the outer rotated figure, ∫_{a}^{b} π(y_{1})^{2} dx

minus,

the volume of the inner rotated figure, ∫_{a}^{b} π(y_{2})^{2} dx.

Disc Integration

The cross section of the figure

looks like a washer:

a circle (r = y_{1}) that has a circular hole (r = y_{2}).

So this method is called the [washer method].

## Example

First find the equation of the tangent line.

f(x) = e^{x}

So f'(x) = e^{x}.

Derivative of e^{x}

So f'(1) = e.

This is the slope of the tangent line.

The tangent line passes through (1, e).

Then the equation of the tangent line

in point-slope form is

y = e(x - 1) + e,

which is y = ex.

Draw y = e^{x}.

Draw the tangent line y = ex.

Then color the region

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

The outer function is y = e^{x}.

It rotates from x = 0 to x = 1.

The inner function is y = ex.

It also rotates from x = 0 to x = 1.

So V = ∫_{0}^{1} π⋅(e^{x})^{2} dx - ∫_{0}^{1} π⋅(ex)^{2} dx.

∫_{0}^{1} π⋅(e^{x})^{2} dx

= π∫_{0}^{1} e^{2x} dx

-∫_{0}^{1} π⋅(ex)^{2} dx

= -π∫_{0}^{1} e^{2}x^{2} dx

Power of a Product

e^{2} is a constant.

So -π∫_{0}^{1} e^{2}x^{2} dx

= -πe^{2}∫_{0}^{1} x^{2} dx.

Solve the integral.

Definite Integral: How to Solve

The integral of e^{2x} is

[1/2]⋅e^{2x}.

Integral of f(ax + b)

Integral of e^{x}

The integral of x^{2} is

[1/3]x^{3}.

Integral of a Polynomial

Put 1 and 0

into [1/2]e^{2x}.

Put 1 and 0

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

[1/2]⋅e^{2⋅1} = e^{2}/2

-[1/2]⋅e^{2⋅0} = -1/2⋅1

[1/3]⋅1^{3} = 1/3

-[1/3]⋅0^{3} = -0

Write the common factor π.

Write e^{2}/2.

Write -1/2⋅1 = -1/2.

See -πe^{2}(1/3 - 0).

π is already used.

So write, -e^{2}/3.

To combine these fractions,

change the fractions like this:

e^{2}/2 = [e^{2}/2]⋅[3/3] = 3e^{2}/6

-1/2 = -[1/2]⋅[3/3] = -3/6

-e^{2}/3 = -[e^{2}/3]⋅[2/2] = -2e^{2}/6.

3e^{2}/6 - 2e^{2}/6 = e^{2}/6

π(e^{2}/6 - 3/6)

= [π/6](e^{2} - 3)

So

[π/6](e^{2} - 3)

is the answer.

## Example

First find the equation of the tangent line.

f(x) = 2√x - 1

So f(x) = 2(x - 1)^{1/2}.

Rational Exponent

Then f'(x) = 2⋅[1/2]⋅(x - 1)^{-1/2}⋅1.

Derivative of a Composite Function

Derivative of a Radical

So f'(2) = 1.

This is the slope of the tangent line.

The tangent line passes through (2, 2).

Then the equation of the tangent line

in point-slope form is

y = 1(x - 2) + 2,

which is y = x.

Draw the given function y = 2(x - 1)^{1/2}.

Draw the tangent line y = x.

Then color the region

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

The outer function is y = x.

It rotates from x = 0 to x = 2.

The inner function is y = 2(x - 1)^{1/2}.

It rotates from x = 1 to x = 2.

So V = ∫_{0}^{2} π⋅x^{2} dx - ∫_{1}^{2} π⋅[2(x - 1)^{1/2}]^{2} dx.

∫_{0}^{2} π⋅x^{2} dx

= π∫_{0}^{2} x^{2} dx

-∫_{1}^{2} π⋅[2(x - 1)^{1/2}]^{2} dx

= -π∫_{1}^{2} 2^{2}(x - 1) dx

-π∫_{1}^{2} 2^{2}(x - 1) dx

= -4π∫_{1}^{2} (x - 1) dx

(Take 2^{2} = 4 out from the integral.)

Solve the integral.

The integral of x^{2} is

[1/3]x^{3}.

The integral of (x - 1) is

[1/2]x^{2} - x.

Put 2 and 0

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

Put 2 and 1

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

[1/3]⋅2^{3} = 8/3

-[1/3]⋅0^{3} = -0

[1/2]⋅2^{2} = 2

[1/2]⋅1^{2} = 1/2

π[8/3 - 0] = 8π/3

Cancel 2 and -2.

(1/2 - 1) = (-1/2)

-4π⋅[-(-1/2)] = -4π⋅[+1/2] = -2π

To solve 8π/3 - 2π,

multiply 3/3 to -2π.

Then 8π/3 - 2π⋅[3/3] = 8π/3 - 6π/3.

8π/3 - 6π/3 = 2π/3

So 2π/3 is the answer.