# 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

### 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
the volume of the outer rotated figure, ∫ab π(y1)2 dx
minus,
the volume of the inner rotated figure, ∫ab π(y2)2 dx.

Disc Integration

The cross section of the figure
looks like a washer:
a circle (r = y1) that has a circular hole (r = y2).

So this method is called the [washer method].

## Example 1

### Solution

First find the equation of the tangent line.

f(x) = ex
So f'(x) = ex.

Derivative of ex

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 = ex.
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 = ex.
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 = ∫01 π⋅(ex)2 dx - ∫01 π⋅(ex)2 dx.

01 π⋅(ex)2 dx
= π∫01 e2x dx

-∫01 π⋅(ex)2 dx
= -π∫01 e2x2 dx

Power of a Product

e2 is a constant.
So -π∫01 e2x2 dx
= -πe201 x2 dx.

Solve the integral.

Definite Integral: How to Solve

The integral of e2x is
[1/2]⋅e2x.

Linear Change of Variable Rule

Integral of ex

The integral of x2 is
[1/3]x3.

Integral of a Polynomial

Put 1 and 0
into [1/2]e2x.

Put 1 and 0
into [1/3]x3.

[1/2]⋅e2⋅1 = e2/2
-[1/2]⋅e2⋅0 = -1/2⋅1

[1/3]⋅13 = 1/3
-[1/3]⋅03 = -0

Write the common factor π.

Write e2/2.
Write -1/2⋅1 = -1/2.

See -πe2(1/3 - 0).
So write, -e2/3.

To combine these fractions,
change the fractions like this:
e2/2 = [e2/2]⋅[3/3] = 3e2/6
-1/2 = -[1/2]⋅[3/3] = -3/6
-e2/3 = -[e2/3]⋅[2/2] = -2e2/6.

3e2/6 - 2e2/6 = e2/6

π(e2/6 - 3/6)
= [π/6](e2 - 3)

So [π/6](e2 - 3) is the answer.

## Example 2

### Solution

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

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 = ∫02 π⋅x2 dx - ∫12 π⋅[2(x - 1)1/2]2 dx.

02 π⋅x2 dx
= π∫02 x2 dx

-∫12 π⋅[2(x - 1)1/2]2 dx
= -π∫12 22(x - 1) dx

-π∫12 22(x - 1) dx
= -4π∫12 (x - 1) dx
(Take 22 = 4 out from the integral.)

Solve the integral.

The integral of x2 is
[1/3]x3.

The integral of (x - 1) is
[1/2]x2 - x.

Put 2 and 0
into [1/3]x3.

Put 2 and 1
into [1/2]x2 - x.

[1/3]⋅23 = 8/3
-[1/3]⋅03 = -0

[1/2]⋅22 = 2
[1/2]⋅12 = 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