Washer Method

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(y₁)² dx
minus,
the volume of the inner rotated figure, ∫_a^b 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].

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 = ∫₀¹ π⋅(e^x)² dx - ∫₀¹ π⋅(ex)² dx.

∫₀¹ π⋅(e^x)² dx
= π∫₀¹ e^2x dx

-∫₀¹ π⋅(ex)² dx
= -π∫₀¹ e²x² dx

Power of a Product

e² is a constant.
So -π∫₀¹ e²x² dx
= -πe²∫₀¹ x² dx.

Solve the integral.

Definite Integral: How to Solve

The integral of e^2x is
[1/2]⋅e^2x.

Linear Change of Variable Rule

Integral of e^x

The integral of x² is
[1/3]x³.

Integral of a Polynomial

Put 1 and 0
into [1/2]e^2x.

Put 1 and 0
into [1/3]x³.

[1/2]⋅e^2⋅1 = e²/2
-[1/2]⋅e^2⋅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).
π is already used.
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.

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 = ∫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

So 2π/3 is the answer.