Washer Method

Washer Method

How to find the volume of the rotated region by using the Washer method: formula, example, and its solution.

Formula

V = [inteval, from a to b, of (pi*y1^2) dx] - [inteval, from a to b, of (pi*y2^2) dx], y1: Outer graph, y2: Inner graph

If the region between two functions is rotated
around the x-axis,
then the cross-sectional area is an annulus.
(= ring shape)

If the outer radius is y1 [= f(x)]
and the inner radius is y2 [= g(x)]

then V = [V1, outer figure] - [V2, inner figure]
= ∫ab πy12 dx - ∫ab πy22 dx.

Disc integration

Example 1

A region is bounded by y = e^x, the tangent line of y = e^x at (1, e), and the y-axis. If the region is rotated around the x-axis, find the volume of the rotated region.

To find the rotate region,
first find the tangent line of y = ex at (1, e).

Set f(x) = ex.
Then f'(x) = ex.

Derivative of ex

Then f'(1) = e.

The tangent line's slope is f'(1) = e.
And the tangent line passes through (1, e).

So the tangent line is y = e(x - 1) + e,
which is y = ex.

Tangent line to a graph

The gray colored region is the region
that is rotated around the x-axis.

y = ex is the outer function.
Its integral interval is [0, 1].

y = ex is the inner function.
Its integral interval is also [0, 1].

So = ∫01 π(ex)2 dx - ∫01 π(ex)2 dx.

Solve the integrals.

Indefinite integration of ex

Definite integration of polynomials

Example 2

A region is bounded by y = 2*sqrt(x - 1), the tangent line of y = 2*sqrt(x - 1) at (2, 2), and the x-axis. If the region is rotated around the x-axis, find the volume of the rotated region.

To find the rotate region,
first find the tangent line of y = 2√x - 1 at (2, 2).

Set f(x) = 2(x - 1)1/2.
Then f'(x) = 2⋅(1/2)(x - 1)-1/2⋅1.

Chain rule in differentiation

Power rule in differentiation (Part 3)

Then f'(2) = 1.

The tangent line's slope is f'(2) = 1.
And the tangent line passes through (2, 2).

So the tangent line is y = 1(x - 2) + 2,
which is y = x.

Tangent line to a graph

The gray colored region is the region
that is rotated around the x-axis.

y = x is the outer function.
Its integral interval is [0, 2].

y = 2(x - 1)1/2 is the inner function.
Its integral interval is [1, 2],
which is different
from the outer function's integral interval.

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

Solve the integrals.

Definite integration of polynomials

To make the denominators the same,
multiply 3/3 by 2.