Steply

Surface area of y = 2x + 1 turned around the x-axis from x = 0 to x…

Step by step, with the reason for every step · Surface area of revolution · Calculus 2

Step by step

The question

Question: find the surface area when y = 2x + 1, from x = 0 to x = 2, turns around the x-axis.

y = 2x + 1

1 The idea

Draw the curve.

Cut the skin into thin bands.

A band goes around a circle of radius f(x): it is 2π f(x) long.

Its width is a tiny piece of the curve, ds, from the arc length lesson.

Not dx: the curve is slanted.

band = 2π f(x) × ds

Add up all the bands: ds is the root of 1 plus f′ squared, times dx.

S = ∫ from 0 to 2 of 2π f(x)√1 + (f′(x))2 dx

2 Find f′(x)

A line: its slope is the number in front of x, 2.

f′(x) = 2

3 Under the root

Square it and add 1: 5.

1 + (f′(x))2 = 1 + 4 = 5
√5 = √5

Careful!

Not dx!

The band is slanted: its width is ds, the root of 1 plus f′ squared, times dx.

Careful: not S = ∫ from 0 to 2 of 2π f(x) dx ✗

4 Integrate

The root is a number here: it waits in front with 2π.

S = 2π√5∫ from 0 to 2 of (2x + 1) dx

The antiderivative: add 1 to each power, divide by the new power.

= 2π√5 [x2 + x ] from 0 to 2

Top minus bottom: 6.

= 2π√5 × 6

So the surface area is 12√5 π square units.

That is about 84.30.

S = 12√5π ≈ 84.30

5 Check

Let's check with a calculator: it adds up 1000 thin bands.

That gives about 84.30, the same as our answer.

Right!

1000 thin bands: ≈ 84.30

Let's look back.

First: the idea.

Then: find f′(x).

Then: under the root.

Then: integrate.

Last: check.

Answer
S = 12√5π ≈ 84.30

Your turn

Try this one yourself, then check:

Find the surface area when y = x + 1
Show the answer
S = 15√2π

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