Worked examples
The question
Question: find the surface area when y = 3/4 × x, from x = 0 to x = 4, turns around the x-axis.
y = 34x
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 4 of 2π f(x)√1 + (f′(x))2 dx
2 Find f′(x)
A line: its slope is the number in front of x, 3/4.
f′(x) = 34
3 Under the root
Square it and add 1: 25/16.
1 + (f′(x))2 = 1 + 916 = 2516
The root of 25/16 is 5/4.
√2516 = 54
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 4 of 2π f(x) dx ✗
4 Integrate
The root is a number here: it waits in front with 2π.
S = 2π54∫ from 0 to 4 of (34x) dx
The antiderivative: add 1 to each power, divide by the new power.
= 2π54 [3x28 ] from 0 to 4
Top minus bottom: 6.
= 2π54 × 6
So the surface area is 15π square units.
That is about 47.12.
S = 15π ≈ 47.12
5 Check
Let's check: the skin is a cone with radius 3 and slanted side 5 (Pythagoras).
A cone's side area is π r ℓ: about 47.12.
The same.
Right!
cone: π r ℓ = π × 3 × 5 ≈ 47.12
Let's look back.
First: the idea.
Then: find f′(x).
Then: under the root.
Then: integrate.
Last: check.
▶ Watch it solved
The question
Question: find the surface area when y = √x, from x = 2 to x = 6, turns around the x-axis.
y = √x
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 2 to 6 of 2π f(x)√1 + (f′(x))2 dx
2 Find f′(x)
√x is x to the one half.
The power rule gives 1/2√x.
f′(x) = 12√x
3 Under the root
Square it: 1/4x.
1 + (f′(x))2 = 1 + 14x
Over one bottom: 1 is 4x/4x.
= 4x + 14x
Careful!
Not dx!
The band is slanted: its width is ds, the root of 1 plus f′ squared, times dx.
Careful: not S = ∫ from 2 to 6 of 2π f(x) dx ✗
4 Simplify
Now f(x) times the root.
= √x × √4x + 14x
The root of a fraction is the root of the top over the root of the bottom.
The root of 4x is 2√x.
= √x × √4x + 12√x
√x and √x cancel: only the root of 4x + 1, over 2, is left.
= √4x + 12
5 Integrate
2π times one half is π: it waits in front.
S = π∫ from 2 to 6 of √4x + 1 dx
u-substitution: u = 4x + 1, so dx is du over 4.
u goes from 9 to 25.
u = 4x + 1, du = 4 dx
x = 2 → u = 9, x = 6 → u = 25
With u: the 1/4 comes out.
√u integrates to two thirds of u to the three halves.
= π4 × 23[ u^(3/2) ] from 9 to 25
Multiply the numbers in front: π over 6.
Then top minus bottom.
= π6(25^(3/2) − 9^(3/2))
= π6(125 − 27)
So the surface area is 49π over 3 square units.
That is about 51.31.
S = 49π3 ≈ 51.31
6 Check
Let's check with a calculator: it adds up 1000 thin bands.
That gives about 51.31, the same as our answer.
Right!
1000 thin bands: ≈ 51.31
Let's look back.
First: the idea.
Then: find f′(x).
Then: under the root.
Then: simplify.
Then: integrate.
Last: check.
▶ Watch it solved
The question
Question: find the surface area when y = 1/3 × x³, from x = 0 to x = 1, turns around the x-axis.
y = 13x3
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 1 of 2π f(x)√1 + (f′(x))2 dx
2 Find f′(x)
The power rule: f′(x) = x².
f′(x) = x2
3 Under the root
Square it and add 1: 1 + x⁴.
1 + (f′(x))2 = 1 + x4
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 1 of 2π f(x) dx ✗
4 Integrate
Put f(x) in front of the root.
S = 2π∫ from 0 to 1 of 13x3√1 + x4 dx
u-substitution: u = 1 + x⁴.
Its derivative has x³ in it, and so does our integral: a good sign.
u goes from 1 to 2.
u = 1 + x4
du = 4x3 dx
x = 0 → u = 1, x = 1 → u = 2
1/3 × x³ dx is 1/3 times x³ dx.
x³ dx is du over 4.
So the number in front is π over 6.
= π6∫ from 1 to 2 of √u du
√u integrates to two thirds of u to the three halves: π over 9 in front.
Then top minus bottom.
= π9(2^(3/2) − 1^(3/2))
So the surface area is π over 9 times (2√2 − 1) square units.
That is about 0.64.
S = π9(2√2 − 1) ≈ 0.64
5 Check
Let's check with a calculator: it adds up 1000 thin bands.
That gives about 0.64, the same as our answer.
Right!
1000 thin bands: ≈ 0.64
Let's look back.
First: the idea.
Then: find f′(x).
Then: under the root.
Then: integrate.
Last: check.
Answer S = π9(2√2 − 1) ≈ 0.64
▶ Watch it solved
