further applications of integration 9. 8.2 area of a surface of revolution in this section, we will...

FURTHER APPLICATIONS FURTHER APPLICATIONS OF INTEGRATIONOF INTEGRATION

9

8.2Area of a Surface

of Revolution

In this section, we will learn about:

The area of a surface curved out by a revolving arc.

FURTHER APPLICATIONS OF INTEGRATION

SURFACE OF REVOLUTION

A surface of revolution is formed when

a curve is rotated about a line.

Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 6.2 and 6.3

AREA OF A SURFACE OF REVOLUTION

We want to define the area of a surface

of revolution in such a way that it corresponds

to our intuition.

If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A .

Let’s start with some

simple surfaces.


CIRCULAR CYLINDERS

The lateral surface area of a circular cylinder

with radius r and height h is taken to be:

A = 2πrh We can imagine

cutting the cylinder and unrolling it to obtain a rectangle with dimensions of 2πrh and h.

CIRCULAR CONES

We can take a circular cone with base radius r

and slant height l, cut it along the dashed line

as shown, and flatten it to form a sector of a

circle with radius and central angle θ = 2πr/l.

CIRCULAR CONES

We know that, in general, the area

of a sector of a circle with radius l and

angle θ is ½ l2 θ.

CIRCULAR CONES

So, the area is:

Thus, we define the lateral surface area of a cone to be A = πrl.

2 21 12 2

2 rA l l rl

l


What about more

complicated surfaces of

revolution?


If we follow the strategy we used with arc

length, we can approximate the original curve

by a polygon.

When this is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area.

By taking a limit, we can determine the exact surface area.

Then, the approximating surface consists

of a number of bands—each formed by

rotating a line segment about an axis.

BANDS

BANDS

To find the surface area, each of these

bands can be considered a portion of

a circular cone.

BANDS

The area of the band (or frustum of a cone)

with slant height l and upper and lower radii r1

and r2 is found by

subtracting the areas of

two cones:

Equation 1

1 1 1

2 1 1 2

( )

( )

A r l l r l

r r l r l

BANDS

From similar triangles, we have:

This gives:

1 1

1 2

l l l

r r

2 1 1 1 1

2 1 1 1

or

( )

r l r l r l

r r l r l

BANDS

Putting this in Equation 1, we get

or

where r = ½(r1 + r2) is the average radius of

the band.

1 2( ) A rl r l

Formula 2

2A rl


Now, we apply this formula

to our strategy.

SURFACE AREA

Consider the surface shown here.

It is obtained by rotating the curve y = f(x), a ≤ x ≤ b, about the x-axis, where f is positive and has a continuous derivative.

SURFACE AREA

To define its surface area, we divide

the interval [a, b] into n subintervals with

endpoints x0, x1, . . . , xn and equal width Δx,

as we did in determining arc length.

SURFACE AREA

If yi = f(xi), then the point Pi(xi, yi) lies

on the curve. The part of the surface between xi–1 and xi

is approximated by taking the line segment Pi–1 Pi and rotating it about the x-axis.

SURFACE AREA

The result is a band with

slant height l = | Pi–1Pi |

and average radius r = ½(yi–1 + yi).

So, by Formula 2, its surface area is:1

12 | |2

i ii i

y yP P

SURFACE AREA

As in the proof of Theorem 2 in Section 8.1,

we have

where xi* is some number in [xi–1, xi].

2

1 1 '( *)i iP P f x x

SURFACE AREA

When Δx is small, we have yi = f(xi) ≈ f(xi*)

and yi–1 = f(xi–1) ≈ f(xi*), since f is continuous.

Therefore,

2* *112 2 ( ) 1 '( )

2i i

i i i i

y yP P f x f x x

SURFACE AREA

Thus, an approximation to what we think

of as the area of the complete surface of

revolution is:

2* *

1

2 ( ) 1 '( )n

i ii

f x f x x

Formula 3

The approximation appears to

become better as n → ∞.

SURFACE AREA

SURFACE AREA

Then, recognizing Formula 3 as a Riemann

sum for the function

we have:

2( ) 2 ( ) 1 '( )g x f x f x

2* *

1

2

lim 2 ( ) 1 '( )

2 ( ) 1 '( )

n

i in

i

b

a

f x f x x

f x f x dx

SURFACE AREA—DEFINITION

Thus, in the case where f is positive and has

a continuous derivative, we define the surface

area of the surface obtained by rotating the

curve y = f(x), a ≤ x≤ b, about the x-axis as:

22 ( ) 1 '( )b

aS f x f x dx

Formula 4

SURFACE AREA

With the Leibniz notation for derivatives,

this formula becomes:

2

2 1b

a

dyS y dx

dx

Formula 5

SURFACE AREA

If the curve is described as x = g(y),

c ≤ y ≤ d, then the formula for surface area

becomes:

2

2 1d

c

dxS y dx

dy

Formula 6

Then, both Formulas 5 and 6 can be

summarized symbolically—using the notation

for arc length given in Section 8.1—as:

2S y ds

Formula 7SURFACE AREA

SURFACE AREA

For rotation about the y-axis, the formula

becomes:

Here, as before, we can use either

or

2S x ds

Formula 8

2

1dy

ds dxdx

2

1dx

ds dydy

SURFACE AREA—FORMULAS

You can remember

these formulas in the following

ways.


Think of 2πy as the circumference of a circle

traced out by the point (x, y) on the curve as

it is rotated about the x-axis.


Think of 2πx s the circumference of a circle

traced out by the point (x, y) on the curve as

it is rotated about the y-axis.

SURFACE AREA

The curve , –1 ≤ x ≤ 1, is an arc

of the circle x2 + y2 = 4 .

Find the area of the surface

obtained by rotating this

arc about the x-axis.

The surface is a portion of a sphere of radius 2.

24y x

Example 1

SURFACE AREA

We have:

2 1 212

2

(4 ) ( 2 )

4

dyx x

dxx

x

Example 1

SURFACE AREA

So, by Formula 5, the surface area is:2

1

1

21 221

1 2

21

1

1

2 1

2 4 14

22 4

4

4 1 4 (2) 8

dyS y dx

dx

xx dx

x

x dxx

dx

Example 1

SURFACE AREA

The arc of the parabola y = x2 from (1, 1)

to (2, 4) is rotated about the y-axis.

Find the area of

the resulting surface.

Example 2

SURFACE AREA

Using y = x2 and dy/dx = 2x,

from Formula 8, we have:

22

1

2 2

1

2

2 1

2 1 4

S x ds

dyx dx

dx

x x dx

E. g. 2—Solution 1

SURFACE AREA

Substituting u = 1 + 4x2, we have du = 8x dx.

Remembering to change the limits

of integration, we have:

17 173 223 554 4

(17 17 5 5)6

S u du u


SURFACE AREA

Using

x = and dx/dy = ,

we have the following solution.

y


12 y

24

1

4

1

4

1

17

5(where 1 4 )

2 2 1

12 1

4

4 1

4

(17 17 5 5)6

u y

dxS xds x dy

dy

y dyy

y dy

udu

E. g. 2—Solution 2SURFACE AREA

SURFACE AREA

Find the area of the surface generated

by rotating the curve y = ex, 0 ≤ x ≤ 1,

about the x-axis.

Example 3

Using Formula 5 with y = ex and dy/dx = ex,

we have:

Example 3

21

0

1 2

0

2

1

2 1

2 1

2 1 (where )

x x

e x

dyS y dx

dx

e e dx

u du u e

SURFACE AREA

SURFACE AREA Example 3

3 1

4

12 4

2 sec (where tan and tan )

2 sec tan ln sec tan (E.g.8, Sec.7.2)

sec tan ln sec tan 2 ln 2 1

d u e

SURFACE AREA

Since tan α = e , we have:

sec2α = 1 + tan α = 1 + e2

Thus,

2 21 ln 1 2 ln 2 1S e e e e

Example 3

further applications of integration 9. 8.2 area of a surface of revolution in this section, we will...

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