our first example is the volume of a solid body. before proceeding, let’s recall that the volume...

Our first example is the volume of a solid body. Before proceeding, let’s recall that the volume of a right cylinder is Ah. Here we use the “right cylinder” in the general sense; the base does not have to be circular, but the sides are perpendicular to the base.Suppose that the solid body extends from height y = c to y = d along the y-axis. Let A(y) be the area of the horizontal cross section at height y.

Volume

h

A y

y

The volume of the

above slice is .A y y

Ac

d

A

h

Volume as the Integral of Cross-Sectional Area Let A(y) be the area of the horizontal cross section at height y of a solid body extending from y = c to y = d. Then

Volume of the Solid Bodyd

c

A y dy

Volume of a Pyramid Calculate the volume V of a pyramid of height 12 m whose base is a square of side 4 m.We need a formula for the horizontal cross section A(y).

y

V A y y

12

2

0

112 1

92 V y u y d dydy u

2

Base 1 0.5

Height 6 12

12

31

129

s

y

ys y

A y y

2

y

12 y

1

2s

d

c

V A y dy

0 122 2

12 0

133

2

3

0

1 1

9 9

1 112

964

3 27 m

u du u du

u

Write in terms of .s y

Write in terms of .A y

Compute the volume V of the solid, whose base is the region between the parabola y = 4 − x2 and the x-axis, and whose vertical cross sections perpendicular to the y-axis are semicircles.

4

24

yA yx y

4

0

42

0

42

42 2

82

4

V y dy

yy

in terms of x y

d

c

V A y dy

Volume of a Sphere: Vertical Cross Sections Use calculus to prove the formula for volume of a sphere.

x

y2 2 2x y r

rr

x

r 2 2y r x

32 2 2

0 0

33 33

2 23

22 2

3 33

4

b

a

rr

V A x dx

xr x dx

r

r x

rr r

QED

CONCEPTUAL INSIGHT Cavalieri’s principle states: Solids with equal cross-sectional areas have equal volume. It is often illustrated convincingly with two stacks of coins

because the volumes V are certainly equal if the cross-sectional areas A(y) are equal.

Our formula d

c

V A y dy includes Cavalieri’s principle,

is constant Total

Given: Density of an object and

Mass

is a fun

Len

cti

gth of an

on of

object

Total M s

,

asb

a

l

x x dx

l

Integral of Mass Density

For example, if ℓ = 10 cm and ρ = 9 g/cm

90 g

Total Mass 9 10 cm gcm

l

DensityNow consider a rod extending along the x-axis from x = a to x = b whose density ρ(x) is a continuous function of x.

x

x

a b

is a function of b

a

x M x dx

is not constant

Total Mass Find the total mass M of a 2-m rod of linear density ρ(x) = 1 + x(2 − x) kg/m, where x is the distance from one end of the rod.

22 3

2 2

0 0

10 kg1 2

33

xM x x dx x x

Total Massb

a

x dx

8 6 12 82 4

3 3

In some situations, density is a function of distance to the origin. For example, in the study of urban populations, it might be assumed that the population density ρ(r) (in people per square kilometer) depends only on the distance r from the center of a city. Such a density function is called a radial density function.

0

Population within a 2 radius R

P r r drR

Total Massb

a

x dx

Computing Total Population The population in a certain city has radial density function ρ(r) = 15(1 + r2)−1/2, where r is the distance from the city center in kilometers and ρ has units of thousands per square kilometer. How many people live in the ring between 10 and 30 km from the city center?

301/ 22

10

301/ 22

10

901901 9011/ 2 1/ 2 1/ 2

1

2

01 101101

2 15 1

30 1

15 1

1,881,825.286

5 2 30

30 901 101 1881.82528 peo6 thousan pd

1

le

2

P r r dr

r r dr

u du u u

u r du rdr

0

Population within a 2 radius R

P r r drR

GRAPHICAL INSIGHT Think of the average value M of a function as the average height of its graph. The region under the graph has the same signed area as the rectangle of height M, because

1Average Value

b

a

f x dxb a

DEFINITION Average Value The average value of an integrable function f (x) on [a, b] is the quantity

.b

a

f x dx M b a

b a

M

0

0

0

1sin

1 1 1cos co 1

2s 1

M xdx

x x

The bushbaby is a small primate with remarkable jumping ability. Find the average speed during a jump if the initial vertical velocity is υ0 = 600 cm/s. Use Galileo’s formula for the height

600

Axis 0 @ 6

&0

22 2 49 90

0 s4

bt h t t

a

Vertical Jump of a Bushbaby

(in centimeters, where g = 980 cm/s2). 20

1

2h t v t gt

6/ 4.9

0

4.9600 980

6t dt

6/9.8 6 / 4.9

0 6/9.8

4.9600 980 600 980

6

4.9 6 6 60

6 9.8 4.9 9.8

4.9 6 6 4.92 2 183.6

300 cm/s

735 0

600 4

6 9.8 4.9

90

6

F x t

t dt t dt

F F F F

F F

2t

To set up our bounds, we need roots!

We can use 0 & the axis, to find the other root.t Speed 'h t

F x

There is an important difference between the average of a list of numbers and the average value of a continuous function. If the average score on an exam is 84, then 84 lies between the highest and lowest scores, but it is possible that no student received a score of 84. By contrast, the Mean Value Theorem (MVT) for Integrals asserts that a continuous function always takes on its average value somewhere in the interval.

For example, the average of f (x) = sin x on [0, π] is 2/π. We have f (c) = 2/π for c = sin−1(2/π) ≈ 0.69. Since 0.69 lies in [0, π], f (x) = sin x indeed takes on its average value at a point in the interval

1 b

a

f c f x dxb a

Theorem 1 Mean Value Theorem for Integrals

If f (x) is continuous on [a, b], then there exists a value c ε [a, b] such that