Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 87 pot

Its density at a depth of h meters from the top is given by 5h milligrams per cubic meter.. As a result, the density of ducks on the pond is given by a function ρx ducks per square meter

(b) Another truffle is made in a hemispherical mold with radius R Layers of different types of chocolate are poured into the mold, one at a time, and allowed to set The number of calories per cubic millimeter varies with x, where x is the depth from the top of the mold The calorie density is given by δ(x) calories/mm3 Write an integral that gives the number of calories in this hemispherical truffle

Top of the mold Hemispherical truffle mold

14 Liquid is being stored in a large spherical tank of radius 2 meters The tank is completely full and has been left standing for a long time A mineral suspended in the liquid is setting Its density at a depth of h meters from the top is given by 5h milligrams per cubic meter Determine the number of milligrams of the mineral contained in the tank

Top

15 A circular pond is 60 meters in diameter and has a bridge running along a diameter

At lunchtime people stand on the bridge and throw bread onto the water to feed the ducks As a result, the density of ducks on the pond is given by a function ρ(x) ducks per square meter, where x is the distance from the bridge How many ducks are on the pond? (We will assume that the bridge itself is very thin so we can ignore its width.)

Notice that we cannot really say that the ducks are continuously distributed on the pond Ducks, after all, are discrete We are making a continuous model of a discrete phenomenon.

16 Let W (t) be the amount of water in a pool at time t, t measured in hours and W measured

in gallons t = 0 corresponds to noon Water is flowing in and out of the pool at a rate given by dWdt = 30 cosπ2t During what time interval between noon and 5:00 p.m (0 ≤ t ≤ 5) is water flowing out of the pool at a rate of 15 gallons an hour or more?

How much water actually has left the pool in this time interval?

17 In the town of Lybonrehc there has been a nuclear reactor meltdown, which released radioactive iodine 131 Fortunately, the reactor has a containment building, which kept the iodine from being released into the air The containment building is hemispherical with a radius of 100 feet The density of iodine in the building was 6 × 10−5(200 − h) g/cubic feet, where h is the height from the floor (in feet) (It ranges from 12 × 10−3 g/cubic feet at the floor to 6 × 10−3g/cubic feet near the top.)

(a) Derive an integral that gives the amount of iodine in the building You must explain your reasoning fully and clearly

(b) Calculate the amount of iodine in the building

18 A spherical star has a radius of 90,000 kilometers The density of matter in the star

is given by ρ(r) = (r+1)K3/2 kilograms per cubic kilometer, where r is the distance (in kilometers) from the star’s center and K is a positive constant

Write out (but do not evaluate) an expression for the total mass of the star Your answer should contain the constant K

19 A substance has been put in a centrifuge We now have a cylindrical sample (radius

3 centimeters, height 4 centimeters) in which density varies with x, the distance (in centimeters) from the central axis If the density is given by ρ(x) mg/cm3, write an integral that gives the total mass of the substance

20 A very thin, lighted pole 10 feet tall is placed upright in a family’s backyard to attract insects to it (where they are electrocuted) At one moment, the density of these insects

is given by ρ(r) =π(r+1)1.3 insects per cubic foot, where r measures the number of feet from the pole

(a) How many insects are within 5 feet of the pole at a height of 10 feet or less? (b) How many insects are within 5 feet of the pole at a height of 10 feet or more?

21 A circus tent has cylindrical symmetry about its center pole The height a distance of

xfeet from the center pole is given by h(x) = 8

1+ x2 16 feet What is the volume enclosed

by the tent of radius 4?

22 At the Three Aces pizzeria, the chef tosses lots of garlic on the pizza The density of garlic varies with x, the distance from the center of the pizza, and is given by

(x3+ 2)2 ounces per square inch of pizza

If the pizza is 14 inches in diameter, and Three Aces cuts six slices from each pizza, how much garlic is on one slice of pizza? (Problem by Andrew Engelward)

23 (a) What is the present value of a single payment of $2000 three years in the future? Assume 5% interest compounded continuously

(b) What is the present value of a continuous stream of income at the rate of $100,000 per year over the next 20 years? Assume 5% interest compounded continuously

By “a continuous stream of income” we mean that we are modeling the situation

by assuming that money is being generated continuously at a rate of $100,000 per year

Begin by partitioning the time interval [0, 20] into n equal pieces Figure out the amount of money generated in the ith interval and pull it back to the present Summing these pull-backs should approximate the present value of the entire income stream

27.2 SLICING TO FIND THE AREA BETWEEN TWO CURVES

Section 27.1 is the heart of the “slicing” discussion It allows us to apply what we know about integration to a very broad array of situations In this section we’ll use the same approach

to enlarge the type of area problems we can deal with

EXAMPLE 27.5 Find an integral that gives the shaded area in Figure 27.10 the area bounded by y = f (x),

y = g(x), and the vertical lines x = a and x = b

x = a

x

x = b

y = g(x)

y = f (x)

Figure 27.10

SOLUTION Chop the area into tall, thin “rectangles” by partitioning the interval [a, b] along the x-axis

into n equal pieces as shown; each piece is of length x =b−an and xi= a + ix.5

x0 x1 x2 x3 x n–1 x n

∆ x

Notice that the height of each rectangle is of the form f (xi) − g(xi), regardless of where the x-axis is in relation to the graphs of f and g To find any vertical distance, just subtract the bottom (lower) y-value from the top (higher) y-value (See Figure 27.11.)

x x

y = 1

y = –3

y = 6

y = 2

6 – 1 = 5

2 – (–3) = 2 + 3 = 5 –3 – (–7) = –3 + 7 = 4

y = –3

y = –7

Figure 27.11

the area of the ith rectangle ≈ (height) · (length)

≈ [f (xi) − g(xi)] · x

5 When we chop up the area into many pieces we get a slew of pseudo-rectangles They don’t have flat tops and flat bottoms,

so they are not really rectangles But each one can be closely approximated by a genuine rectangle, one end of which lies on the curve y = f (x) and the other end on the curve y = g(x) For the remainder of this section, for ease of discussion we will say that

the area of the region ≈

n

 i=1 [f (xi) − g(xi)] · x

the area of the region = lim

n→∞

n

 i=1 [f (xi) − g(xi)]x

=

b

a [f (x) − g(x)] dx

x = b

x = a

y = g(x)

y = g(x)

≈f(x i ) – g(x i)

y = f (x)

y = f (x)

Figure 27.12

EXAMPLE 27.6 Find an integral that gives the shaded area in Figure 27.13, the area bounded by x = q(y),

x = r(y), and the horizontal lines y = c and y = d

x = q(y)

x = r(y)

x

y d

c

Figure 27.13

SOLUTION Chop the area horizontally into long, narrow rectangles by partitioning the interval [c, d]

along the y-axis into n equal pieces as shown; each piece is of height y =d−cn Notice that the length of each rectangle is of the form q(yi) − r(yi), regardless of where the y-axis

is in relation to the graphs of q and r To find any horizontal distance, we subtract the x-value on the left (the smaller one) from the x-x-value on the right (the larger one) (See Figure 27.14.)

x = 3

x = 3

–2 + 5 = 3

x = –5 x = –2

y

2

}∆y

d

c

y n

y3

y2

y1

y0

.

Figure 27.14

the area of the ith rectangle ≈ (length)· (height)

≈ [q(yi) − r(yi)] · y

x = q(y)

x = q(y)

x

y d

c

≈ q(y i ) – r (y i)

Figure 27.15

So,

the area of the region ≈

n

 i=1 [q(yi) − r(yi)] · y

the area of the region = lim

n→∞

n

 i=1 [q(yi) − r(yi)] · y

=

d

EXERCISE 27.5 Find the area bounded above by y = x(4 − x) and below by

Answers are given at the end of the section.

EXERCISE 27.6 Which of the following expressions gives the area bounded on the left by the y-axis, on the

right by y = ln x, below by the x-axis, and above by the line y = 1? (There is more than one correct answer.)

y

1

y = lnx

Figure 27.16

(a)01(1 − ln x) dx (b)011 dx +1e(1 − ln x) dx

Answers are given at the end of the section.

Using Slicing to Help with Some Definite Integrals—A Slick Trick

In this subsection we show you a nice byproduct of combining the idea of slicing with that

of inverse functions.6By the end of this section you will be able to evaluate definite integrals such as1eln x dx,01arctan x dx, and00.5arcsin x dx These integrals look intractable because we don’t know antiderivatives of ln x, arctan x, and arcsin x As we’ll soon see, a change of perspective will work wonders for us.7

EXAMPLE 27.7 Evaluate1eln x dx

SOLUTION Let’s begin by drawing a picture of the region whose area is given by the integral above

Evaluating1eln x dx is equivalent to finding the area of the shaded region We’re accus-tomed to slicing the area into vertical rectangles However, we don’t know an antiderivative

of ln x, so let’s try a change of perspective We’ll look at the very same region but this time slice the region horizontally into “rectangles” as shown in Figure 27.18(b), chopping the interval [0, 1] along the y-axis

y = lnx or x = e y

(e, 1)

y

1 1

(b) (a)

change of perspective

y = lnx (e, 1)

y

1 1

Figure 27.17

We partition the interval [0, 1] into n equal pieces and label as shown Each piece is of length y where y =1−0n

6 It will be helpful to try Exercise 27.6 from the preceding page before proceeding.

7 Generally these integrals are evaluated using a technique called integration by parts, a technique derived from the Product

The area of the ith rectangle ≈ (length) · (height).

The height of each rectangle is y; the length is given by

e − (the x-value on the y = ln x curve)

y = lnx

1 = y n

i

y3

y2

y1

0 = y0

.

}∆y

x = e

x i = e yi

x = e y

e – x i = e – e yi

Figure 27.18

We want to write y = ln x in the form x = r(y)

yi= ln xi is equivalent to xi= eyi Therefore,

the area of the ith rectangle ≈ (length) · (height)

≈ (e − eyi) · y

the total area ≈

n

 i=1 (e − eyi) · y

the total area = lim

n→∞

n

 i=1 (e − eyi) · y

=

1

0 (e − ey) dy This integral is simple to evaluate

1

0 (e − ey) dy = (ey − ey)



1

In Summary

The picture associated with the original integral in Example 27.7 enabled us to phrase the problem as “find the area under y = ln x from x = 1 to x = e.” Slicing the area horizontally instead of vertically was the key to a new outlook The problem then became “find the area between x = e and x = eyfrom y = 0 to y = 1.”

EXAMPLE 27.8 Express the shaded area in Figure 27.19 on the following page as an integral (or sum or

difference of integrals) and evaluate

y

π 4

π 4

(1, )

y = arctan x

y = –x

Figure 27.19 SOLUTION If we chop vertically we must split up the area into two regions because the height of the

rectangles requires two different descriptions For x in the interval [−π

4, 0] the rectangles have height [1 − (−x)] = 1 + x, while for x in the interval [0, 1] they have height (1 − arctan x)

height = –x height = – arctan x

x

y

y =

π 4

π 4 π

4

π 4

Figure 27.20

the shaded area =

0

−π4 (1 + x) dx +

1

0 (1 − arctan x) dx The first of these integrals is not a problem for us to evaluate (It’s the area of a triangle.) Evaluating the second integral, however, requires finding an antiderivative of arctan x; this information is not at our fingertips Let’s see if a change in perspective will help us out

x

y

π 4

π 4

(1, )

y = arctan x

x = tan y

y = –x

or

x = –y

Figure 27.21

Suppose we chop up the region horizontally We must express the equations

y = arctan x and y = −x in the form x = q(y) and x = r(y) respectively by solving each one for x

y = arctan x is equivalent to x = tan y

If we chop horizontally we can use one integral to describe the entire area, as the length of the rectangles is given by (tan yi− (−yi))throughout the region

the area =

1

0 (tan y − (−y)) dy

This integral in y is easier to handle than the one above involving arctan x

1

0 (tan y + y) dy =

1

0

sin y

1

0

y dy

In the first integral, let u = cos y, du = − sin y dy The first integral becomes

y=0

−du



 y=1

y=0

= − ln | cos y|



 y=1

y=0

Therefore,

1

0

sin y cos ydy +

1

0 y dy = − ln | cos y|



 1

0+y

2 2



 1

0

= − ln | cos 1| + ln | cos 0| +1

= − ln | cos 1| + ln 1 +1

2

= − ln | cos 1| +1

2. The area is − ln | cos 1| +12

Answers to Selected Exercises

Exercise 27.5

Draw a picture for yourself

(a)13x(4 − x) − 3 dx = · · · =43

(b)5

−1



x(4 − x) − (−5) dx = · · · = 10223

(c) Find the x-coordinates of the points of intersection x = −1 and x = 6

6

−1x(4 − x) − (−x − 6) dx =−16 (−x2+ 5x + 6) dx = · · · = 5716

Exercise 27.6

(b), (c), and (d) are correct (a), (e), and (f) are incorrect

(See Figure 27.22 on the following page.)

(b) corresponds to the sum of the two shaded areas (c) uses the fact that y = ln x is equivalent to x = ey (d) corresponds to the area of the speckled rectangle minus that of the striped region

x y

1

y = lnx

y = l

(e, 1)

(b)

x y

1

y = lnx or x = e y

(e, 1)

(c)

x y

1

y = l

(d)

Figure 27.22

P R O B L E M S F O R S E C T I O N 2 7 2

1 Find the area bounded by the curves y = ex, y = 1 − x, and x = 1

2 Find the area bounded by y = 2 − x2and y = x

3 Find the area bounded above by y = −x + 6, below by y = x2+ 1, and on the left by the y-axis

4 Find the area bounded below by the x-axis, on the left by the y-axis, and above by

y = x2+ 1 and y = −x + 6

5 Find the area in the first quadrant bounded by y = arcsin x, y = π/2, and x = 0 (Hint:

To get an exact answer it will be simplest to integrate with respect to y.)

6 (a) Let A be the area between the cosine and the sine curves between x = −π/4 and

x = π/4

i Sketch the graphs of sin x and cos x, shading the region A described above

ii Write a definite integral giving the area A

(b) Let B be the area between the cosine and the sine curves between x = 0 and

x = 2π

i Sketch the graphs of sin x and cos x, shading the region B described above

ii Write a sum of the definite integrals giving the area B Notice that over a certain interval the sine curve lies above the cosine curve, while on other intervals it lies below it Therefore, it is necessary to split up the interval and write a sum of integrals

7 Write an integral (or the sum or difference of integrals) that gives the area of the region shaded on the following page