Application Definite Integration

27C H A P T E RApplying the Denite Integral:Slice and Conquer

27.1 FINDING MASS WHEN DENSITY VARIES

Up to this point weve used the denite integral to

nd the signed area under the graph of f and

nd the net change in amount when given a rate function.

In this chapter we will see how the idea of slicing that led us to the limit denitionof the denite integral can help us apply our knowledge about denite integrals to othersituations.

EXAMPLE 27.1 A geologist is working with a rectangular block-shaped chunk of sedimentary rock whoseheight is 3 meters, width is 4 meters, and length is 7 meters. A certain mineral is uniformlydistributed throughout the rock sample with a density of 5 milligrams per cubic meter. Howmany milligrams of the mineral are in the sample?

SOLUTION To calculate the number of milligrams given the density we simply multiply volume bydensity. The volume of the rock is length width height = (7 m) (4 m) (3 m) = 84 m3.Therefore,

number of grams of the mineral = 84 m3 5 mgm3

= 420 mg.

Notice that the units of volume cancel to leave an answer in milligrams.

EXAMPLE 27.2 The geologist is now looking at another mineral in the same sedimentary rock sample. Thismineral had begun to settle as the rock was being formed, so its density varies with depth.The density is given by (h) = 1 + 0.1h2 milligrams per cubic meter, where h is the depth

827

828 CHAPTER 27 Applying the Denite Integral: Slice and Conquer

below the top surface of the sample.1 Notice that as the depth h increases, so does the densityof the mineral in the rock sample. We want to nd the number of milligrams of this mineralin the sample.

SOLUTION Compare this problem to the rst one. The difculty is that the density is not constant; itvaries from (0) = 1 mg/m3 at the top of the sample to (3) = 1.9 mg/m3 at the bottom. Ifwe were to multiply the total volume of the rock sample by (0), we would get 84 mg; if wewere to multiply the total volume of the rock sample by (3), we would get 159.6 mg. Theformer number is much too small and the latter is much too big. The actual number of gramsof the mineral must be somewhere between these two extremes. How can we improve uponour under- and overestimates?

At any xed depth the density is constant. Suppose we divide the sample into horizontalslabs, slices in which the density doesnt vary much. In each slice we can approximate thenumber of milligrams of the mineral; well add up the estimates in each slice to approximatethe total number of milligrams. The ner the slices, the less the density will vary within aslice and the better our approximation will be.

Notice that this strategy is the same one we successfully applied to the problem ofnding the area under a curve. In the area problem our dilemma was that the function fvaried with x over the x-interval [a, b]. To deal with this we chopped the interval [a, b]into n equal pieces, on each piece approximated the function by a constant, used that toapproximate the area on the subinterval, and summed these areas. We found the area underthe curve by letting the number of subintervals grow without bound. Here our dilemmais that the density varies with h, so well partition the h-interval [0, 3] into pieces, therebypartitioning the rock sample into thin horizontal slabs in which density doesnt vary greatly.

Approximating the Number of Milligrams of Mineral in the Sample

Suppose we divide the h-interval [0, 3] into six equal subintervals. Using the notation wedeveloped for Riemann sums, these are [h0, h1] through [h5, h6], where the height of eachsubinterval is h = 0.5 m and hi = ih for i = 0, 1, . . . , 6. This chops the entire rocksample into six slabs, each of which has height 1/2 meter, width 4 meters, and length 7meters, as shown in Figure 27.1.

h0 = 0h1 = .5h2 = 1h3 = 1.5h4 = 2h5 = 2.5h6 = 3

h0 = 0h1 = .5h2 = 1h3 = 1.5h4 = 2h5 = 2.5h6 = 3

7 m

3 m

4 m

h = 0.5 m {

Figure 27.1

We can estimate the number of milligrams of mineral in each slice. For example, in thetop slice, the density varies from (0) = 1 + 0.1(0)2 = 1 mg/m3 at the top surface to(0.5) = 1 + 0.1(0.5)2 = 1.025 mg/m3 at the bottom surface. The volume of the slice is

1 The Greek letter , pronounced rho, is often used in the sciences to represent density.

27.1 Finding Mass When Density Varies 829

(4 m)(7 m)(h m) = (4)(7)(0.5) m3 = 14 m3. The density increases with h, so we have thefollowing estimates for the number of milligrams of mineral in this top slice.

underestimate: (0) (volume of slice) = 1mgm3

14 m3 = 14 mg

overestimate: (0.5) (volume of slice) = 1.025mgm3

14 m3 = 14.35 mgWe follow the same procedure for each of the ve other slices. Then, by adding all

the underestimates for the separate slices, we obtain an underestimate for the total numberof milligrams in the sample. Similarly, adding the overestimates on all six slices gives anoverestimate for the total.

underestimate =5

i=0(hi) [28 h]

=5

i=0(1 + 0.1h2i ) [28 h]

= 103.25 mg

overestimate =6

i=1(hi) [28 h]

=6

i=1(1 + 0.1h21) [28 h]

= 115.85 mgNotice that our method of slicing horizontally along the axis of the independent variableh ensured that the density would not vary as much within each slice as it had in the rocksample as a whole; this is what brought our estimates closer together. To slice verticallywould have defeated the purpose of slicing, because the density in each slice would stillhave varied between (0) and (3), the same variance as in the entire rock sample.

KEY NOTION The choice of how to slice is at the heart of this kind of problem. This choiceis determined by the density function. In this case, the density varies with height, so weneed to slice in a way that keeps the height (and hence the density) approximately constantwithin each slice.

Computing the Exact Number of Milligrams in the Sample

We can improve upon our estimates by partitioning the h-interval [0, 3] into ner and nersubintervals; this corresponds to cutting thinner and thinner horizontal slabs of the rocksample. As we let the number of subintervals grow without bound, we expect the differencebetween the overestimates and the underestimates to tend toward zero, just as it did in thearea problem.

Chop the h-interval [0, 3] into n equal subintervals each of height h = 3n

m and labelh0, h1, h2, . . . , hn as shown. hi = (i) 3n for i = 0, 1, . . . , n. This corresponds to slicing therock sample horizontally into n slabs, each 7 meters long by 4 meters wide by h meterstall; the volume of each slice is (area) (thickness) = 28h m3.


h ha

hi

hn

47

Figure 27.2

Lets look at a generic slice, say the ith slice.(the # of mg

in the ith slice

)(

estimate of density of

the mineral in the slice

) (volume of the slice)

(hi)mgm3

(28h m3) (using (hi) gives an overestimate) (hi) 28h mg

Thus, we obtain the following sums to estimate the total number of milligrams of the mineralin the sample.

underestimate =n1i=0

(hi) [28 h] =n1i=0

28(hi) h

overestimate =n

i=1(hi) [28 h] =

ni=1

28(hi) h

These sums are Riemann sums. The difference between the overestimate and the un-derestimate is 28[(hn) (h0)] h. This can be written as 28[(3) (0)] h or28(0.9)

(3n

)= 75.6

n. As n grows without bound this difference tends toward zero; the limit

of the Riemann sums is the denite integral 3

0 (h) 28 dh.We now calculate.

number of milligrams in sample = limn

ni=1

(hi) [28 h] (= limn

n1i=0

(hi) [28 h])

= 3

0(h) 28 dh (Both sums approach the

value of the integral.)

= 3

0(1 + 0.1h2) 28 dh

= 28[h + 0.1h

3

3

] 3

0

= 28 [(3 + 0.9) (0)]= 109.2 mg

The strategy of slicing, approximating, summing, and then taking a limit of the sumallowed us to arrive at an exact answer.


Observe the following.

We did not simply integrate (h). There was a 28 in the integrand as well. Lets parsethe integral.

We are summing (h) 28 dh from h = 0 to h = 3,corresponding to (

density

in mg/vol.

) (area of slice) (thickness of slice)

or

(density in mg/vol.) volume of a slice.The limits of integration, while not explicit in the summation notation of the Riemannsum, are implicitly there. As n , h1 0 and hn1 3. The limits of integrationare always determined by the endpoints of the interval being chopped up.

EXAMPLE 27.3 When a meteorite crashes into the earth, debris is scattered nearby. Suppose that the densityof debris is modeled by (r) = 1

1+r2 kilograms per square meter, where r is the number ofmeters from the center of the meteorites impact. What is the mass of the debris that lieswithin 10 meters of the center of impact?

SOLUTION In order to nd mass, we need to multiply density by area. Density is not constant, so wewill need to use a slicing approach to break the problem down into regions where the densityis approximately constant.

KEY NOTION Just as in Example 27.2, we must slice in a way that keeps the densityapproximately constant within each slice. In this case density depends on r , the distancefrom the center of impact. Accordingly, we subdivide the r-interval [0, 10] into n equalsubintervals each of length 10

n= r m; we label r0, r1, r2, . . . , rn as shown. The slices that

result from this are concentric circular rings. Unlike in the rock sample example, in thissituation the area of the slices varies.

r1 rn1 ri r

ri

rr0

r20 r3

=

rn

1010

=

Figure 27.3

The mass in the ith ring (the approximate density in the ring) (the area of the ith ring).What is the area of the ith ring?

Intuitive Approach Think of the ith ring as made of yarn. If the ring is thin enough, itsinner and outer circumferences are approximately equal. To approximate the area, clip theyarn and unbend it, laying it out as a long narrow rectangle. The length of the rectangle is


approximately the circumference of the ring and the width is r . The area of the ring isapproximately the area of the rectangle, which is approximately 2ri r .

r

2ri

Figure 27.4

Alternative ApproachThe ith ring has an outer radius of ri and an inner radius of ri r . We can compute thearea of this ring by subtracting the area of a disk of radius (ri r) from the area of a diskof radius ri.

area of ith ring = area of larger disk area of smaller disk= r2i (ri r)2

= [r2i (r2i 2rir + (r)2)

]=

[2rir (r)2

]We will be constructing a Riemann sum and then taking the limit as the thickness ofeach slice (r) approaches zero. When r is very small, (r)2 is much smaller still.(Consider for instance that if r = 0.001, then (r)2 = 0.000001.) In the limit the (r)2term is insignicant when compared2 with rr . We use the approximation: area of slice 2rir .

In Exercise 27.1 you will show that approximating the area of the ith ring in this wayis valid by arriving at the same integral (using a different labeling system) without makingany approximation in the area of the ith slice.

Now we estimate the mass of the meteorite debris that lies in each ring-shaped slice.

the mass contained in the ith slice (area of slice)(approximate density in slice)

(2rir) m2 (

1

1 + r2i

)kg

m2

2rir1 + r2i

kg

The total mass contained within 10 meters of the center can be approximated by summingthe mass lying in the individual slices; well use a right-hand sum.

total mass limn

ni=1

2ri1 + r2i

r

To get the exact mass, let n grow without bound. In the limit the Riemann sum becomes adenite integral in which the limits of integration are the endpoints of the r-interval beingchopped up.

2 This is the same line of reasoning we used in our discussion of the Product Rule.


total mass = limn

ni=1

2ri1 + r2i

r

= 10

0

2r

1 + r2 dr

= 10

0

2r dr

1 + r2 Substitute u = 1 + r2, so du = 2r dr .

= r=10r=0

du

u

= [ln u] r=10

r=0

= ln(1 + r2)r=10

r=0= [ln 101 ln 1]= ln 101 14.50 kg

EXERCISE 27.1 In this exercise you will verify that the approximation we made in Example 27.3 whenlooking at the area of the ith ring is valid. Partition the r-interval [0, R] into n equalsubintervals each of length r = R

ncentimeters. In this exercise, let ri be the midpoint

of the ith interval, i = 1, . . . , n. That is, r1 = 12r , r2 = r1 + r , r3 = r1 + 2r , an so on,as shown in Figure 27.5.

ri = 12r + (i 1)r for i = 1, 2, . . . , n

The ith ring has an outer radius of ri + 12r and an inner radius of ri 12r .

r1 rn

ri r

ri

r20 . . .r4r3

=

R

r12

12

ri + r12

Figure 27.5

(a) Show that the area of the ith ring is exactly 2rir .

(b) Conclude that the mass of the debris described in Example 27.3 is given by 100 (r)2r dr .

The solution is given at the end of the section.


EXERCISE 27.2 A potter is using a spray gun to apply a cobalt glaze to a plate of radius R centimeters.3 Thedensity of glaze varies with r , the distance from the center of the plate, and is given by (r)mg/cm2. Write an expression that gives the amount of glaze on the plate.

The answer is given at the end of the section.

EXAMPLE 27.4 A farmer plants a crop of wheat on a circular plot with radius 80 meters. A straight irrigationpipe 160 meters long runs down the center of the plot. Due to a drought, his yield varieswith x, the distance from the irrigation pipe. Suppose his yield at a distance x from the pipeis given by (x) kg/m2. What is the farmers yield from the whole plot?

SOLUTION

KEY NOTION We need to slice this disk so that (x) is approximately constant within eachslice. Since the density of yield varies with x, the distance from the irrigation pipe, we needto keep x approximately constant within each slice. This means that we need to slice intolong, thin strips, as shown in Figure 27.6.

12 i i

8080

xi

xix

x x

irrigation pipe

Figure 27.6

We can nd the number of kilograms yield on one side of the pipe and double it to getthe total yield. Well assume for the sake of our model that the irrigation pipe itself has nothickness.

yield in the ith strip (

the approximate density of

yield in the strip

)(

the approximate area

of the strip

) (xi) ix where i is the length of the ith strip.

This expression looks a bit different from the ones we have seen above because it hastwo variables in it. The x indicates that we will eventually integrate with respect to x.Therefore, we need to express i in terms of xi. From Figure 27.6 we see that 80, xi, and12i, can be related using the Pythagorean Theorem. Well leave the completion of thisproblem for you as an exercise.

Observe: Although the regions in Examples 27.3 and 27.4 are both circles, the shapes of theslices are different. The choice of how to slice is determined by the variable of the densityfunction, not by the shape of the overall region.

EXERCISE 27.3 Show that the crop yield described in Example 27.4 is given by 4 80

0 (x)

6400 x2 dx.If you are off by a factor of 2, be sure that youve looked at the whole eld, not just half ora quarter of the eld.

3 Cobalt turns blue when red. The shade of blue varies with the density of the application of cobalt.


The method of slicing allows us to calculate total mass in many different situationswhere the density is not constant. It is an example of the approach we have used repeatedly.To tackle a problem we start with a simpler situation and apply the strategy used there toapproximate the quantity we are looking for. We construct a generic approximation anduse a limiting process to arrive at an exact answer.

Divide and ConquerSlicing Strategy

1. Determine the independent variablethe variable upon which the density depends.

2. Chop along the axis of this independent variable in order to keep density approximatelyconstant within each slice. The endpoints of the interval you chop will determine thelimits of integration.

3. Determine the shape and then the volume/area/length of a generic ith slice; this mayvary with i.

4. Approximate the mass of each slice by multiplying volume/area/length by density.4

5. Write a Riemann sum by adding up the masses of the individual slices. There shouldbe only one variable in this sum; the variable should be the same one determined instep 1.

6. Take the limit as the number of slices increases without bound in order to obtain adenite integral.

7. Evaluate the integral.

Common Errors of the NoviceTry to Avoid Them

The biggest calamity is to have sliced incorrectly. After establishing a way of slicing,check carefully to make sure that the density does not change much within a slice.

Alternatively, to get the slices to begin with, x the independent variable in thedensity formula. This gives you a cut. Change the independent variable an iota (a verysmall amount); this gives you another cut. Youve just carved out a slice for yourself.

Suppose, as in Example 27.2, that the relevant interval of the independent variable is[0, 3]. Beginners often have an intense desire to see the limits of integration appearexplicitly in their Riemann sum. Thus, a typical error in writing the Riemann sum is towrite

3i=0. When you write this you are summing up exactly four terms; to be useful

to you in arriving at an exact answer, your Riemann sum should have n terms.But the biggest logical problem is what comes next. Writing limn

3i=0 doesnt

make sense since there is no n in sight. The alternative, lim33

i=0, is nonsense,because 3 is a constant and therefore glued in place on the number line; the number 3cant pack up and march off to innity.

To arrive at an exact answer youll want to write a generic Riemann sum with nterms:

ni=1

(or, alternatively,n1

i=0 , a generic left-hand Riemann sum with n terms), so you canlet n increase without bound.

4 The units of the density will indicate whether to multiply by volume or area or length.


Try your hand at slicing and approximating the mass of the ith slice by working throughthe exercise below.

EXERCISE 27.4 Suppose the density of an object varies with height. Density = (h) grams per cubiccentimeter. Your task is to write an integral giving the total mass of the object. For eachof the objects listed, do the following.

(a) Describe a typical slice.

(b) Approximate the mass of the ith slice. Express the mass in terms of hi.

(c) Write a generic Riemann sum approximating the total mass of the object.

(d) Write an integral giving the total mass.

Object 1: A right circular cylinder of height 20 centimeters and radius 5 centimeters

Object 2: A right circular cone of height 20 centimeters and base of radius 5 centimeters

Object 3: A sphere of radius 5 centimeters

Answers

Object 1

(a) Partition [0, 20] into n equal pieces. This slices the cylinder into coin-shaped disksof thickness h. Each disk has radius 5 and volume 25h.

(b) mass of ith disk density volume (hi) 25h

55

20slice:

{h}h

hi

mass (hi) 25 h

Figure 27.7

(c) total mass n0 (hi) 25h(d) total mass = limn n0 (hi) 25h = 200 (h)25 dhObject 2

(a) Partition [0, 20] into n equal pieces. This slices the cone into approximately coin-shaped disks of thickness h. The radius of each disk varies with h. (The slices arenot perfect disks, but they become increasingly disk-like as m increases.)

(b) mass of ith disk density volume (hi) (ri)2h. We must get ri in termsof hi.


55

h}

for h smallhi hi

20 hiri

ri

}

hslice:

Figure 27.8

Using similar triangles we know that

20 hi20

= ri5

so

ri = 20 hi20

5 = 20 hi4

= 5 14hi.

mass of ith slice (hi)(5 14hi)2h(c) total mass n0 (hi)(5 14hi)2h(d) total mass = limn n0 (hi)(5 14hi)2h = 200 (h)(5 14h)2 dhObject 3

(a) The height of the sphere is 10 centimeters. Partition [0, 10] into n equal pieces.This slices the sphere into essentially coin-shaped disks of thickness h. Theradius of each disk varies with h. The slices are not perfect disks, but they becomeincreasingly disk-like as n increases.

(b) mass of ith disk density volume (hi) (ri)2h. We must get ri in termsof hi.

55

5

5 hi

hi 5

ri

ri

ri

hi

hi

hi

i th slice: (a)

(b) (c)

h

Notice that(hi 5)

2 = (5 hi)2,

although hi 5 5 hi.

Figure 27.9

Using the Pythagorean Theorem (see Figure 27.9, parts b and c), we know

(5 hi)2 + r2i = 25r2i = 25 (5 hi)2.


the mass of the ith disk (hi) [25 (5 hi)2]h(c) total mass n0 (hi) [25 (5 hi)2]h(d) total mass = limn n0 (hi) [25 (5 hi)2]h

= 100 (h) [25 (5 h)2] dh = 100 (h) (10h h2) dhAnswers to Selected Exercises

Exercise 27.1

(a)area of ith ring =

(ri + r

2

)2

(ri r

2

)2

= [r2i + rir +

(r

2

)2]

[r2i rir +

(r

2

)2]

= [r2i + rir +

(r

2

)2 r2i + rir

(r

2

)2]

= (2rir)(b) In Example 27.3, R = 10. The amount of debris in the ith interval (ri)2rir .

Therefore the total mass ni=1 (ri)2rir .Total mass = limn

ni=1 (ri)2rir =

100 (r)2r dr .

Exercise 27.2

Partition the r-interval [0, R] into n equal subintervals each of length r = Rn

. Letri = ir , i = 1, . . . , n.

The amount of glaze in the ith ring (ri) 2rir .Therefore, the total amount of glaze ni=1 (ri)2rir .Total amount of glaze = limn ni=1 (ri)2rir = R0 (r)2r dr .

P R O B L E M S F O R S E C T I O N 2 7 . 1

1. A cylinder 80 centimeters tall with a 10-centimeter radius is lled with a compressiblesubstance. The density of this substance is given by (h) grams per cubic centimeter,where h is the height (in centimeters) from the bottom of the cylinder. Write anexpression for the total mass of the substance in the cylinder.

2. A city is in the shape of a rectangle 4 miles wide by 6 miles long. A river runs throughthe middle of the city, parallel to the 6 mile-long sides. People prefer to live nearer thewater, so the density of people is given by (x) = 10,000 800x people per squaremile, where x is the distance from the river. (You may ignore the width of the river inthis problem.)

(a) Show in a sketch how you will need to slice up the region.

(b) What is the area of the ith slice?

(c) What is the approximate population in the ith slice?

(d) Write a Riemann sum to estimate the total population of the city.


(e) Calculate the exact population by taking the limit of the Riemann sum and evalu-ating the resulting denite integral.

3. Traditionally, when a college football team seems certain to receive a bid to play inthe post-season Orange Bowl, fans begin to throw oranges onto the eld. Suppose thatat one point during a game, the number of oranges per square yard between the goalline and the 30-yard line is given by (x) = 30x3 oranges per square yard, where x isthe number of yards from the goal line. If the eld is 160 feet (160/3 yards) wide, howmany oranges lie between the goal line and the 30-yard line?

4. (a) A farmer has planted corn on a rectangular plot of land 800 meters by 1000 meters.A straight stream runs alongside one of the long borders of the plot, and the farmersirrigation system is such that his yield decreases with the distance from the stream.Suppose his yield is given by f (x) = 50 0.3x ears of corn per square meter,where x is the distance from the stream in meters. What is the farmers yield fromthe plot?

(b) A second farmer plants his corn in a circular plot with radius 80 meters and he hasa centralized irrigation system located in the middle of his eld. His yield dropswith the distance from the center of the eld. Suppose his yield is also given byf (x) = 50 0.3x ears of corn per square meter, this time x being the distancefrom the center of the eld. What is the farmers yield from this plot?

5. Consider a box of cereal with raisins. The box is 5 centimeters deep, 25 centimeterstall, and 16 centimeters wide. The raisins tend to fall toward the bottom; assume theirdensity is given by (h) = 4

h+10 raisins per cubic centimeter, where h is the heightabove the bottom of the box. How many raisins are in the box?

6. The density of dart holes on an old dartboard is given by (r) = 1010(r2+1)2 holes per

square inch, where r is the distance, in inches, from the center of the board. If theboard is a circle with diameter 20 inches, nd the total number of holes in the board.

7. A beam of light is shining onto a screen creating a disk of radius 50 centimeters. Theintensity of light is brightest at the center and diminishes away from the center. If theintensity of light at a distance r from the center of the beam is given by f (r) = 150

20+r2watt/square cm, nd the total wattage of the beams image on the screen.

8. A coastal town is in the shape of a 7-mile by 2-mile rectangle, with one of the 7-milesides along the coast. In this town people want to live near the beach and the populationdensity at a distance x from the coast is given by (x) = 4000 2000x people persquare mile.

(a) Write a general Riemann sum that approximates the total population of the town.

(b) Use your answer to part (a) to write a denite integral that represents the totalpopulation of the town and evaluate the integral.

9. In Example 27.3, the density within a circle depended on the distance from the center,so we sliced the circle into concentric circular rings and found the area of a slice bytwo different methods.


Now suppose that the density within a sphere is given by (r), where r is thedistance from its center.

(a) Describe the slices that would be used when approximating the total mass.

(b) Find the volume of each slice by a method analogous to the rst method used inExample 27.3. (You will need to know that the volume of a sphere of radius r is4r3

3 .)

(c) Explain both geometrically and numerically why it is reasonable to approximatethe volume of the ith slice by 4r2i r .

10. A rectangular meadow is 100 meters. A straight irrigation pipe 400 meters long runsdown the center of the meadow dividing it lengthwise in half. The density of wildowersin the meadow varies with x, the distance from the irrigation pipe.

(a) If the density is given by g(x) owers per square meter, write an integral givingthe number of wildowers in the meadow. (Hint: Take advantage of symmetry.)

(b) If the density is given by g(x) = 501+x2 owers per square meter, how many wild-

owers are in the meadow?

11. (a) Suppose that the density of organisms in a certain petri dish varies with the distancefrom the center of the dish. The density at a distance x centimeters from the centeris given by f (x) organisms per square centimeter. The petri dish is 18 centimetersin diameter.

i. Write an integral that gives the number of organisms in the dish.ii. Find the number of organisms in the dish if f (x) = 100ex2 organisms per

square centimeter.

(b) Suppose that the density of organisms in a certain petri dish varies with the distancefrom a strip of nutrients running along the diameter of the dish. The density at adistance x centimeters from the line of nutrients is given by f (x) organisms persquare centimeter. The petri dish is 18 centimeters in diameter.

i. How will you slice up the petri dish?ii. Approximate the number of organisms in the ith slice.

iii. Write a Riemann sum approximating the total number of organisms in thepetri dish.

iv. Write an integral that gives the number of organisms in the dish.

12. Suppose that the density of a planet of mass in a gaseous planet is given by the function(r) = 40000

1+.0001r3 kilograms per cubic kilometer, where r is the number of kilometersfrom the center of the planet. Find the total mass of the planet if it has a radius of 8000kilometers.

13. A chocolate trufe is a wonderfully decadent chocolate concoction. Trufes tend to bespherical or hemispherical.

(a) Consider a trufe made by dipping a round hazelnut into various chocolates,building up a delicious spherical delicacy. The number of calories per cubicmillimeter varies with x, where x is the distance from the center of the hazelnut.If (x) gives the calories/mm3 at a distance x millimeters from the center, writean integral that gives the number of calories in a trufe of radius R.


(b) Another trufe is made in a hemispherical mold with radius R. Layers of differenttypes of chocolate are poured into the mold, one at a time, and allowed to set. Thenumber of calories per cubic millimeter varies with x, where x is the depth fromthe top of the mold. The calorie density is given by (x) calories/mm3. Write anintegral that gives the number of calories in this hemispherical trufe.

Top of the mold

Hemispherical truffle mold

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

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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 theducks. As a result, the density of ducks on the pond is given by a function (x) ducksper square meter, where x is the distance from the bridge. How many ducks are on thepond? (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 thepond. Ducks, after all, are discrete. We are making a continuous model of a discretephenomenon.

16. Let W(t) be the amount of water in a pool at time t , t measured in hours and W measuredin gallons. t = 0 corresponds to noon. Water is owing in and out of the pool at a rategiven by dW

dt= 30 cos (2 t). During what time interval between noon and 5:00 p.m.

(0 t 5) is water owing 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 releasedradioactive iodine 131. Fortunately, the reactor has a containment building, which keptthe iodine from being released into the air. The containment building is hemisphericalwith a radius of 100 feet. The density of iodine in the building was 6 105(200 h)g/cubic feet, where h is the height from the oor (in feet). (It ranges from 12 103g/cubic feet at the oor to 6 103 g/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 staris given by (r) = K

(r+1)3/2 kilograms per cubic kilometer, where r is the distance (inkilometers) from the stars center and K is a positive constant.

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

19. A substance has been put in a centrifuge. We now have a cylindrical sample (radius3 centimeters, height 4 centimeters) in which density varies with x, the distance (incentimeters) from the central axis. If the density is given by (x) mg/cm3, write anintegral that gives the total mass of the substance.

20. A very thin, lighted pole 10 feet tall is placed upright in a familys backyard to attractinsects to it (where they are electrocuted). At one moment, the density of these insectsis given by (r) = 1.3

(r+1) insects per cubic foot, where r measures the number of feetfrom 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 ofx feet from the center pole is given by h(x) = 8

1+ x216feet. 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 ofgarlic varies with x, the distance from the center of the pizza, and is given by

g(x) = x(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 $100,000per year over the next 20 years? Assume 5% interest compounded continuously.By a continuous stream of income we mean that we are modeling the situationby assuming that money is being generated continuously at a rate of $100,000 peryear.

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

Application Definite Integration

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