Unit #14 - Center of Mass, Improper Integrals

Some problems and solutions selected or adapted from Hughes-Hallett Calculus.

Computing Totals With Slices

1. Find the mass of a rod of length 10 cm, with lineal density δ(x) = e−x g/cm, where x is the distance in cm fromthe left end of the rod.

Since the density is e−x g/cm, the mass of each thin slices of the rod will be e−x ·∆x. Units: (g/cm) × cm = g.

Total mass =

∫ 10

0

e−x dx = −e−x∣∣∣100

= 1− e−10 ≈ 0.99995 g

2. A rod is 2 meters long. At a distance x meters from its left end, the density of the rod is given by δ(x) = 2 + 6xg/cm.

(a) Write a Riemann sum approximating the total mass of the rod. (Do not evaluate this sum.)

(b) Find the exact mass by converting the sum into an integral and evaluating it.

(a) Note that the bar is 2 meters long, but the units of density are in g/cm. For consistency, we will measure everying incm.

The density is changing as we move along the rod, so we can’t simply multiply (density)×(length) to get the totalmass. Instead, we’ll need to slice the rod up into thin pieces (say ∆x wide each), and find the mass on each sliceseparately; on each slice, because it is thin, the density will be roughly constant.

Mass of a slice = density(x) · length of slice

= (2 + 6x)[ g/cm] · (∆x)[ cm]

= (2 + 6x) ∆x[g]

The total mass will be given by the sum of the mass of all slices, or

Total mass of the rod ≈∑i

(2 + 6xi)∆x

(b) By letting the slice width ∆x → 0, we conver the Riemann sum to the integral below. Note that the limits ofintegration are x = 0 and x = 200, because we are measuring x in cm, and the bar is 2 m = 200 cm long.

Total mass of the rod = intx=200x=0 (2 + 6x) dx

=

(2x+ 6

x2

2

) ∣∣∣2000

= [2(200 + 3(200)2]− [0 + 0]

= 120, 400 g

(or very heavy! That’s 120.4 kg!)

3. Find the mass of the block 0 ≤ x ≤ 10, 0 ≤ y ≤ 3, 0 ≤ z ≤ 1, whose density δ, is given by

δ = 2− z mass units/unit volume, for 0 ≤ z ≤ 1.

1

4. Circle City, a typical metropolis, is densely populated near its center, and its population gradually thins out towardthe city limits. In fact, its population density is 10, 000(3 − r) people/ square mile at distance r miles from thecenter.

(a) Assuming that the population density at the city limits is zero, find the radius of the city.

(b) What is the total population of the city?

5. The density of oil in a circular oil slick on the surface of the ocean at a distance r meters from the center of theslick is given by δ(r) = 50/(1 + r) kg/ m2.

(a) If the slick extends from r = 0 to r = 10,000 m, find a Riemann sum approximating the total mass of oil in theslick.

(b) Find the exact value of the mass of oil in the slick by turning your sum into an integral and evaluating it.

(c) Within what distance r is half the oil of the slick contained?

This question is very similar to the one done in the lecture notes.

(a) Cut the circular spill into thin circular ribbons, each ribbon being a small ∆r thick/wide. When you unroll one ofthese ribbons, they will be 2πr (circle circumference) long, and ∆r wide, giving an area of

slice area = 2πr∆r

On each of these slices, the radius is essentially constant (because ∆r is small), so we can determine the amount ofoil in the slice by

mass of oil in slice (kg) = oil density (kg/m2)× area m2

=

(50

1 + r

)(2πr∆r)

so total mass of oil =∑ 50

1 + r(2πr) ∆r

2

(b) Converting the Riemann sum into an integral,

Total mass of oil =

∫ r=10,000

r=0

50

1 + r(2πr) dr

= 100π

∫ 10,000

0

r

r + 1dr

You can evaluate this integral with a substitution (let w = r+ 1), but a shorter way is to use a quick trick to simplifyit. We note that the problem with the integral is the r + 1 in the denominator can’t be simplified from the currentform. However, if we had r + 1 in the numerator as well, we could do some cancelling of terms. To obtain this r + 1in the numerator, we can both add and subtract 1 from the numerator.

100π

∫ 10,000

0

r

r + 1dr = 100π

∫ 10,000

0

r+1− 1

r + 1dr

Regrouping, = 100π

∫ 10,000

0

(r + 1

r + 1

)− 1

r + 1dr

= 100π

∫ 10,000

0

1− 1

r + 1dr

Now integrating, = 100π (r − ln |r + 1|)∣∣∣10,0000

= 100π(10, 000− ln(10, 001)))

≈ 3.14× 106 kg

(c) To determine the radius where half the oil is contained, what we are asking is “What upper limit of integration wouldgive us half the integral value of the whole oil slick?” I.e. for what radius R is

100π

∫ R

0

r

r + 1dr =

(1

2

)3.14× 106?

For simplicity, we’ll use π for the 3.14.A

To solve this, we re-use our antiderivative from before, and get

100π (r − ln |r + 1|)∣∣∣R0

=

(1

2

)π · 106

(r − ln |r + 1|)∣∣∣R0

=

(1

2

)104

(R− ln |R+ 1|)− (0− ln(1)) = 5× 103

R− ln |R+ 1| = 5× 103

This is not an equation that is easy to solve by hand (actually, it’s not even possible to solve by hand!) Fortunately,we have some tools in our toolkit that can help.

• Guessing and checking, to get a rough answer, and/or

• using Newton’s method from the first half of the course to solve the equation.

Both these techniques would be used in practice by scientists and engineers. Note: on a test or exam, we would giveyou warning that these techniques would be required when we posed the question. Here the problem arose naturally,and it’s a good concept check to see if you remembered how you would deal with this type of problem, without aprompt.

Since the whole slick is 10,000 m in radius, we could try R = 5000 as a reasonable starting point:

R = 5000 : 5000− ln(5000 + 1) ≈ 4992

which is pretty close to 5×103. If we guess just 8 higher (since the R term has more impact than the ln(R+ 1) term),

R = 5008 : 5008− ln(5008 + 1) ≈ 4999.5

3

which is very close to 5× 103. Based on this, we can say that half the oil would be contained in a circle of radius 5008m (compared to the whole slick in 10,000 m radius).

Using Newton’s method to refined this value further to 5008.519 m, but that kind of accuracy is a little silly given therelatively low precision of the input values. The guess-and-check 5008 m is likely accurate enough for most uses.

6. The soot produced by a garbage incinerator spreads out in a circular pattern. The depth, H(r), in millimeters, ofthe soot deposited each month at a distance r kilometers from the incinerator is given by H(r) = 0.115e−2r.

(a) Write a definite integral giving the total volume of soot deposited within 5 kilometers of the incinerator eachmonth.

(b) Evaluate the integral you found in part (a), giving your answer in cubic meters.

Center of Mass

7. A point mass of 2 grams located 3 centimeters to the left of the origin and a point mass of 5 grams located 4centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system.

8. Find the center of mass of a system containing three point masses of 5 g, 3 g, and 1 g located respectively atx = −10, x = 1, and x = 2.

9. A rod with density δ(x) = 2 + sin(x) g/cm lies on the x-axis between x = 0 and x = π. Find the center of mass ofthe rod.

4

The center of mass is given by

x =

∫ π0x (2 + sin(x)) dx∫ π

0(2 + sin(x)) dx

.

Evaluating the numerator of x, ∫ π

0

x (2 + sin(x)) dx =

∫ π

0

(2x+ x sin(x)) dx

(The second part of integral,∫x sin(x) dx, requires integration by parts: see earlier Units.)∫ π

0

(2x+ x sin(x)) dx = x2 − x cos(x) + sin(x)∣∣∣π0

= (π2 − π cos(π) + sin(π))− (0− 0 + sin(0))

= π2 + π

Evaluating the denominator of x, ∫ π

0

(2 + sin(x)) dx = (2x− cos(x))∣∣∣π0

= (2π − cos(π))− (0− cos(0))

= (2π + 1) + 1

= 2π + 2

This gives the overall center of mass as

x =π2 + π

2π + 2=π(π + 1)

2(π + 1)=π

2.

Note that this means the center of mass is exactly in the middle of the rod:x = π2 is halfway between the ends of x = 0

and x = π. If you graphed the density of the rod, given by δ(x) = 2 + sin(x), you would see that it is symmetric, so thiswould match our result.

10. A rod of length 1 meter has density δ(x) = 1 + kx2 grams/meter, where k is a positive constant. The rod is lyingon the positive x-axis with one end at the origin.

(a) Find the center of mass as a function of k.

(b) Show that the center of mass of the rod satisfies 0.5 < x < 0.75.

11. A rod of length 2 meters and density δ(x) = 3− e−x kilograms per meter is placed on the x-axis with its ends atx = ±1.

(a) Will the center of mass of the rod be on the left or right of the origin? Explain.

(b) Find the coordinate of the center of mass.

5

12. A metal plate, with constant density 2 g/cm2, has a shape bounded by the curve y = x2 and the x-axis, with0 ≤ x ≤ 1 and x, y in cm.

(a) Find the total mass of the plate.

(b) Sketch the plate, and decide, on the basis of the shape, whether x is less than or greater than 1/2.

(c) Find x.

6

13. A metal plate, with constant density 5 g/cm2, has a shape bounded by the curve y =√x and the x-axis, with

0 ≤ x ≤ 1 and x, y in cm.

(a) Find the total mass of the plate.

(b) Find x and y.

7

14. An isosceles triangle with uniform density, altitude a, and base b is placed in the xy-plane as in the diagram below.

Show that the center of mass is at x = a/3, y = 0. Hence show that the center of mass is independent of thetriangle’s base.

The triangle is symmetric about the x axis, so the center of mass in the y direction is y = 0.

To find x, we treat the density of the plate as a constant ρ kg/m2. We take vertical slices, which by similar triangles will

all have widths w =b

a(a− x), so

Area of small strip ≈ b

a(a− x) ∆x m2

Mass of small strip ≈ ρ[b

a(a− x) ∆x

]kg

8

We then use the center of mass formula (sum of moments, divided by sum of masses) to find the x center of mass:

x =

∑ximi∑mi

=

∫ x=ax=0

ρx ba (a− x)dx∫ x=ax=0

ρ ba (a− x)dx

cancelling the constant ρ’s and ba terms, =

∫ a0x(a− x)dx∫ a

0(a− x)dx

Taking each integral separately, ∫ x=a

x=0

x(a− x)dx =

∫ a

0

ax− x2 dx

=ax2

2− x3

3

∣∣∣a0

=a3

6∫ x=a

x=0

(a− x)dx = ax− x2

2

∣∣∣a0

=a2

2

Taking the ration of these two integral values,

x =a3/6

a2/2=a

3

The center of mass will always lie at 1/3 of the distance away from the base of the triangle.

Improper Integrals

15.

∫ ∞1

1

5x+ 2dx

9

∫ ∞1

1

5x+ 2dx = lim

b→∞

∫ b

1

1

5x+ 2dx

By substitution (w = 5x+ 2): = limb→∞

1

5ln(5x+ 2)

∣∣∣b1

= limb→∞

1

5(ln(5b+ 2)− ln(7))

Since ln(x)→∞ as x→∞, the limit limb→∞

ln(5b+ 2) does not converge, so the integral does not converge.

16.

∫ ∞1

1

(x+ 2)2dx

∫ ∞1

1

(x+ 2)2dx = lim

b→∞

∫ b

1

1

(x+ 2)2dx

By substitution (w = x+ 2): = limb→∞

−1

x+ 2

∣∣∣b1

= limb→∞

−1

b+ 2︸ ︷︷ ︸→0

+1

1 + 2=

1

3

The integral converges, and its value is1

3.

17.

∫ ∞0

xe−x2

dx

∫ ∞0

xe−x2

dx = limb→∞

∫ b

0

xe−x2

dx

By substitution (w = −x2): = limb→∞

−1

2e−x

2∣∣∣b0

= limb→∞

−1

2e−b

2

︸ ︷︷ ︸→0

−−1

2e0 =

1

2

The integral converges, and its value is1

2.

18.

∫ 0

−∞

ex

1 + exdx

∫ 0

−∞

ex

1 + exdx = lim

b→−∞

∫ 0

b

ex

1 + exdx

By substitution (w = 1 + ex): = limb→−∞

ln(1 + ex)∣∣∣0b

= ln(1 + e0)− limb→−∞

ln(1 + eb︸︷︷︸→0

)

= ln(2)− ln(1)︸ ︷︷ ︸=0

= ln(2)

The integral converges, and its value is ln 2.

10

19.

∫ 4

0

1√16− x2

dx

This integral requires remembering the derivative of arcsin:d

dxarcsinx =

1√1− x2

.

The function is undefined at x = 4, so we need a limit for that limit of integration.

∫ 4

0

1√16− x2

dx = limb→4−

∫ b

0

1√16− x2

dx limb→4−

∫ b

0

1

4√

1− (x/4)2dx

By substitution w = x/4: =4

4limb→4−

arcsin(x/4)∣∣∣b0

= limb→4−

arcsin(b/4)︸ ︷︷ ︸→π/2

− arcsin(0)︸ ︷︷ ︸=0

=π

2

The integral converges to π/2.

20.

∫ π/2

π/4

sinx√cosx

dx

Note that the denominator becomes undefined when cosx = 0, which occurs at x = pi/2 on the interval of integration.Thus we need to use a limit to “sneak up” on that end of the integral.

∫ π/2

π/4

sinx√cosx

dx = limb→π

2−

∫ b

π/4

sinx√cosx

dx

By substitution (w = cosx): = limb→π

2−−2√

cosx∣∣∣bπ/4

= limb→π

2−−2√

cos b︸ ︷︷ ︸→0

+2√

cos(π/4) = 21

21/4= 23/4

21.

∫ ∞1

1

x2 + 1dx

∫ ∞1

1

x2 + 1dx = lim

b→∞

∫ b

1

1

x2 + 1dx

By recalling d/dx arctanx = 1/(1 + x2): = limb→∞

arctanx∣∣∣b1

= limb→∞

arctan b︸ ︷︷ ︸→π/2

− arctan 1 =π

2− π

4=π

4

The integral converges to the value π/4.

22.

∫ ∞2

1

x lnxdx

11

∫ ∞2

1

x lnxdx = lim

b→∞

∫ b

2

1

x lnxdx

By substitution (w = ln(x)): = limb→∞

ln(lnx)∣∣∣b2

= limb→∞

ln(ln b)︸ ︷︷ ︸→∞

− ln(ln 2)

The limit does not exist, so the integral cannot converge.

23.

∫ 1

0

lnx

xdx

Note that the function is undefined at x = 0, because both lnx and 1x are undefined there. This means the integral is

improper, and we need to use limits for the lower limit of integration.

∫ 1

0

lnx

xdx = lim

b→0+

∫ 1

b

lnx

xdx

By substitution (w = lnx): = limb→0+

1

2(lnx)2

∣∣∣1b

=1

2(ln 1)2 − 1

2limb→0+

(ln b)2︸ ︷︷ ︸→∞

Since the limit does not converge, the integral does not converge either.

24.

∫ π

0

1√xe−√x dx

The function is undefined at x = 0, so we need to use a limit for the lower limit of integration.

∫ π

0

1√xe−√x dx = lim

b→0

∫ π

b

1√xe−√x dx

By substitution (w =√x): = lim

b→0+−2e−

√x∣∣∣πb

= −2e−√π − lim

b→0+−2e−

√b︸ ︷︷ ︸

→−2e0=−2

= 2− 2e−√π

The integral converges to the value 2− 2e−√π.

25. Removed - Required table of integrals.

26. Given that

∫ ∞−∞

e−x2

dx =√π, calculate the exact value of

∫ ∞−∞

e−(x−a)2/b dx

12

We can approach with two substitutions. First, let w = (x−a). This doesn’t affect this limits,since as x→ ±∞, w → ±∞as well.

w = (x− a) and dw = dx

So

∫ ∞−∞

e−(x−a)2/b dx =

∫ ∞−∞

e−w2/b dw

Before making the next substitution, it will help to group the exponent factors inside the square:∫ ∞−∞

e−w2/b dw =

∫ ∞−∞

e−( w√

b)2dw

We can now do a second substitution, t = w/√b, dt =

dw√b:∫ ∞

−∞e−w

2/b dw =

∫ ∞−∞

e−( w√

b)2dw

=√b

∫ ∞−∞

e−t2

dt

However, for definite integrals, variable names don’t matter, so∫ ∞−∞

e−t2

dt =

∫ ∞−∞

e−x2

dx =√π

Therefore,

∫ ∞−∞

e−(x−a)2/b dw =

√b

∫ ∞−∞

e−t2

dt =√b√π

27. The rate, r, at which people get sick during an epidemic of the flu can be approximated by r = 1000te−.5t, wherer is measured in people/day and t is measured in days since the start of the epidemic.

(a) Sketch a graph of r as a function of t.

(b) When are people getting sick fastest?

(c) How many people get sick altogether?

(a) Since we’re looking at t ≥ 0, both t and e−0.5t will be positive, so r(t) will always be positive on our graph. To get astarting point, at t = 0, r = 0. As t→∞, the exponential factor will be the most important, so as e−0.5t → 0, so toor(t)→ 0. As we’ll see in part (b), r(t) has only one critical point, so this leads to a shape like the one below:

(b) People are getting sickest fastest when the rate is largest. Since we actually have a formula for the rate, we can findits maximal value by looking for critical points (where r′(t) = 0):

r′ = 1000 · 1 · e−0.5t + 1000 · t · (−0.5)e−0.5t

Setting r′ = 0: 0 = 1000 · 1 · e−0.5t + 1000 · t · (−0.5)e−0.5t

Divide by 1000e−0.5t: 0.5t = 1

t = 2

13

So there is a single critical point at t = 2. We can tell this is a global maximum by looking at the signs of r′(t) =1000e−0.5t1− 0.5t:

• for all t < 2, 1− 0.5t > 0, so r′ > 0, and r is increasing

• for all t > 2, 1− 0.5t < 0, so r′ < 0, and r is decreasing

At t = 2, r(2) = 1000(2)e−1 ≈ 735, so at its worst, the disease is spreading by approximately 735 people per day.

(c) The number of sick people over the whole infection is the integral of the rate from the start time (t = 0) to the endtime t→∞, or

Total Sick =

∫ ∞0

1000te−0.5t dt

= limb→∞

1000

∫ b

0

te−0.5t dt

By parts, with u = t, dv = e−0.5t dt: = 1000

(limb→∞

te−0.5t

−0.5

∣∣∣b0−∫ b

0

e−0.5t

−0.5dt

)

= 1000

(limb→∞

−te−0.5t

0.5− e−0.5t

0.25dt∣∣∣b0

)

= 1000

limb→∞

−be−0.5b

0.5︸ ︷︷ ︸→0- see below

− e−0.5b

0.25︸ ︷︷ ︸→0

− [0− e0

0.25

]= 1000/0.25 = 4000

Over the course of the disease, 4000 people will have been infected. 4000 also represents the area under the r(t) graph.

Note: The limit limb→∞

−be−0.5b can be more easily studied by writing it as limb→∞

− b

e0.5b. Since any positive exponential

function will eventually be much larger than a linear function (think of their respective graphs), the denominator willdominate this expression. This leads to a limit of zero.

Alternatively, you can think of this function having the same shape as the graph, shown in part (a). If you considerlarger and larger values of t (or b), the function approaches zero asymptotically.

28. Suppose a function h is defined by

h(x) =

1

x

√x− 1

16if 0 < x ≤ 4

1

x2if x > 4.

Consider the following integrals:

(i)

∫ 2

0

h(x)dx (iii)

∫ ∞4

h(x)dx

(ii)

∫ 4

2

h(x)dx (iv)

∫ ∞0

h(x)dx

For each integral, determine the following:

• Is the integral improper?

• If improper, does it diverge?

• If it is convergent, or a proper integral, what numerical value does the integral converge to?

For easier integration, we write the function h(x) using powers as follows:

h(x) =

{x−1/2 − 1

160 < x ≤ 4

x−2 x > 4

and note that its domain is the positive real numbers (all x > 0).

14

(i)

∫ 2

0

h(x) dx =

∫ 2

0

(x−1/2− 1

16) dx since x < 4 for all x in the interval 0 < x ≤ 2. Since the integrand gets unbounded

as x→ 0+, i.e. limx→0+

h(x) =∞, the integral

∫ 2

0

h(x) dx is an improper integral.

Moreover, ∫ 2

0

h(x) dx =

∫ 2

0

(x−1/2 − 1

16) dx

= limb→0+

∫ 2

0

(x−1/2 − 1

16) dx

= limb→0+

[2x1/2 − (

1

16)x

] ∣∣∣2b

= limb→0+

[(2(2)1/2 − 1/8]− [2b1/2 − (1

16)b]

= [2(2)1/2 − 1/8] + 0 + 0

= 2√

2− 1/8

Since this limit is finite, the integral

∫ 2

0

h(x) dx converges.

Summary: (i) is an improper integral that converges.

(ii)

∫ 4

2

h(x) dx is not an improper integral, because the limits of integration are finite and h(x) is also finite over the

interval of integration. It is just an ordinary definite integral.

Evaluating it, ∫ 4

2

h(x) dx =

∫ 4

2

(x−1/2 − 1

16) dx

= [2x1/2 − (1

16)x]∣∣∣42

= [4− 1/4]− [2 · 21/2 − 1/8]

= 31/8− 2√

2 ≈ 1.0466

Summary: (ii) is an proper integral (which converges, but that’s no surprise because so do all proper integrals).

(iii)

∫ ∞4

h(x) dx =

∫ ∞4

x−2 dx (since x > 4 for all x in the interval (4,∞)). Since the interval from 4 to infinity is

infinitely long,

∫ ∞4

h(x) dx is an improper integral.∫ ∞4

h(x) dx =

∫ ∞4

x−2 dx

= limb→∞

∫ b

4

x−2 dx

= limb→∞

[−x−1]∣∣∣b4

= limb→∞

−1

b︸ ︷︷ ︸→0

+1

4

=1

4

Summary: (iii) the integral

∫ ∞4

h(x) dx is improper, and converges to the value1

4.

(iv) ∫ ∞0

h(x) dx =

∫ 2

0

h(x) dx+

∫ 4

2

h(x) dx+

∫ ∞4

h(x) dx

15

All three of these integrals converge, so we can conclude that the overall integral

∫ ∞0

h(x) dx also converges.

We can be even precise and compute a value for the integral:∫ ∞0

h(x) dx = (2√

2− 1/8)︸ ︷︷ ︸∫ 20

+ (31/8− 2√

2)︸ ︷︷ ︸∫ 42

+ 1/4︸︷︷︸∫∞4

= 4∫ ∞0

h(x) dx is an improper integral because the interval from 0 to infinity is infinitely long, but it converges to the

finite value 4.

Summary: (iv) the integral

∫ ∞0

h(x) dx is improper and converges.

16