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**How come in question 2A the radius of the cylindrical shell isnt y sub k + 1? Contrast that to the question 1B in the fall2012 midterm, where the cylindrical shell in the horizontal slicing method adds a +2 to x sub k to account for the axis of rotation? **

Fall 2012 Exam 1

Fall 2012 Exam 1

1. (a) Let R be the region in the xy-plane bounded by the curve y = x3 , the y-axis, and y = 5. Find the

volume obtained when R is revolved around the y-axis. (Please evaluate your integral to give a

numerical answer, and include a sketch to indicate how you are slicing.)

Solution.

Let us slice the region horizontally. That is, we will slice the interval [0,5] on the y-axis into n

slices. We label the points on the y-axis as y0 = 0, y1 , y2 , . . . , yn = 5, as in the picture below. Now we revolve the k-th slice (the red one). We get a solid that looks like a coin. To compute its

volume, we need to know its thickness and radius. The thickness is ?y. To compute the radius,

we look back at our graph: it should be the x-coordinate of the point (R,yk ) on the graph! This

means that

1

yk = R3 ?? R = yk3

Putting everything together, we have

1 volume of revolved k-th slice ? ?(radius)2 (thickness) = ?(yk3 )2 ?y

APPROX

Summing over all slices, we get our Riemann sum:

n

X

k=0 1 2 ?yk3 ?y Finally, taking the limit as n goes to infinity, we get the following expression for the volume of

the solid:

Z 5

2

?y 3 dy

0 Note that the bounds go from 0 to 5, just like the interval we sliced.

Let us evaluate this integral.

Z 5

0 2

5

2

3 5

3

?y 3 dy = [ ?y 3 ]5x=0 = ?(5 3 ) ? 0 = 3?5 3

5

5 (b) Let S be the region in the plane that lies above the x-axis, below the curve y = cos x, and between

x = ? ?2 and x = ?2 .

i. Write an integral that gives the volume that is swept out by revolving the region S around the

vertical line x = ?2. (Please include a sketch to indicate how you are slicing.)

Solution.

We first sketch the graph for the curve y = cos x, ? ?/2 ? x ? ?/2, with the revolution axis

x = ?2, a vertical line passing through (?2,0). We shade the region that is enclosed by this

graph and x-axis. Depending on the way we want to slice we will have either of the following

pictures. (In order to help us see the revolving radius, we also select one slice and find out

the radiuses that we shall need.) vertical slicing horizontal slicing slicing vertically along x: If we slice vertically, then when revolving a slice around the line

x = 2 we will get a hollow cylinder (a paper towel tube). As always we cut it and flatten it

to see a rectangle.

The width of the rectangle is the circumference of the cylinder, which

is 2? times the radius of revolution xk + 2 (the distance |xk ? (?2)|

between xk and ?2). The height is the height of the red strip, which is

cos xk .

Multiplying the thickness ?x, we get that the volume of the k-th slice hollow cylinder is

k th volume = 2? (xk + 2) cos xk ?x. 2 The definite integral for the total volume shall be

Z x= ?

2 2? (x + 2) cos x dx.

x=? ?

2 slicing horizontally along y: If we slice horizontally, then when revolving a slice around the

line x = 2 we will get a annulus (a washer or a CD). The area of such shape is the area of

the outside circle minus the area of the inside circle. (Formula: ?(Rk2 ? rk2 ).)

The inside radius of the washer is ? arccos yk + 2 , the distance between the closer end of the red strip to the axis, this is the distance

between x1 (yk ) = ? arccos yk and ?2. The outside radius of the washer

is arccos yk + 2 , the distance between the further end of the red strip

to the axis, this is the distance between x2 (yk ) = arccos yk and ?2.

Multiplying the thickness ?y, we get that the volume of the k-th slice washer is

k th volume = ?[(arccos yk + 2)2 ? (? arccos yk + 2)2 ]?y.

The definite integral for the total volume shall be

Z y=1

?[(arccos y + 2)2 ? (? arccos y + 2)2 ] dy.

y=0 ii. Evaluate your integral to give a numerical answer for the volume of revolution in part i.

Please simplify your answer.

Solution.

with vertical slice: It may be easier if we break it into two integrals.

Z ?

2 Z ?

2 2? (x + 2) cos x dx = 2?[

??

2 ??

2 3 Z ?

2 x cos x dx + 2 cos x dx]

??

2 For the first integral we are going to use integration by parts:

Z ?

2 x cos x dx

??

2 Z ?

2 x sin x|? ? ? = 2 (Here take u = x, dv = cos x dx ? v = sin x) sin x dx

??

2 ? = ?

2 ? x sin x|?2 ? + cos x|?2 ?

2

2

?

?

[( )(1) ? (? )(?1)] + [(0) ? (0)] = 0

2

2 = (Indeed, we could have notice this earlier because x cos x is an odd function.)

The Second integral is much easier:

Z ?

2 ??

2 ? 2 cos x dx = 2 sin x|?2 ? = 2[(1) ? (?1)] = 4

2 We plug them back and see

Z ?

2 Z

2? (x + 2) cos x dx = 2?[ ??

2 ?

2 Z ?

2 2 cos x dx] = 2?[0 + 4] = 8? . x cos x dx + ??

2 ??

2 with horizontal slice: Let us first simplify it:

Z y=1 ?[(arccos y + 2)2 ? (? arccos y + 2)2 ] dy = y=0 Z y=1 ?(8 arccos y) dy.

y=0 This is the area under the graph h(y) = 8? arccos y from y = 0 to y = 1. However, when

we compute the definite integral by Riemann sum, we can also try to slice it the other way

(vertically to horizontally). In other words, from the picture h

this the same as the area between the graph of y = cos 8?

and the h-axis from height h = 0

to height h = 8? arccos 0 = 4. So we can compute the integral by the following integral Z h=4 cos

h=0 h

dh = 8?

8? Z u= ?

2 u= ? cos u du = 8? sin u|u=02 = 8? .

u=0 4 2. Bill has been growing a giant pumpkin for the Minnesota State Fair. The pumpkin is spherical with a

radius of 8 feet. Because of gravity, the moisture inside the pumpkin tends to sink to the bottom, so the

density of the pumpkin varies; the density is given by ?(h) pounds/cubic foot, where h is height above

the ground (the pumpkin sits on the ground). Write a definite integral that gives the mass of

the pumpkin.

Please include a sketch and describe how you are slicing. Please also define all your

notation carefully, including your ?h, your hk and the interval you are slicing.

Solution.

A sketch of a sphere with horizontal slices is adequate here - we want to slice horizontally so that

distance from the ground is constant.

It is perfectly fine to represent the pumpkin with the equation x2 + y 2 = 64, which many students

did. However, it?s important to note that y 6= h because y is a vertical component centered in the

middle of the pumpkin (y = 0 corresponds to the horizontal center line 8 feet off the ground), while h

is height (h = 0 corresponds to the ground). I will answer this problem using either variable orientation.

First, let?s describe our notation: The interval we are working with depends on choice of whether we

are working in h or y. For y: [?8,8], for h:[0, 16].

16

Accordingly, ?y = 16

n , ?h = n

hk = 0 + k?h = k?h, yk = ?8 + k?y

Determining the mass of the kth slice is the volume of the kth slice multiplied by the density of the

kth slice. It?s especially important to note that a single slice of the pumpkin is a disc, and this is what

our entire integral/mass calculation focuses!

If we are working in terms of y, we don?t need to adjust y to represent the radius of the pumpkin in the

volume portion of the equation, but we will need to adjust y to represent height, since ? is a function

which we can only plug height into:

q

mass of kth slice = ?( 64 ? yk2 )2 ?(yk + 8)?y

If we are working in terms of h, we don?t need to adjust our input into the density function, but we

do need to adjust h for it to represent our radius (since we are using the equation of a circle centered

at the origin, not sitting on the ground!).

p

mass of kth slice?( 64 ? (hk ? 8))2 ?(hk )?h

The Riemann sum of either (not necessary to the problem, but useful nonetheless and a good habit to

get into, looks like this:

n

q

X

?( 64 ? yk2 )2 ?(yk + 8)?y

k=1 or n

X p

?( 64 ? (hk ? 8)2 )2 ?(hk )?h k=1 Which gives us a final integral of either:

Z 8

Z

?

(64 ? y 2 )?(y + 8)dy or ?

?8 0 5 16 (64 ? (h ? 8)2 )?(h)dh 3. On the surface of a circular pond of radius 100m there are millions of Duckweeds(1) . Duckweeds grow

close to shore, and the density of duckweeds is given by ?(x) plants/square meter, where x is the

distance in meters from the edge of the pond. Write an integral that gives the total number of

duckweed plants within 10m of the shore. Please include a sketch that shows how you are

slicing.

Solution. We slice in rings, so that slices have roughly constant density. The radius of the kth ring is 100 ? xk ,

and its thickness is ?x = 10

n , so the number of plants in the kth slice is

2?(100 ? xk )?(xk )?x.

Thus, the total number of plants within 10m of the edge of the pond is

Z 10

2?(100 ? x)?(x)dx.

0 The above is in terms of x, the distance from the edge of the pond. If we instead wanted to do things

in terms of r, the distance from the center of the pond, then we could use the relation x = 100 ? r to

see that the number of plants in the kth slice is

2?rk ?(100 ? rk )?r.

In terms of r, the total number of plants within 10m of the edge of the pond is

Z 100

2?r?(100 ? r)dr

90 which we can show is equivalent to the integral in terms of x using the substitution x = 100 ? r.

(1) These plants are also called ?water lentils?, because each tiny plant is about the size of a lentil floating on the water. They

are an important food source for waterfowl, and important as water purifiers, since they grow quickly and absorb excess nitrogen

and phosphates. 6 4. Parts (a), (b) and (c) of this question are unrelated, so please read carefully.

(a) A Chevrolet Corvette accelerates from rest on a test course. Its speed is recorded at two second

intervals as follows:

Time (s)

0

2

4

6

8 Speed (m/s)

0

12

21

27

30 i. If you decide to use the trapezoid rule to estimate the distance traveled by the car, will you

expect to get an over- or under-estimate? Please write one sentence of explanation. You don?t

need to carry out this computation.

Solution.

If we plot the first few values of the speed, we see that the graph of speed versus time is

concave down. (That is, the speed increases, but more slowly as time passes.) Since concave

down functions lie above their secant lines, and since the trapezoid rule approximates the

function on each slice by a secant, we expect the trapezoid rule to give an underestimate.

ii. Use Simpson?s rule to estimate the distance the Corvette traveled in the first eight seconds.

Solution. The easiest solution is to remember that the Simpson?s Rule can be computed by:

S2n =

1

y0 + 4y1 + 2y2 + . . . + 2y2n?2 + 4y2n?1 + y2n ?x

6 In this case, the values of yi are the values in the right column of the table. We need to

remember that, for example y0 and y2 are the left and right hand endpoints of the first slice,

and that y1 is the midpoint of the first slice ? Simpson?s rule requires about twice as much

data as the other two rules. This tells us that our slice width her is ?x = 4 seconds (a lot of

students mistakenly think it?s 2.) We get: S2(2) =

1

0 + 4(18) + 2(21) + 4(27) + 30 4 = 152

6 (b) Give an upper bound for |f (x)| on the interval from x = ?3 to x = 3, where

f (x) = 2 cos(2x) + sin x

.

2 + x2 Your bound doesn?t have to be the very best, but it should be reasonable, and it should be clear how

you arrived at this bound. Let?s get practical!

Solution. We can bound the function as follows: 2 cos(2x) + sin x 1 |f (x)| = = |2 cos(2x) + sin x| 2 + x2 2 + x2 1 ? 2 |cos(2x)| + | sin x| 2 + x2 7 Notice that since we?re trying to find an upper bound for the function, we want to find a number

that is bigger than the thing in parentheses ? 3 will clearly do, since the cos and sin terms are

never bigger than 1, and the worst imaginable case is that they happen to both be 1 at the same

time (even though this never happens, we?re being practical here!). At the same time, we?re

hoping to find a number that?s bigger than the fraction, which means we really care about how

small the denominator gets. The function 2 + x2 is never smaller than 2, since x2 is positive, and

so this is a reasonable value to take. In other words: 1 2 cos(2x) + sin x |f (x)| = = |2 cos(2x) + sin x| 2 + x2 2 + x2 1 ? 2 |cos(2x)| + | sin x| 2 + x2 1

? 3

2 (c) Your classmate, George, used Simpson?s Rule with n = 5 slices to estimate the definite integral of

Z b

some function,

f (x) dx. He computed the appropriate bound on the error, using the formula

a M(4) (b ? a)5

|S2n ? I| ?

, and comes to you with the following question:

180(2n)4

?When I used n = 5 slices, the error bound is 1/10. But I actually need my estimate to be within

1

± 10000

of the actual value. How many slices should I use??

You don?t need to give George the very best n that will work, but give him some reasonable value

you are sure of, and explain why it will work. Let?s get practical!

Solution. George knows that when he uses 5 slices, his error is no bigger than his bound of

M(4) (b ? a)5

1

.

=

180(2 · 5)4

10

He wants to know what n should be to get his bound to be less than:

M(4) (b ? a)5

1

?

.

180(2 · n)4

10000

Dividing the first equation by the second, the M(4) (b ? a5 )/180 cancels, and we get (think about

why does the sign change!):

(2n)4

10000

?

4

(2 · 5)

10 or n? 107/4

.

2

2 We are allowed to be practical here, so let?s take n a little bigger, say n = 102 = 50. Note that

another way to solve this problem entirely is to be practical from the start: George wants to

1

improve his error bound by a factor of 1000

. We can see from the formula for the error bound

that when Geoge increases the number of sliced n by a factor of 10, his error bound will improve

1

by a factor of 1014 = 10000

, which is ten times better than we need! So using n = 50 in place of

n = 5 will do fine!

8 5. Do the following integrals converge or diverge? Please give complete mathematical justification, either

by evaluating or by using an appropriate comparison. If you evaluate, you must use limits explicitly

in your writeup. If you use a comparison, you must explain carefully, and include the inequalities you

are using.

Z ? (a)

1 1 + cos2 x

dx

x + x2 Solution.

The numerator of the integrand can be bounded above and below: starting with

0 ? cos2 x ? 1

we can add 1 to get

1 ? 1 + cos2 x ? 2.

We can therefore see that the numerator and denominator are always positive, so our integrand

is positive. Also,

2

2

1 + cos2 x

?

? 2.

x + x2

x + x2

x

But

Z ?

Z ?

2

1

dx

=

2

dx

2

2

x

x

0

0

is just a multiple of a p-integral with p = 2 > 1, so it converges. Since our original integrand

is non-negative and is bounded above by a function with convergent integral, the comparison

theorem tells us that

Z ?

1 + cos2 x

dx

x + x2

1

converges as well.

Z ?

0.05 + 0.01 sin x

?

(b)

dx

x x

1

Solution.

?1 ? sin x ? 1

so

?0.01 ? 0.01 sin x ? 0.01

and

0.04 ? 0.05 + 0.01 sin x ? 0.06

Our integrand is therefore positive, and

0.05 + 0.01 sin x

0.06

?

? ?

x x

x x

The integral

Z

1 ? 1

? dx

x x converges since it is a p-integral with p = 3/2, so by the Comparison Theorem our original integral

converges. 9 Z ? (c)

1 (1 + ? ?

x)3 ( 1 + x)3

dx

x? Solution. If you want to figure out whether you should be proving this integral converges or

diverges, you can look at the highest powers of x in the numerator and denominator of the

integrand to see that as x becomes large the integrand behaves like

x3/2 x3/2

1

= ??3

x?

x

which has divergent integral from 1 to ? since ? ? 3 < 1. Please note that this is not a valid

solution to this problem, just a trick for you to figure out how to proceed.

We therefore want to try and prove that the integral diverges; in order to do so using the comparison theorem, we need to give a lower bound for the integrand. Since

(1 +

and ? We see that ? x)3 ?

3 1+x ? ? ? x x 3 3 ?

x3/2 x3/2

1

x)3 ( 1 + x)3

?

= ??3

?

x

x?

x

and the function on the right has a divergent integral by the p-test. By the comparison theorem,

the original integral converges.

(1 + ? 10

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