Step-by-step solution file
How come in question 2A the radius of the cylindrical shell isnt
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
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.)
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
yk = R3 ?? R = yk3
Putting everything together, we have
1 volume of revolved k-th slice ? ?(radius)2 (thickness) = ?(yk3 )2 ?y
Summing over all slices, we get our Riemann sum:
k=0 1 2 ?yk3 ?y Finally, taking the limit as n goes to infinity, we get the following expression for the volume of
?y 3 dy
0 Note that the bounds go from 0 to 5, just like the interval we sliced.
Let us evaluate this integral.
?y 3 dy = [ ?y 3 ]5x=0 = ?(5 3 ) ? 0 = 3?5 3
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.)
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.
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
?[(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.
with vertical slice: It may be easier if we break it into two integrals.
2 Z ?
2 2? (x + 2) cos x dx = 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:
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 ?
[( )(1) ? (? )(?1)] + [(0) ? (0)] = 0
2 = (Indeed, we could have notice this earlier because x cos x is an odd function.)
The Second integral is much easier:
2 ? 2 cos x dx = 2 sin x|?2 ? = 2[(1) ? (?1)] = 4
2 We plug them back and see
2? (x + 2) cos x dx = 2?[ ??
2 Z ?
2 2 cos x dx] = 2?[0 + 4] = 8? . x cos x dx + ??
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
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
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.
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].
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:
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!).
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:
?( 64 ? yk2 )2 ?(yk + 8)?y
k=1 or n
?( 64 ? (hk ? 8)2 )2 ?(hk )?h k=1 Which gives us a final integral of either:
(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
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
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
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:
8 Speed (m/s)
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.
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:
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) =
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
2 (c) Your classmate, George, used Simpson?s Rule with n = 5 slices to estimate the definite integral of
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:
?When I used n = 5 slices, the error bound is 1/10. But I actually need my estimate to be within
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
180(2 · 5)4
He wants to know what n should be to get his bound to be less than:
M(4) (b ? a)5
180(2 · n)4
Dividing the first equation by the second, the M(4) (b ? a5 )/180 cancels, and we get (think about
why does the sign change!):
(2 · 5)
10 or n? 107/4
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
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
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
Z ? (a)
1 1 + cos2 x
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,
1 + cos2 x
x + x2
x + x2
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
1 + cos2 x
x + x2
converges as well.
0.05 + 0.01 sin x
?1 ? sin x ? 1
?0.01 ? 0.01 sin x ? 0.01
0.04 ? 0.05 + 0.01 sin x ? 0.06
Our integrand is therefore positive, and
0.05 + 0.01 sin x
1 ? 1
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
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
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
and ? We see that ? x)3 ?
3 1+x ? ? ? x x 3 3 ?
x)3 ( 1 + x)3
and the function on the right has a divergent integral by the p-test. By the comparison theorem,
the original integral converges.
(1 + ? 10
This question was answered on: Feb 21, 2020
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