Subtract the Hole from the Whole.

Posted on December 6, 2016 by Lin McMullin

Sometimes I think textbooks are too rigorous. Behind every Riemann sum is a definite integral. So, authors routinely show how to solve an application of integration problem by developing the method starting from the Riemann sum and proceeding to an integral that give the result that is summarized in a “formula.” There is nothing wrong with that except that often the formula is all the students remember and are lost when faced with a similar situation that the formula does not handle. .

The volume of solid figure problems are developed from the idea that if a solid figure has a regular cross-section (that is, when cut perpendicular to a line, each face is similar – in the technical sense – to all the others). They are all squares, or equilateral triangles, or whatever. The last shape considered is usually a “washer”, that is, an annulus or two concentric circles. This is formed by revolving the region between two curves around a line. Authors develop a formula for such volumes: $\displaystyle \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$ .

Now there is nothing wrong with that, but I like to give the students their chance to show off. They can usually figure out the answer without Riemann sums. Here is my suggestion. After students have had some practice with circular cross-sections (“Disk” method”) I give them a series of three volumes to find.

Example 1: The curve $f\left( x \right)=\sin \left( \pi x \right)$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure.

Solution: $\displaystyle V=\int_{0}^{1}{\pi {{\left( \sin \left( \pi x \right) \right)}^{2}}dx}=\frac{\pi }{4}$

Example 2: The curve $g\left( x \right)=8{{x}^{3}}$ on the interval [0, ½] is revolved around the x-axis to form a solid figure. Find the volume of this figure.

Solution: $\displaystyle V=\int_{0}^{1/2}{\pi {{\left( 8{{x}^{3}} \right)}^{2}}dx=}\frac{\pi }{14}$

These they find easy. Then, leaving the first two examples in plain view, I give them:

Example 3: The region in the first quadrant between the graphs of $f\left( x \right)=\sin \left( \pi x \right)$ and $g\left( x \right)=8{{x}^{3}}$ is revolved around the x-axis. Find the volume of the resulting figure.

A little thinking and (rarely) a hint and they have it. $\displaystyle V=\frac{\pi }{4}-\frac{\pi }{14}$

What did they do? Easy, they subtracted the hole from the whole. We discuss this and why they think it is correct. We try one or two others. And now they are set to do any “washer” method problem without another formula to memorize.

Extensions:

1. In symbols, when rotation around a horizontal line, if R(x) is the distance from the curve farthest from the line of rotation and r(x) the distance from the closer curve to the line of rotation the result can be summarized in the formula

$\displaystyle V = \int_{a}^{b}{\pi {{\left( R\left( x \right) \right)}^{2}}dx}-\int_{a}^{b}{\pi {{\left( r\left( x \right) \right)}^{2}}dx}$ .

Notice, that I like to keep the $\pi$ inside the integral sign so that each integrand looks like the formula for the area of a circle. What the students need to know is to subtract the volume hole from the outside volume. With that idea and the disk method they can do any volume by washers problem.

2. You should show the students how this equation above can be rearranged into the formula in their books,

$\displaystyle V = \pi \int_{a}^{b}{{{\left( R\left( x \right) \right)}^{2}}-{{\left( r\left( x \right) \right)}^{2}}dx}$ .

This is so that they understand that the formulas are the same, and not think you’ve forgotten to tell them something important. It is also a good exercise in working with the notation. (see MPAC 5 – Notational fluency)

3. Next discuss what ${{\left( \pi R\left( x \right) \right)}^{2}}-\pi {{\left( r\left( x \right) \right)}^{2}}$ is the area of and how it relates to this problem. See if the students can understand what the textbook is doing; what shape the book is using.. Discuss the Riemann sum approach. (MPAC 1 Reasoning with definitions and theorems, and MPAC 5 Notational fluency)

4. With the idea of subtracting the “hole” try a problem like this. Example 4: The region in the first quadrant between x-axis and the graphs of $f\left( x \right)=\sqrt{x}$ and $g\left( x \right)=\sqrt{2x-4}$ is revolved around the x-axis. Find the volume of the resulting figure.

Solution: $\displaystyle V=\int_{0}^{4}{\pi {{\left( \sqrt{x} \right)}^{2}}dx}-\int_{2}^{4}{\pi {{\left( \sqrt{2x-4} \right)}^{2}}dx}=4\pi$

(Notice the limits of integration.)

Traditionally, this is done by the method of cylindrical shells, but you don’t need that. You could divide the region into two parts with a vertical line at x = 2 and use disks on the left and washers on the right, but you don’t need to do that either. Just subtract the hole from the whole.

Good Question 8 – or not?

Posted on January 5, 2016 by Lin McMullin

Today’s question is not a good question. It’s a bad question.

But sometimes a bad question can become a good one.

This one leads first to a discussion of units, then to all sorts of calculus.

Here’s the question a teacher sent me this week taken from his textbook:

The normal monthly rainfall at the Seattle-Tacoma airport can be approximated by the model $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$ , where R is measured in inches and t is the time in months, t = 1 being January. Use integration to approximate the normal annual rainfall. Hint: Integrate over the interval [0,12].

Of course, with the hint it’s not difficult to know what to do and that makes it less than a good question right there. The answer is $\displaystyle \int_{0}^{12}{R(t)dt=37.4736}$ inches. You could quit here and go on to the next question, but …

Then a student asked. “If R is in inches shouldn’t be in units of the integral be inch-months, since the unit of an integral is the unit of the integrand times the units of the independent variable?” Well, yes, they should. So, what’s up with that?

Also, the teacher figured that the integral of a rate is an amount and our answer is an amount, so why isn’t the integrand a rate?

The only answer I could come up with is that the statement “R is measured in inches” is incorrect; R should be measured in inches /month. The opening phrase “normal monthly rainfall” also seems to point to the correct units for R being inches/month.

Problem solved; or maybe does this lead to a different concern?

The teacher pointed out that R(6) = 0.7658 inches is a reasonable answer for the amount of rain in June whereas $\displaystyle \int_{0}^{6}{R(t)dt=}20.4786$ is not.

If R is a rate, then the amount of rain that falls in June (t = 6) is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$ .

From here on we will assume that R is a rate with units of inches/month. Here are the individual monthly rates calculated with a CAS.

The total amount of rainfall (second line above) appears be R(1) + R(2) + R(3) + … +R(12) = 37.4742. This is very close to the amount calculated by integration.

The slight difference of 0.0006 is not a round off error.

Remember, behind every definite integral there is a Riemann sum!

Again, the units are the problem. Why does the sum of the monthly rates seem to give the total amount? The reason is that the terms of the sequence above are actually the values of a right-side Riemann sum of the rate, R(t), over the interval [0,12] with 12 equal subdivisions of width 1 (month) each with the 1’s left out as 1’s often are. Therefore, their sum should come close to the total yearly rainfall, but it is really just an approximation of it.

The actual total for any month, n, is given by $\displaystyle \int_{n-1}^{n}{r(t)}dt$ . For example the amount of rain that falls in June is given by $\displaystyle \int_{5}^{6}{R(t)dt}=0.9890$ inches.

Here is the sequence of the actual monthly rainfall values in inches, and their sum.

This agrees with the integral. Why? Because one of the properties of integrals tell us that $\displaystyle \sum\limits_{n=1}^{12}{\int_{n-1}^{n}{r(t)dt}}=\int_{0}^{12}{r(t)dt}$ .

Another instructive thing with this integral is this: The function $R=3.121+2.399\sin \left( 0.524t+1.377 \right)$ is periodic with a period of $\frac{2\pi }{0.524}\approx 11.9908\approx 12$ . So the sine function takes on (almost) all its values in a year, as you would expect. Since the sine values all but cancel each other out

$\displaystyle \int_{0}^{12}{3.121+2.399\sin \left( 0.524t+1.377 \right)dt}\approx \int_{0}^{12}{3.121dt=3.121\left( 12-0 \right)=37.452}$ . Close!

The total rainfall divided by 12 is $\frac{37.452}{12}=3.121$ this must be close to the average rainfall each month. The average rainfall is $\displaystyle \frac{1}{12}\int_{0}^{12}{R\left( t \right)dt}=3.1228$ inches. Close, again!

So, there you have it. Is this a good question or not? We considered all these concepts while working not just with an equation but with numbers from a poorly stated problem:

Reading and interpreting words.
Unit analysis
Integration by technology
Realizing that a pretty good approximation is not correct, due again to units.
A Riemann sum approximation in a real situation that comes very close to the value by integration
Using a property of a periodic function to greatly simplify an integral
Finding average value two ways

So, it turned out to be a sunny day in Seattle.

January 2016

Posted on January 1, 2016 by Lin McMullin

HAPPY NEW YEAR

– a few more days off and then back to school!

If you haven’t already, it is time to start integration. The posts on starting integration were listed in December since I try to stay ahead of things so you’ll have time to prepare. With that in mind here are past posts on applications of integration – area, volume, average value, improper integrals, accumulation, and some enrichment for BC classes on parametric equations. This will take AB courses well into February. The February postings will be for the BC topics on sequences and series.

The four featured post below are the most popular from this list.

Applications of integration: area, volume, average value of a function, Accumulation and functions defined by integrals.

January 2, 2013 Integration by Parts – 1

January 4, 2013 Integration by Parts – 2

January 7, 2013 Area Between Curves

A series on visualizing solid figures:

November 13, 2014 Visualizing Solid Figures 1 Physical models of solid figures
November 16, 2014 Visualizing Solid Figures 2 Solids with regular cross sections
November 19, 2014 Visualizing Solid Figures 3 Volume by “Washers”
November 22, 2014 Visualizing Solid Figure 4 Volume by “Shells”
November 25, 2014 Visualizing Solid Figures 5 Volume by “Shells” and “Washers” compared

January 9, 2013 Volume of Solids with Regular Cross-sections

January 11, 2013 Volumes of Revolution

January 14, 2013 Why You Never Need Cylindrical Shells

January 25, 2014 Improper Integrals and proper areas.

January 16, 2013 Average Value of a Function

February 6, 2013: Logarithms

January 19, 2013 Most Triangles are Obtuse! What is the probability that a triangle picked at random will be acute? An average value problem solved by a tenth grader.

A series on ROULETTES some special parametric curves (BC topic – enrichment):

Accumulation – On the exams; not in many textbooks

January 21, 2103 Accumulation: Need an Amount? Accumulation 1: If you need an amount, look around for a rate to integrate.
January 23, 2013 AP Accumulation Questions Accumulation 2: AP Exam Rate/Accumulation Questions
January 26, 2013 Graphing with Accumulation 1 Accumulation 3: Graphing Ideas in Accumulation – Increasing and decreasing
January 28, 2013 Graphing with Accumulation: Accumulation 4: Graphing Ideas in Accumulation – Concavity
January 30, 2013 Stamp Out Slope-intercept Form! Accumulation 5: Lines
February 2, 2014 Accumulation and Differential Equations Accumulation 6: Differential equations
February 4, 2014 Painting a Point Accumulation 7: An application (of paint)

Good Question 7 – 2009 AB 3

Posted on October 5, 2015 by Lin McMullin

Another in my occasional series on Good Questions to teach from. This is the Mighty Cable Company question from the 2009 AB Calculus exam, number 3

This question presented students with a different situation than had been seen before. It is a pretty standard “in-out” question, except that what was going in and out was money. Students were told that the Mighty Cable Company sold its cable for $120 per meter. They were also told that the cost of the cable varied with its distance from the starting end of the cable. Specifically, the cost of producing a portion of the cable x meters from the end is $6\sqrt{x}$ dollars per meter. Profit was defined as the difference between the money the company received for selling the cable minus the cost of producing the cable.

Students had a great deal of trouble answering this question. (The mean was 1.92 out of a possible 9 points. Fully, 36.9% of students earned no point; only 0.02% earned all 9 points.) This was probably because they had difficulty in interpreting the question and translating it into the proper mathematical terms and symbols. Since economic problems are not often seen on AP Calculus exams, students needed to be able to use the clues in the stem:

The $120 per meter is a rate. This should be deduced from the units: dollars per meter.
The cost of producing the portion of cable x meters from one end cable is also a rate for the same reason. In economics this is called the marginal cost; the students did not need to know this term.
The profit is an amount that is a function of x, the length of the cable.

Part (a): Students were required to find the profit from the sale of a 25-meter cable. This is an amount. As always, when asked for an amount, integrate a rate. In this case integrate the difference between the rate at which the cable sells and the cost of producing it.

$\displaystyle P(25)=\int_{0}^{25}{\left( 120-6\sqrt{x} \right)dx}=\$2500$

$\displaystyle P(25)=120(25)-\int_{0}^{25}{6\sqrt{x}\ dx}=\$2500$

Part (b): Students were asked to explain the meaning of $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ in the context of the problem. Since the answer is probably not immediately obvious, here is the reasoning involved.

This is the integral of a rate and therefore, gives the amount (of money) needed to manufacture the cable. This can be found by a unit analysis of the integrand: $\displaystyle \frac{\text{dollars}}{\text{meter}}\cdot \text{meters}=\text{dollars}$ .

Let C be the cost of production, so $\displaystyle \frac{dC}{dx}=6\sqrt{x}$ , and therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}=C\left( 30 \right)-C\left( 25 \right)$ by the Fundamental Theorem of Calculus (FTC).

Therefore, $\displaystyle \int_{25}^{30}{6\sqrt{x}dx}$ is the difference in dollars between the cost of producing a cable 30 meters long, C(30), and the cost of producing a cable 25 meters long, C(25). (Another acceptable is that the integral is the cost in dollars of producing the last 5 meters of a 30 meter cable.)

Part (c): Students were asked to write an expression involving an integral that represents the profit on the sale of a cable k meters long.

Part (a) serves as a hint for this part of the question. Here the students should write the same expression as they wrote in (a) with the 25 replaced by k.

$\displaystyle P(k)=\int_{0}^{k}{\left( 120-6\sqrt{x} \right)dx}$

$\displaystyle P(k)=120k-\int_{0}^{k}{6\sqrt{x}\ dx}$

Part d: Students were required to find the maximum profit that can be earned by the sale of one cable and to justify their answer. Here they need to find when the rate of change of the profit (the marginal profit) changes from positive to negative.

Using the FTC to differentiate either of the answers in part (c) or by starting fresh from the given information:

$\displaystyle \frac{dP}{dx}=120-6\sqrt{x}$

$\displaystyle \frac{dP}{dx}=0$ when x = 400 and P(400)= $16,000.

Justification: The maximum profit on the sale of one cable is $16,000 for a cable 400 meters long. For $0<x<400,{P} '(x)>0$ and for $x>400,{P} '(x)<0$ therefore, the maximum profit occurs at x = 400. (The First Derivative Test).

Once students are familiar with in-out questions, this is a good question to challenge them with. The actual calculus is not that difficult or unusual but concentrating on the translation of the unfamiliar context into symbols and calculus ideas is different. Show them how to read the hints in the problem such as the units.

Steel cable or steel wire rope as it is called also has some interesting geometry in its construction. You can find many good illustrations of this, such as the ones below, by Googling “steel wire rope.”

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Good Question 6: 2000 AB 4

Posted on August 25, 2015 by Lin McMullin

Another of my favorite questions from past AP exams is from 2000 question AB 4. If memory serves it is the first of what became known as an “In-out” question. An “In-out” question has two rates that are working in opposite ways, one filling a tank and the other draining it.

In subsequent years we saw a question with people entering and leaving an amusement park (2002 AB2/BC2), sand moving on and off a beach (2005 AB 2), another tank (2007 AB2), an oil leak being cleaned up (2008 AB 3), snow falling and being plowed (2010 AB 1), gravel being processed (2013 AB1/BC1), and most recently water again flowing in and out of a pipe (2015 AB1/BC1). The in-between years saw rates in one direction only but featured many of the same concepts.

The questions give rates and ask about how the quantity is changing. As such, they may be approached as differential equation initial value problems, but there is an easier way. This easier way is that a differential equation that gives the derivative as a function of a single variable, t, with an initial point $\left( {{t}_{0}},y\left( {{t}_{0}} \right) \right)$ always has a solution of the form

$y\left( t \right)=y\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{{y}'\left( x \right)dx}$ .

This is sometimes called the “accumulation equation.” The integral of a rate of change ${y}'\left( t \right)$ gives the net amount of change over the interval of integration $[{{t}_{0}},t]$ . When this is added to the initial amount the result is an expression that gives the amount at any time t.

In a motion context, this same idea is that the position at any time t, is the initial position plus the displacement:

$\displaystyle s\left( t \right)=s\left( {{t}_{0}} \right)+\int_{{{t}_{0}}}^{t}{v\left( x \right)dx}$ where $v\left( t \right)={s}'\left( t \right)$

The scoring standard gave both forms of the solution. The ease of the accumulation form over the differential equation solution was evident and subsequent standards only showed this one.

2000 AB 4

The question concerned a tank that initially contained 30 gallons of water. We are told that water is being pumped into the tank at a constant rate of 8 gallons per minute and the water is leaking out at the rate of $\sqrt{t+1}$ gallons per minute.

Part a asked students to compute the amount of water that leaked out in the first three minutes. There were two solutions given. The second solves the problem as an initial value differential equation:

Let L(t) be the amount that leaks out in t minutes then

$\displaystyle \frac{dL}{dt}=\sqrt{t+1}$

$L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C$

$L\left( 0 \right)=\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C=0$ since nothing has leaked out yet, so C = -2/3

$L\left( t \right)=\frac{2}{3}{{\left( t+1 \right)}^{3/2}}-\frac{2}{3}$

$L\left( 3 \right)=\frac{14}{3}$

The first method, using the accumulation idea takes a single line:

$\displaystyle L\left( 3 \right)=\int_{0}^{3}{\sqrt{t+1}dt}=\left. \frac{2}{3}{{\left( t+1 \right)}^{3/2}} \right|_{0}^{3}=\frac{2}{3}{{\left( 4 \right)}^{3/2}}-\frac{2}{3}{{\left( 1 \right)}^{3/2}}=\frac{14}{3}$

I think you’ll agree this is easier and more direct.

Part b asked how much water was in the tank at t = 3 minutes. We have 30 gallons to start plus 8(3) gallons pumped in and 14/3 gallons leaked out gives 30 + 24 – 14/3 = 148/3 gallons.

This part, worth only 1 point, was a sort of hint for the next part of the question.

Part c asked students to write an expression for the total number of gallons in the tank at time t.

Following part b the accumulation approach gives either

$\displaystyle A\left( t \right)=30+8t-\int_{0}^{t}{\sqrt{x+1}dx}$ or

$\displaystyle A\left( t \right)=30+\int_{0}^{t}{\left( 8-\sqrt{x+1} \right)dx}$ .

The first form is not a simplification of the second, but rather the second form is treating the difference of the two rates, in minus out, as the rate to be integrated.

The differential equation approach is much longer and looks like this:

$\displaystyle \frac{dA}{dt}=8-\sqrt{t+1}$

$A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+C$

$A\left( 0 \right)=30=8(0)-\frac{2}{3}{{\left( 0+1 \right)}^{3/2}}+C$ , so $C=\frac{92}{3}$

$A\left( t \right)=8t-\frac{2}{3}{{\left( t+1 \right)}^{3/2}}+\frac{92}{3}$

Again, this is much longer. In recent years when asking student to write an expression such as this, the directions included a phrase such as “write an equation involving one or more integrals that gives ….” This pretty much leads students away from the longer differential equation initial value problem approach.

Part d required students to find the time when in the interval $0\le t\le 120$ minutes the amount of water in the tank was a maximum and to justify their answer. The usual method is to find the derivative of the amount, A(t), set it equal to zero, and then solve for the time.

${A}'\left( t \right)=8-\sqrt{t+1}$

Notice that this is the same regardless of which of the three forms of the expression for A(t) you start with. Thus, an excellent example of the Fundamental Theorem of Calculus used to find the derivative of a function defined by an integral. Or you could just start here without reference to the forms above: the overall rate in the rate in minus the rate out.

${A}'\left( t \right)=0$ when t = 63

This is a maximum by the First Derivative Test since for 0 < t < 63 the derivative of A is positive and for 63 < t <120 the derivative of A is negative.

There is an additional idea on this part of the question in the Teaching Suggestions below.

I like this question because it is a nice real (as real as you can hope for on an exam) situation and for the way the students are led through the problem. I also like the way it can be used to compare the two methods of solution. Then the way they both lead to the same derivative in part d is nice as well. I use this one a lot when working with teachers in workshops and summer institutes for these very reasons.

Teaching Suggestions

Certainly, have your students work through the problem using both methods. They need to learn how to solve an initial value problem (IVP) and this is good practice. Additionally, it may help them see how and when to use one method or the other.
Be sure the students understand why the three forms of A(t) in part c give the same derivative in part d. This makes an important connection with the Fundamental theorem of Calculus.
Like many good AP questions part d can be answered without reference to the other parts. The question starts with more water being pumped in than leaking out. This will continue until the rate at which the water leaks out overtakes the rate at which it is being pumped in. At that instant the rate “in” equals the rate “out” so you could start with $8=\sqrt{t+1}$ . After finding that t = 63, the answer may be justified by stating that before this time more water is being pumped in than is leaking out and after this time the rate at which water leaks out is greater than the rate at which it is pumped in, so the maximum must occur at t = 63.
And as always, consider the graph of the rates.

I used this question as the basis of a lesson in the current AP Calculus Curriculum Module entitled Integration, Problem Solving and Multiple Representations © 2013 by the College Board. The lesson gives a Socratic type approach to this question with a number of questions for each part intended to help the teacher not only work through this problem but to bring out related ideas and concepts that are not in the basic question. The module is currently available at AP sponsored workshops and AP Summer Institutes. Eventually, it will be posted at AP Central on the AB and BC Calculus Home Pages.

Half-full and Half-empty

Posted on January 16, 2015 by Lin McMullin

A thought experiment:

Suppose you had a container with a rectangular base whose length runs from x = a to x = b, with a width of one inch. The container has four vertical rectangular sides. You put a piece of, say, plastic into the container which fits snugly along the bottom and four sides. The top of the piece is irregular and has the equation y = f(x). If the plastic were to melt, how high up the sides would the melted plastic rise?

One way to think about this is to consider the final level, L. When melted, the plastic above the final level must fill in the part below, leaving a rectangle with the same area as that under the original function’s levels. (The one-inch width will remain the same and not affect the outcome.)

So the original area is $\displaystyle \int_{a}^{b}{f\left( x \right)dx}$ and the final area is $L\left( b-a \right)$ . Since these are the same, we can write an equation and solve it for L.

$\displaystyle L\left( b-a \right)=\int_{a}^{b}{f\left( x \right)dx}$

$\displaystyle L=\frac{\int_{a}^{b}{f\left( x \right)dx}}{b-a}$

But that’s the equation for the average value of a function!

What a surprise!

Well, not a surprise for you, the teacher. This might be a good way to sneak up on the average value of a function idea for your students while giving them a good visual idea of the concept.

Visualizing Solid Figures 5

Posted on November 25, 2014 by Lin McMullin

To end this series of posts on visualizing solid figures, we will look at a problem that relates to how volumes of solid figures are formed. It has 5 parts which will be presented first. Then the solution will be given.

Consider the region in the first quadrant bounded by the graphs of the parabola $y={{x}^{2}}$ and the line $y=9$ both for $0\le x\le 3$ .

This region is revolved around the y-axis to form a solid figure.

Use the disk/washer method to find the volume of this figure.
Use the method of cylindrical shells to find the volume of this figure.
Use the disk/washer method to find a number j , such that when x = j the volume of the figure is one-half that of the original figure.
Use the method of cylindrical shells to find a number k , such that when x = k the volume of the figure is one-half that of the original figure.
The answers to parts a) and b) should be the same, but the answers to parts c) and d) are different. Explain why they are different.

Solutions:

1. The volume by the disk/washer method is

$\displaystyle V=\int_{0}^{9}{\pi {{x}^{2}}dy}=\int_{0}^{9}{\pi ydy}=\frac{81}{2}\pi \approx 127.235$

2. The volume by the method of cylindrical shells is

$\displaystyle \int_{0}^{3}{2\pi x\left( 9-y \right)dx}=\int_{0}^{3}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{2}\pi \approx 127.235$

3. The value is be found by solving the equation for j:

$\displaystyle \int_{0}^{{{j}^{2}}}{\pi y\,dy}=\frac{81}{4}\pi$ , so $j\approx 2.52269$

4. The value is be found by solving the equation for k:

$\displaystyle \int_{0}^{k}{2\pi x\left( 9-{{x}^{2}} \right)dx}=\frac{81}{4}\pi$ and $k\approx 1.62359$

5. The reason the values are not the same is this. Think of the revolved parabola as a bowl. If you pour water into the bowl until it is half full, the bowl looks like the figure below. The water is pooled in the bottom of the bowl as you would expect. This is what happens when using the disk/washer method. The washers stack up starting in the bottom of the bowl until it is half full.

On the other hand, the method of cylindrical shells sort of wraps the water in layers (the shells) around the y-axis. Picture the water being sprayed on the y-axes and frozen there. Each new layer (shell) increases the amount and you end up half of the total volume arranged as a cylinder with a rounded (paraboloid shaped) bottom as shown in the figure below. Both bowls contain the same amount of water, arranged differently.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Category Archives: Integration Applications

Subtract the Hole from the Whole.

Good Question 8 – or not?

January 2016

HAPPY NEW YEAR

Good Question 7 – 2009 AB 3

Good Question 6: 2000 AB 4

Half-full and Half-empty

Visualizing Solid Figures 5

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