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1 GALILEO, Universit Sstem of Georgia GALILEO Open Learning Materials Mathematics Open Tetbooks Mathematics Spring 05 Armstrong Calculus Michael Tiemeer Armstrong State Universit, Jared Schlieper Armstrong State Universit, Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Tiemeer, Michael and Schlieper, Jared, "Armstrong Calculus" (05) Mathematics Open Tetbooks Book This Open Tetbook is brought to ou for free and open access b the Mathematics at GALILEO Open Learning Materials It has been accepted for inclusion in Mathematics Open Tetbooks b an authorized administrator of GALILEO Open Learning Materials For more information, please contact affordablelearninggeorgia@usgedu

2 Open Tetbook Armstrong State Universit UNIVERSITY SYSTEM OF GEORGIA Jared Schlieper, Michael Tiemeer Armstrong Calculus

3 J A R E D S C H L I E P E R & M I C H A E L T I E M E Y E R C A L C U L U S

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5 Copright 05 Jared Schlieper and Michael Tiemeer Licensed to the public under Creative Commons Attribution-Noncommercial-ShareAlike 40 International License

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7 Preface A free and open-source calculus First and foremost, this tet is mostl an adaptation of two ver ecellent open-source tetbooks: Active Calculus b Dr Matt Boelkins and AP E X Calculus b Drs Gregor Hartman, Brian Heinold, Tro Siemers, Dimplekumar Chalishajar, and Jennifer Bowen Both tets can be found at Dr Boelkins also has a great blog for open source calculus at The authors of this tet have combined sections, eamples, and eercises from the above two tets along with some of their own content to generate this tet The impetus for the creation of this tet was to adopt an open-source tetbook for Calculus while maintaining the tpical schedule and content of the calculus sequence at our home institution Several fundamental ideas in calculus are more than 000 ears old As a formal subdiscipline of mathematics, calculus was first introduced and developed in the late 600s, with ke independent contributions from Sir Isaac Newton and Gottfried Wilhelm Leibniz Mathematicians agree that the subject has been understood rigorousl since the work of Augustin Louis Cauch and Karl Weierstrass in the mid 800s when the field of modern analsis was developed, in part to make sense of the infinitel small quantities on which calculus rests Hence, as a bod of knowledge calculus has been completel understood b eperts for at least 50 ears The discipline is one of our great human intellectual achievements: among man spectacular ideas, calculus models how objects fall under the forces of gravit and wind resistance, eplains how to compute areas and volumes of interesting shapes, enables us to work rigorousl with infinitel small and infinitel large quantities, and connects the varing rates at which quantities change to the total change in the quantities themselves

8 6 While each author of a calculus tetbook certainl offers her own creative perspective on the subject, it is hardl the case that man of the ideas she presents are new Indeed, the mathematics communit broadl agrees on what the main ideas of calculus are, as well as their justification and their importance; the core parts of nearl all calculus tetbooks are ver similar As such, it is our opinion that in the st centur an age where the internet permits seamless and immediate transmission of information no one should be required to purchase a calculus tet to read, to use for a class, or to find a coherent collection of problems to solve Calculus belongs to humankind, not an individual author or publishing compan Thus, the main purpose of this work is to present a new calculus tet that is free In addition, instructors who are looking for a calculus tet should have the opportunit to download the source files and make modifications that the see fit; thus this tet is open-source Because the tet is free and open-source, an professor or student ma use and/or change the electronic version of the tet for no charge Presentl, a pdf cop of the tet and its source files ma be obtained b download from Github (insert link here!!) This work is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 40 Unported License The graphic that appears throughout the tet shows that the work is licensed with the Creative Commons, that the work ma be used for free b an part so long as attribution is given to the author(s), that the work and its derivatives are used in the spirit of share and share alike, and that no part ma sell this work or an of its derivatives for profit, with the following eception: it is entirel acceptable for universit bookstores to sell bound photocopied copies to students at their standard markup above the coping epense Full details ma be found b visiting or sending a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 9404, USA

9 7 Acknowledgments We would like to thank Affordable Learning Georgia for awarding us a Tetbook Transformation Grant, which allotted a twocourse release for each of us to generate this tet Please see for more information on this initiative to promote student success b providing affordable tetbook alternatives We will gladl take reader and user feedback to correct them, along with other suggestions to improve the tet Jared Schlieper & Michael Tiemeer

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11 Contents 0 Introduction to Calculus 0 Wh do we stud calculus? 0 How do we measure velocit? 5 Limits 3 The notion of limit 3 Limits involving infinit 4 3 Continuit 49 4 Epsilon-Delta Definition of a Limit 59 Derivatives 67 Derivatives and Rates of Change 67 The derivative function 83 3 The second derivative 95 4 Basic derivative rules 07 5 The product and quotient rules 9 6 The chain rule 35 7 Derivatives of functions given implicitl 49 8 Derivatives of inverse functions 6 3 Applications of Derivatives 77 3 Related rates 77 3 Using derivatives to identif etreme values of a function 9 33 Global Optimization Applied Optimization 5 35 The Mean Value Theorem 5 36 The Tangent Line Approimation 3 37 Using Derivatives to Evaluate Limits Newton s Method 57 4 Integration 65 4 Determining distance traveled from velocit 65 4 Riemann Sums The Definite Integral 9 44 The Fundamental Theorem of Calculus, Part I The Fundamental Theorem of Calculus, part II 33

12 0 Contents 46 Integration b Substitution More with Integrals Topics of Integration Integration b Parts Trigonometric Integrals Trigonometric Substitution Partial Fractions Improper Integrals Numerical Integration Other Options for Finding Antiderivatives 47 6 Applications of Integration Using Definite Integrals to Find Volume Volume b The Shell Method Arc Length and Surface Area Densit, Mass, and Center of Mass Phsics Applications: Work, Force, and Pressure An Introduction to Differential Equations Separable differential equations Hperbolic Functions 5 7 Sequences & Series 59 7 Sequences 59 A A Short Table of Integrals 539

13 Chapter 6 Applications of Integration 6 Using Definite Integrals to Find Volume Motivating Questions In this section, we strive to understand the ideas generated b the following important questions: How can we use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular ais? In what circumstances do we integrate with respect to instead of integrating with respect to? What adjustments do we need to make if we revolve about a line other than the - or -ais? Introduction Just as we can use definite integrals to add up the areas of rectangular slices to find the eact area that lies between two curves, we can also emplo integrals to determine the volume of certain regions that have cross-sections of a particular consistent shape As a ver elementar eample, consider a clinder of radius and height 3, as pictured in Figure 69 While we know that we can compute the area of an circular clinder b the formula V = πr h, if we think about slicing the clinder into thin pieces, we see that each is a clinder of radius r = and height (thickness) Hence, the volume of a representative slice is 3 V slice = π Letting 0 and using a definite integral to add the volumes of the slices, we find that Figure 6: A right circular clinder Moreover, since 3 0 V = 3 0 π d 4π d = π, we have found that the volume of the clinder is 4π The principal problem of interest in

14 434 Chapter 6 Applications of Integration our upcoming work will be to find the volume of certain solids whose cross-sections are all thin clinders (or washers) and to do so b using a definite integral To that end, we first consider another familiar shape in Preview Activit 6: a circular cone Preview Activit 6 3 = f() Figure 6: The circular cone described in Preview Activit 6 5 Consider a circular cone of radius 3 and height 5, which we view horizontall as pictured in Figure 6 Our goal in this activit is to use a definite integral to determine the volume of the cone (a) Find a formula for the linear function = f () that is pictured in Figure 6 (b) For the representative slice of thickness that is located horizontall at a location (somewhere between = 0 and = 5), what is the radius of the representative slice? Note that the radius depends on the value of (c) What is the volume of the representative slice ou found in (b)? (d) What definite integral will sum the volumes of the thin slices across the full horizontal span of the cone? What is the eact value of this definite integral? (e) Compare the result of our work in (d) to the volume of the cone that comes from using the formula Vcone = 3 πr h The Volume of a Solid of Revolution A solid of revolution is a three dimensional solid that can be generated b revolving one or more curves around a fied ais For eample, we can think of a circular clinder as a solid of revolution: in Figure 69, this could be accomplished b revolving the line segment from (0, ) to (3, ) about the -ais Likewise, the circular cone in Figure 6 is the solid of revolution generated b revolving the portion of the line = from = 0 to = 5 about the -ais It is particularl important to notice in an solid of revolution that if we slice the solid perpendicular to the ais of revolution, the resulting cross-section is circular We consider two eamples to highlight some of the natural issues that arise in determining the volume of a solid of revolution Eample Find the volume of the solid of revolution generated when the region R bounded b = 4 and the -ais is revolved about the -ais Solution First, we observe that = 4 intersects the -ais at the points (, 0) and (, 0) When we take the region R that lies between the curve and the -ais on this interval and revolve it about the -ais, we get the three-dimensional solid pictured in Figure 63

15 6 Using Definite Integrals to Find Volume 435 Taking a representative slice of the solid located at a value that lies between = and =, we see that the thickness of such a slice is (which is also the height of the clinder-shaped slice), and that the radius of the slice is determined b the curve = 4 Hence, we find that V slice = π(4 ), since the volume of a clinder of radius r and height h is V = πr h Using a definite integral to sum the volumes of the representative slices, it follows that V = π(4 ) d It is straightforward to evaluate the integral and find that the volume is V = 5 5 π 4 = 4 Figure 63: The solid of revolution in Eample For a solid such as the one in Eample, where each crosssection is a clindrical disk, we first find the volume of a tpical cross-section (noting particularl how this volume depends on ), and then we integrate over the range of -values through which we slice the solid in order to find the eact total volume Often, we will be content with simpl finding the integral that represents the sought volume; if we desire a numeric value for the integral, we tpicall use a calculator or computer algebra sstem to find that value The general principle we are using to find the volume of a solid of revolution generated b a single curve is often called the disk method Disk Method If = r() is a nonnegative continuous function on [a, b], then the volume of the solid of revolution generated b revolving the curve about the -ais over this interval is given b V = b a π [r()] d Eample Find the volume of the solid formed b revolving the curve = /, from = to =, about the -ais Solution Since the ais of rotation is vertical, we need to convert the function into a function of and convert the -bounds to -bounds Since = / defines the curve, we rewrite it as = / The bound = corresponds to the -bound =, and the bound = corresponds to

16 436 Chapter 6 Applications of Integration = / the -bound = / Thus we are rotating the curve = /, from = / to = about the -ais to form a solid The curve and sample differential element are sketched in Figure 64-(a), with a full sketch of the solid in Figure 64-(b) We integrate to find the volume: R() = / V = π / d = π / = π units 3 (a) A different tpe of solid can emerge when two curves are involved, as we see in the following eample (b) Figure 64: Sketching the solid in Eample 4 r() R() Figure 65: At left, the solid of revolution in Eample 3 At right, a tpical slice with inner radius r() and outer radius R() Eample 3 Find the volume of the solid of revolution generated when the finite region R that lies between = 4 and = + is revolved about the -ais Solution First, we must determine where the curves = 4 and = + intersect Substituting the epression for from the second equation into the first equation, we find that + = 4 Rearranging, it follows that + = 0, and the solutions to this equation are = and = The curves therefore cross at (, 0) and (, ) When we take the region R that lies between the curves and revolve it about the -ais, we get the three-dimensional solid pictured at left in Figure 65 Immediatel we see a major difference between the solid in this eample and the one in Eample : here, the three-dimensional solid of revolution isn t solid in the sense that it has open space in its center If we slice the solid perpendicular to the ais of revolution, we observe that in this setting the resulting representative slice is not a solid disk, but rather a washer, as pictured at right in Figure 65 Moreover, at a given location between = and =, the small radius r() of the inner circle is determined b the curve = +, so r() = + Similarl, the big radius R() comes from the function = 4, and thus R() = 4 Thus, to find the volume of a representative slice, we compute the volume of the outer disk and subtract the volume of the inner disk Since πr() πr() = π[r() r() ], it follows that the volume of a tpical slice is V slice = π[(4 ) ( + ) ] Hence, using a definite integral to sum the volumes of the respective

17 6 Using Definite Integrals to Find Volume 437 slices across the integral, we find that V = π[(4 ) ( + ) ] d Evaluating the integral, the volume of the solid of revolution is V = 08 5 π The general principle we are using to find the volume of a solid of revolution generated b a single curve is often called the washer method Washer Method If = R() and = r() are nonnegative continuous functions on [a, b] that satisf R() r() for all in [a, b], then the volume of the solid of revolution generated b revolving the region between them about the -ais over this interval is given b V = b a π[r() r() ] d Activit 6 In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated ais In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid (a) The region S bounded b the -ais, the curve =, and the line = 4; revolve S about the -ais (b) The region S bounded b the -ais, the curve =, and the line = ; revolve S about the -ais (c) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (d) The finite region S bounded b the curves = + and = + 4; revolve S about the -ais (e) The region S bounded b the -ais, the curve =, and the line = ; revolve S about the -ais How does the problem change considerabl when we revolve about the -ais? Revolving about the -ais As seen in Activit 6, problem (e), the problem changes considerabl when we revolve a given region about the -ais Foremost, this is due to the fact that representative slices now have

18 438 Chapter 6 Applications of Integration thickness, which means that it becomes necessar to integrate with respect to Let s consider a particular eample to demonstrate some of the ke issues Eample 4 Find the volume of the solid of revolution generated when the finite region R that lies between = and = 4 is revolved about the -ais = r() R() Figure 66: At left, the solid of revolution in Eample 4 At right, a tpical slice with inner radius r() and outer radius R() Solution We observe that these two curves intersect when =, hence at the point (, ) When we take the region R that lies between the curves and revolve it about the -ais, we get the three-dimensional solid pictured at left in Figure 66 Now, it is particularl important to note that the thickness of a representative slice is, and that the slices are onl clindrical washers in nature when taken perpendicular to the -ais Hence, we envision slicing the solid horizontall, starting at = 0 and proceeding up to = Because the inner radius is governed b the curve =, but from the perspective that is a function of, we solve for and get = = r() In the same wa, we need to view the curve = 4 (which governs the outer radius) in the form where is a function of, and hence = 4 Therefore, we see that the volume of a tpical slice is V slice = π[r() r() ] = π[ 4 ( ) ] Using a definite integral to sum the volume of all the representative slices from = 0 to =, the total volume is = V = =0 [ π 4 ( ) ] d It is straightforward to evaluate the integral and find that V = 7 5 π Activit 6 In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated ais In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid (a) The region S bounded b the -ais, the curve =, and the line = ; revolve S about the -ais (b) The region S bounded b the -ais, the curve =, and the line = 4; revolve S about the -ais (c) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (d) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (e) The finite region S bounded b the curves = ( ) and =

19 6 Using Definite Integrals to Find Volume 439 ; revolve S about the -ais Revolving about horizontal and vertical lines other than the coordinate aes Just as we can revolve about one of the coordinate aes ( = 0 or = 0), it is also possible to revolve around an horizontal or vertical line Doing so essentiall adjusts the radii of clinders or washers involved b a constant value A careful, well-labeled plot of the solid of revolution will usuall reveal how the different ais of revolution affects the definite integral we set up Again, an eample is instructive Eample 5 Find the volume of the solid of revolution generated when the finite region S that lies between = and = is revolved about the line = Solution Graphing the region between the two curves in the first quadrant between their points of intersection ((0, 0) and (, )) and then revolving the region about the line =, we see the solid shown in Figure 67 Each slice of the solid perpendicular to the ais of revolution is a washer, and the radii of each washer are governed b the curves = and = But we also see that there is one added change: the ais of revolution adds a fied length to each radius In particular, the inner radius of a tpical slice, r(), is given b r() = +, while the outer radius is R() = + Therefore, the volume of a tpical slice is [ V slice = π[r() r() ] = π ( + ) ( + ) ] - r() R() Finall, we integrate to find the total volume, and V = 0 [ π ( + ) ( + ) ] d = 7 5 π Figure 67: The solid of revolution described in Eample 5 Activit 6 3 In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated ais In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid For each prompt, use the finite region S in the first quadrant bounded b the curves = and = 3 (a) Revolve S about the line = (b) Revolve S about the line = 4 (c) Revolve S about the line = (d) Revolve S about the line = 5

20 440 Chapter 6 Applications of Integration Volumes of Other Solids Given an arbitrar solid, we can approimate its volume b cutting it into n thin slices When the slices are thin, each slice can be approimated well b a general right clinder Thus the volume of each slice is approimatel its cross-sectional area thickness (These slices are the differential elements) B orienting a solid along the -ais, we can let A( i ) represent the cross-sectional area of the i th slice, and let i represent the thickness of this slice (the thickness is a small change in ) The total volume of the solid is approimatel: Volume = n i= [ ] Area thickness n A( i ) i i= Recognize that this is a Riemann Sum B taking a limit (as the thickness of the slices goes to 0) we can find the volume eactl Volume B Cross-Sectional Area The volume V of a solid, oriented along the -ais with cross-sectional area A() from = a to = b, is V = b a A() d Figure 68: Orienting a pramid along the -ais in Eample 6 Eample 6 Find the volume of a pramid with a square base of side length 0 in and a height of 5 in Solution There are man was to orient the pramid along the - ais; Figure 68 gives one such wa, with the pointed top of the pramid at the origin and the -ais going through the center of the base Each cross section of the pramid is a square; this is a sample differential element To determine its area A(), we need to determine the side lengths of the square When = 5, the square has side length 0; when = 0, the square has side length 0 Since the edges of the pramid are lines, it is eas to figure that each cross-sectional square has side length, giving A() =

21 6 Using Definite Integrals to Find Volume 44 () = 4 We have 5 V = 4 d 0 = = in in 3 We can check our work b consulting the general equation for the volume of a pramid (see the back cover under Volume of A General Cone ): area of base height 3 Certainl, using this formula from geometr is faster than our new method, but the calculus-based method can be applied to much more than just cones Summar In this section, we encountered the following important ideas: We can use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular ais b taking slices perpendicular to the ais of revolution which will then be circular disks or washers If we revolve about a vertical line and slice perpendicular to that line, then our slices are horizontal and of thickness This leads us to integrate with respect to, as opposed to with respect to when we slice a solid verticall If we revolve about a line other than the - or -ais, we need to carefull account for the shift that occurs in the radius of a tpical slice Normall, this shift involves taking a sum or difference of the function along with the constant connected to the equation for the horizontal or vertical line; a well-labeled diagram is usuall the best wa to decide the new epression for the radius

22 44 Chapter 6 Applications of Integration Eercises Terms and Concepts ) T/F: A solid of revolution is formed b revolving a shape around an ais ) In our own words, eplain how the Disk and Washer Methods are related 7) 05 = cos 3) Eplain the how the units of volume are found in the integral: if A() has units of in, how does A() d have units of in 3? Problems In Eercises 4 7, a region of the Cartesian plane is shaded Use the Disk/Washer Method to find the volume of the solid of revolution formed b revolving the region about the -ais 05 5 In Eercises 8-, a region of the Cartesian plane is shaded Use the Disk/Washer Method to find the volume of the solid of revolution formed b revolving the region about the -ais = 8) 05 = = 4) 05 = = 3 3 = 3 9) 5) 0 = 5 0 = 5 0) 5 6)

23 6 Using Definite Integrals to Find Volume 443 8) A right circular cone with height of 0 and base radius of 5 ) 05 = cos (Hint: Integration B Parts will be necessar, twice First let u = arccos, then let u = arccos ) In Eercises 7, a region of the Cartesian plane is described Use the Disk/Washer Method to find the volume of the solid of revolution formed b rotating the region about each of the given aes ) Region bounded b: =, = 0 and = Rotate about: (a) the -ais (b) = (c) the -ais (d) = 3) Region bounded b: = 4 and = 0 Rotate about: (a) the -ais (c) = (b) = 4 (d) = 4) The triangle with vertices (, ), (, ) and (, ) Rotate about: (a) the -ais (b) = (c) the -ais (d) = 5) Region bounded b = + and = Rotate about: 9) A skew right circular cone with height of 0 and base radius of 5 (Hint: all cross-sections are circles) 5 0 0) A right triangular cone with height of 0 and whose base is a right, isosceles triangle with side length ) A solid with length 0 with a rectangular base and triangular top, wherein one end is a square with side length 5 and the other end is a triangle with base and height of 5 5 (a) the -ais (b) = (c) = ) Region bounded b = / +, =, = and the -ais Rotate about: 0 (a) the -ais (c) = (b) = 7) Region bounded b =, = and = Rotate about: (a) the -ais (c) the -ais (b) = 4 (d) = In Eercises 8, a solid is described Orient the solid along the -ais such that a cross-sectional area function A() can be obtained, then find the volume of the solid

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25 6 Volume b The Shell Method Volume b The Shell Method Motivating Questions In this section, we strive to understand the ideas generated b the following important questions: Is there other methods to use a definite integral to find the volume of a three-dimensional solid or a solid of revolution that results from revolving a two-dimensional region about a particular ais? In what circumstances do we integrate with respect to instead of integrating with respect to? When is it better to use the Shell Method as oppossed to the Washer Method? Introduction Often a given problem can be solved in more than one wa A particular method ma be chosen out of convenience, personal preference, or perhaps necessit Ultimatel, it is good to have options The previous section introduced the Disk and Washer Methods, which computed the volume of solids of revolution b integrating the cross-sectional area of the solid This section develops another method of computing volume, the Shell Method Instead of slicing the solid perpendicular to the ais of rotation creating cross-sections, we now slice it parallel to the ais of rotation, creating shells The Preview Activit 6 introduces a situation where using the Washer Method from Section 6 becomes ver tedious Preview Activit 6 Consider the function f () = 3, whose graph is in Figure 69 Our goal in this activit is to use a definite integral to determine the volume of the solid formed b revolving the region bounded b f () and = 0 about the -ais (a) Using the Washer Method, find an epression for the inner and outer radii of a slice (b) Set up a definite integral to find the volume If ou tr to evaluate the integral, what do ou notice that happens? (c) Find where the local maimum occurs (d) Use the results of past c) to split the solid into two pieces Set up two definite integrals to find the volume of the original solid Figure 69: The circular cone described in Preview Activit 6 Consider Figure 60-(a), where the region shown rotated around the -ais forming the solid shown in Figure 60-(b) A small slice of the region is drawn in Figure 60-(a), parallel to the ais of rotation When the region is rotated, this thin slice forms a clindrical shell, as pictured in Figure 60-(c) The

26 446 Chapter 6 Applications of Integration (a) (b) = + previous section approimated a solid with lots of thin disks (or washers); we now approimate a solid with man thin clindrical shells To compute the volume of one shell, first consider the paper label on a soup can with radius r and height h What is the area of this label? A simple wa of determining this is to cut the label and la it out flat, forming a rectangle with height h and length πr Thus the area is A = πrh; see Figure 6-(a) Do a similar process with a clindrical shell, with height h, thickness, and approimate radius r Cutting the shell and laing it flat forms a rectangular solid with length πr, height h and depth Thus the volume is V πrh ; see Figure 6-(b) (We sa approimatel since our radius was an approimation) B breaking the solid into n clindrical shells, we can approimate the volume of the solid as V = n πr i h i i, i= (c) where r i, h i and i are the radius, height and thickness of the i th shell, respectivel This is a Riemann Sum Taking a limit as the thickness of the shells approaches 0 leads to a definite integral Figure 60: Introducing the Shell Method The Shell Method h r cut here (a) πr A = πrh h Let a solid be formed b revolving a region R, bounded b = a and = b, around a vertical ais Let r() represent the distance from the ais of rotation to (ie, the radius of a sample shell) and let h() represent the height of the solid at (ie, the height of the shell) The volume of the solid is b V = π r()h() d a h r πr cut here V πrh h (b) Figure 6: Determining the volume of a thin clindrical shell Special Cases: ) When the region R is bounded above b = f () and below b = g(), then h() = f () g() ) When the ais of rotation is the -ais (ie, = 0) then r() = Let s practice using the Shell Method

27 6 Volume b The Shell Method 447 Eample Find the volume of the solid formed b rotating the region bounded b = 0, = /( + ), = 0 and = about the -ais Solution This is the region used to introduce the Shell Method in Figure 60, but is sketched again in Figure 6 for closer reference A line is drawn in the region parallel to the ais of rotation representing a shell that will be carved out as the region is rotated about the -ais (This is the differential element) The distance this line is from the ais of rotation determines r(); as the distance from to the -ais is, we have r() = The height of this line determines h(); the top of the line is at = /( + ), whereas the bottom of the line is at = 0 Thus h() = /( + ) 0 = /( + ) The region is bounded from = 0 to =, so the volume is h() = + V = π 0 + d This requires substitution Let u = +, so du = d We also change the bounds: u(0) = and u() = Thus we have: = π u du = π ln u = π ln 78 units 3 Note: in order to find this volume using the Disk Method, two integrals would be needed to account for the regions above and below = / }{{} r() Figure 6: Graphing a region in Eample With the Shell Method, nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, as demonstrated net Eample Find the volume of the solid formed b rotating the triangular region determined b the points (0, ), (, ) and (, 3) about the line = 3 Solution The region is sketched in Figure 63 along with the differential element, a line within the region parallel to the ais of rotation The height of the differential element is the distance from = to = +, the line that connects the points (0, ) and (, 3) Thus h() = + = The radius of the shell formed b the differential element is the distance from to = 3; that is, it is r() = 3 The 3 = + h() } {{ } r() 3 Figure 63: Graphing a region in Eample

28 448 Chapter 6 Applications of Integration Figure 64: Graphing a region in Eample -bounds of the region are = 0 to =, giving V = π (3 )() d 0 ( = π 6 ) d 0 ( = π 3 ) = 4 3 π 466 units3 Activit 6 In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated ais In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid It is not necessar to evaluate the integrals ou find (a) The region S bounded b the -ais, the curve =, and the line = ; revolve S about the -ais (b) The region S bounded b the -ais, the curve =, and the line = 4; revolve S about the -ais (c) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (d) The finite region S bounded b the curves = ( ) and = ; revolve S about the -ais (e) How do ou answers to this activit compare to the results of Activt 6? When revolving a region around a horizontal ais, we must consider the radius and height functions in terms of, not 3 = } {{ } h() r() Figure 65: Graphing a region in Eample 3 Eample 3 Find the volume of the solid formed b rotating the region given in Eample about the -ais Solution The region is sketched in Figure 65 with a sample differential element and the solid is sketched in Figure 66 (Note that the region looks slightl different than it did in the previous eample as the bounds on the graph have changed) The height of the differential element is an -distance, between = and = Thus h() = ( ) = + 3 The radius is the distance from to the -ais, so r() = The bounds of the region are = and = 3, leading to the integral

29 6 Volume b The Shell Method [ ( V = π + 3 )] d 3 = π [ + 3 ] d [ = π ] 3 4 [ 9 = π 4 7 ] = 0 3 π 047 units3 3 Figure 66: Graphing a region in Eample 3 Activit 6 In each of the following questions, draw a careful, labeled sketch of the region described, as well as the resulting solid that results from revolving the region about the stated ais In addition, draw a representative slice and state the volume of that slice, along with a definite integral whose value is the volume of the entire solid It is not necessar to evaluate the integrals ou find (a) The region S bounded b the -ais, the curve =, and the line = ; revolve S about the -ais (b) The region S bounded b the -ais, the curve =, and the line = 4; revolve S about the -ais (c) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (d) The finite region S in the first quadrant bounded b the curves = and = 3 ; revolve S about the -ais (e) The finite region S bounded b the curves = ( ) and = ; revolve S about the -ais At the beginning of this section it was stated that it is good to have options The net eample finds the volume of a solid rather easil with the Shell Method, but using the Washer Method would be quite a chore Eample 4 Find the volume of the solid formed b revolving the region bounded b = sin() and the -ais from = 0 to = π about the -ais Solution The region and the resulting solid are given in Figure 67 and Figure 68 respectivel The radius of a sample shell is r() = ; the height of a sample shell is h() = sin(), each from = 0 to = π Thus the volume of the solid is π V = π sin() d 0 r() {}}{ h() π Figure 67: Graphing a region in Eample 4 π

30 450 Chapter 6 Applications of Integration π Figure 68: Graphing a region in Eample 4 This requires Integration B Parts Set u = and dv = sin() d; we leave it to the reader to fill in the rest We have: [ = π cos() [ = π π + sin() π 0 [ ] = π π + 0 π 0 + π = π 974 units 3 ] 0 ] cos() d Note that in order to use the Washer Method, we would need to solve = sin() for, requiring the use of the arcsine function We leave it to the reader to verif that the outside radius function is R() = π arcsin() and the inside radius function is r() = arcsin() Thus the volume can be computed as π 0 [ (π arcsin()) (arcsin()) ] d This integral isn t terrible given that the arcsin () terms cancel, but it is more onerous than the integral created b the Shell Method Summar In this section, we encountered the following important ideas: We can use a definite integral to find the volume of a three-dimensional solid of revolution that results from revolving a two-dimensional region about a particular ais b taking slices parallel to the ais of revolution which will then be clindrical shells If we revolve about a vertical line and slice perpendicular to that line, then our shells are vertical and of thickness This leads us to integrate with respect to If we revolve about a horizontal line and slice parallel to that line, then our shells are horizontal and of thickness This leads us to integrate with respect to, as opposed to with respect to when we slice a solid verticall Let a region R be given with -bounds = a and = b and -bounds = c and = d Horizontal Ais Vertical Ais Washer Method Shell Method b ( π R() r() ) d d π r()h() d a c d ( π R() r() ) b d π r()h() d c a

31 6 Volume b The Shell Method 45 Eercises Terms and Concepts ) T/F: A solid of revolution is formed b revolving a shape around an ais = ) T/F: The Shell Method can onl be used when the Washer Method fails 8) 05 = 3) T/F: The Shell Method works b integrating cross sectional areas of a solid 4) T/F: When finding the volume of a solid of revolution that was revolved around a vertical ais, the Shell Method integrates with respect to Problems In Eercises 5 8, a region of the Cartesian plane is shaded Use the Shell Method to find the volume of the solid of revolution formed b revolving the region about the -ais 05 In Eercises 9, a region of the Cartesian plane is shaded Use the Shell Method to find the volume of the solid of revolution formed b revolving the region about the -ais 3 = 3 9) 3 = 3 5) 0 = 5 0 = 5 0) 5 6) ) 05 = cos 7) 05 = cos

32 45 Chapter 6 Applications of Integration = ) 05 = 05 In Eercises 3 8, a region of the Cartesian plane is described Use the Shell Method to find the volume of the solid of revolution formed b rotating the region about each of the given aes 3) Region bounded b: =, = 0 and = Rotate about: (a) the -ais (b) = (c) the -ais (d) = 4) Region bounded b: = 4 and = 0 Rotate about: (a) = (b) = (c) the -ais (d) = 4 5) The triangle with vertices (, ), (, ) and (, ) Rotate about: (a) the -ais (b) = (c) the -ais (d) = 6) Region bounded b = + and = Rotate about: (a) the -ais (b) = (c) = 7) Region bounded b = / +, =, = and the -ais Rotate about: (a) the -ais (b) = (c) = 8) Region bounded b =, = and = Rotate about: (a) the -ais (b) = (c) the -ais (d) = 4

33 63 Arc Length and Surface Area Arc Length and Surface Area Motivating Questions In this section, we strive to understand the ideas generated b the following important questions: How can a definite integral be used to measure the length of a curve? How can a definite integral be used to measure the surface area of a solid of revolution? Introduction Earl on in our work with the definite integral, we learned that if we have a nonnegative velocit function, v, for an object moving along an ais, the area under the velocit function between a and b tells us the distance the object traveled on that time interval Moreover, based on the definition of the definite integral, that area is given precisel b b a v(t) dt Indeed, for an nonnegative function f on an interval [a, b], we know that b a f () d measures the area bounded b the curve and the -ais between = a and = b, as shown in Figure 69 Through our upcoming work in the present section and chapter, we will eplore how definite integrals can be used to represent a variet of different phsicall important properties In Preview Activit 6, we begin this investigation b seeing how a single definite integral ma be used to represent the area between two curves Preview Activit 63 = f() a A = b f() d a Figure 69: The area between a nonnegative function f and the -ais on the interval [a, b] b In the following, we consider the function f () = over the interval [, ] Our goal is to estimate the length of the curve (a) Graph f () over the interval [, ] Label the points on the curve that correspond to =,, 0, /, and (b) Draw the secant line connecting the points (, f ( ) and (, f ( )) Use the distance formula to find the length of the secant line from = to = (c) Repeat drawing a secant line between the remaining points starting with (, f ( )) and (0, f (0)) For each line segment, use the distance formula to find the length of the segment (d) Add the distances together to get an approimation to the length of the curve (e) How can we improve our approimation? Write a Riemann sum that will give an improvement to our approimation

34 454 Chapter 6 Applications of Integration Finding the length of a curve f L 0 3 Figure 60: At left, a continuous function = f () whose length we seek on the interval a = 0 to b = 3 At right, a close up view of a portion of the curve h In addition to being able to use definite integrals to find the volumes of solids of revolution, we can also use the definite integral to find the length of a portion of a curve We use the same fundamental principle: we take a curve whose length we cannot easil find, and slice it up into small pieces whose lengths we can easil approimate In particular, we take a given curve and subdivide it into small approimating line segments, as shown at left in Figure 60 To see how we find such a definite integral that measures arc length on the curve = f () from = a to = b, we think about the portion of length, L slice, that lies along the curve on a small interval of length, and estimate the value of Lslice using a well-chosen triangle In particular, if we consider the right triangle with legs parallel to the coordinate aes and hpotenuse connecting two points on the curve, as seen at right in Figure 60, we see that the length, h, of the hpotenuse approimates the length, L slice, of the curve between the two selected points Thus, L slice h = ( ) + ( ) B algebraicall rearranging the epression for the length of the hpotenuse, we see how a definite integral can be used to compute the length of a curve In particular, observe that b removing a factor of ( ), we find that L slice = = ( ) + ( ) ( ) ( ) + ( ) ( ) + ( ) ( ) Furthermore, as n and 0, it follows that d d = f () Thus, we can sa that L slice + f () Taking a Riemann sum of all of these slices and letting n, we arrive at the following fact

35 63 Arc Length and Surface Area 455 Arc Length Given a differentiable function f on an interval [a, b], the total arc length, L, along the curve = f () from = a to = b is given b L = b a + f () d Eample Find the arc length of f () = 3/ from = 0 to = 4 Solution We begin b finding f () = 3 / Using the formula, we find the arc length L as 4 ( ) 3 L = + 0 / d 4 = d 4 ( = + 9 ) / 0 4 d ( + 9 ) 3/ 4 4 = = 8 7 ( 0 3/ A graph of f is given in Figure 6 0 ) 907units Figure 6: A graph of f () = 3/ from Eample Eample Find the arc length of f () = 8 ln from = to = Solution This function was chosen specificall because the resulting integral can be evaluated eactl We begin b finding f () = /4 / The arc length is ( L = + 4 ) d = d

36 456 Chapter 6 Applications of Integration Figure 6: A graph of f () = 8 ln from Eample = d ( = 4 + ) d ( = 4 + ) d ( ) = 8 + ln = 3 + ln 07 units 8 A graph of f is given in Figure 6; the portion of the curve measured in this problem is in bold Eample 3 Find the length of the sine curve from = 0 to = π + cos 0 π/4 3/ π/ 3π/4 3/ π Table 6: A table of values of = + cos to evaluate a definite integral in Eample 3 Solution This is somewhat of a mathematical curiosit; in Activit 45 (b) we found the area under one hump of the sine curve is square units; now we are measuring its arc length The setup is straightforward: f () = sin and f () = cos Thus π L = 0 + cos d This integral cannot be evaluated in terms of elementar functions so we will approimate it with Simpson s Method with n = 4 Table 6 gives + cos evaluated at 5 evenl spaced points in [0, π] Simpson s Rule then states that π 0 + cos d π 0 ( ) + 4 3/ + () + 4 3/ = 3898 Using a computer with n = 00 the approimation is L 380; our approimation with n = 4 is quite good Activit 63 Each of the following questions somehow involves the arc length along a curve (a) Use the definition and appropriate computational technolog to determine the arc length along = from = to = (b) Find the arc length of = 4 on the interval 0 4 Find this value in two different was: (a) b using a definite integral, and (b) b using a familiar propert of the curve (c) Determine the arc length of = e 3 on the interval [0, ] (d) Will the integrals that arise calculating arc length tpicall be ones

37 63 Arc Length and Surface Area 457 that we can evaluate eactl using the First FTC, or ones that we need to approimate? Wh? (e) A moving particle is traveling along the curve given b = f () = 0 +, and does so at a constant rate of 7 cm/sec, where both and are measured in cm (that is, the curve = f () is the path along which the object actuall travels; the curve is not a position function ) Find the position of the particle when t = 4 sec, assuming that when t = 0, the particle s location is (0, f (0)) Surface Area of Solids of Revolution We have alread seen how a curve = f () on [a, b] can be revolved around an ais to form a solid Instead of computing its volume, we now consider its surface area We begin as we have in the previous sections: we partition the interval [a, b] with n subintervals, where the i th subinterval is [ i, i+ ] On each subinterval, we can approimate the curve = f () with a straight line that connects f ( i ) and f ( i+ ) as shown in Figure 63 (a) Revolving this line segment about the -ais creates part of a cone (called the frustum of a cone) as shown in Figure 63 (b) The surface area of a frustum of a cone is a i i+ b π length average of the two radii R and r The length is given b L; we use the material just covered b arc length to state that L + f (c i ) i { r L {}}{ R for some c i in the i th subinterval The radii are just the function evaluated at the endpoints of the interval That is, ạ i i+ b R = f ( i+ ) and r = f ( i ) Thus the surface area of this sample frustum of the cone is approimatel Figure 63: Establishing the formula for surface area π f ( i) + f ( i+ ) + f (c i ) i Since f is a continuous function, the Intermediate Value Theorem states there is some d i in [ i, i+ ] such that f (d i ) = f ( i) + f ( i+ ) ; we can use this to rewrite the above equation as π f (d i ) + f (c i ) i

38 458 Chapter 6 Applications of Integration Summing over all the subintervals we get the total surface area to be approimatel Surface Area n π f (d i ) + f (c i ) i, i= which is a Riemann Sum Taking the limit as the subinterval lengths go to zero gives us the eact surface area, given in the following Ke Idea Surface Area of a Solid of Revolution Let f be differentiable on an open interval containing [a, b] where f is also continuous on [a, b] ) The surface area of the solid formed b revolving the graph of = f (), where f () 0, about the -ais is b Surface Area = π f () + f () d a ) The surface area of the solid formed b revolving the graph of = f () about the -ais, where a, b 0, is b Surface Area = π + f () d a When revolving = f () about the -ais, the radii of the resulting frustum are i and i+ ; their average value is simpl the midpoint of the interval In the limit, this midpoint is just π Figure 64: Revolving = sin on [0, π] about the -ais Eample 4 Find the surface area of the solid formed b revolving = sin on [0, π] around the -ais, as shown in Figure 64 Solution The setup is relativel straightforward; we have the surface area SA is: π SA = π sin + cos d 0 = π + u du π/4 = π sec 3 θ dθ π/4 = π The integration above is nontrivial, utilizing Substitution, Trigonometric Substitution, and Integration b Parts

39 63 Arc Length and Surface Area 459 Eample 5 Find the surface area of the solid formed b revolving the curve = on [0, ] about: ) the -ais ) the -ais Solution ) The integral is straightforward to setup: SA = π + () d 0 Like the integral in Eample 4, this requires Trigonometric Substitution = π ( (8 3 + ) ) sinh () 0 = π ( 8 ) 5 sinh units The solid formed b revolving = around the -ais is graphed in Figure 65-(a) ) Since we are revolving around the -ais, the radius of the solid is not f () but rather Thus the integral to compute the surface area is: SA = π + () d 0 This integral can be solved using substitution Set u = + 4 ; the new bounds are u = to u = 5 We then have (a) = π 5 u du 4 = π u3/ = π ( 5 ) units The solid formed b revolving = about the -ais is graphed in Figure 65-(b) (b) Figure 65: The solids used in Eample 5 This last eample is a famous mathematical parado Eample 6 Consider the solid formed b revolving = / about the -ais on [, ) Find the volume and surface area of this solid (This shape, as graphed in Figure 66, is known as Gabriel s Horn since it looks like a ver long horn that onl a supernatural person, such as an angel, could

40 460 Chapter 6 Applications of Integration pla) Solution We have: To compute the volume it is natural to use the Disk Method V = π d b = lim π b d ( ) b = lim π b ( = lim π ) b b = π units Figure 66: A graph of Gabriel s Horn Gabriel s Horn has a finite volume of π cubic units Since we have alread seen that objects with infinite length can have a finite area, this is not too difficult to accept We now consider its surface area The integral is straightforward to setup: SA = π + / 4 d Integrating this epression is not trivial We can, however, compare it to other improper integrals Since < + / 4 on [, ), we can state that π d < π + / 4 d The improper integral on the left diverges Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel s Horn has infinite surface area Hence the parado : we can fill Gabriel s Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it Somehow this parado is striking when we think about it in terms of volume and area However, we have seen a similar parado before, as referenced above We know that the area under the curve = / on [, ) is finite, et the shape has an infinite perimeter Strange things can occur when we deal with the infinite Summar In this section, we encountered the following important ideas: To find the area between two curves, we think about slicing the region into thin rectangles If, for instance, the area of a tpical rectangle on the interval = a to = b is given b A rect = (g() f ()), then the eact area of the region is given b the definite integral b A = (g() f ()) d a

41 63 Arc Length and Surface Area 46 The shape of the region usuall dictates whether we should use vertical rectangles of thickness or horizontal rectangles of thickness We desire to have the height of the rectangle governed b the difference between two curves: if those curves are best thought of as functions of, we use horizontal rectangles, whereas if those curves are best viewed as functions of, we use vertical rectangles The arc length, L, along the curve = f () from = a to = b is given b b L = a + f () d

42 46 Chapter 6 Applications of Integration Eercises Terms and Concepts ) T/F: The integral formula for computing Arc Length was found b first approimating arc length with straight line segments ) T/F: The integral formula for computing Arc Length includes a square root, meaning the integration is probabl eas Problems In Eercises 3, find the arc length of the function on the given integral 3) f () = on [0, ] 4) f () = 8 on [, ] In Eercises 0 4, find the surface area of the described solid of revolution 0) The solid formed b revolving = on [0, ] about the -ais ) The solid formed b revolving = on [0, ] about the -ais ) The solid formed b revolving = 3 on [0, ] about the -ais 3) The solid formed b revolving = on [0, ] about the -ais 4) The sphere formed b revolving = on [, ] about the -ais 5) f () = 3 3/ / on [0, ] 6) f () = 3 + on [, 4] 7) f () = 3/ 6 on [0, 9] 8) f () = ( e + e ) on [0, ln 5] 9) f () = 5 + on [, ] 53 0) f () = ln ( sin ) on [π/6, π/] ) f () = ln ( cos ) on [0, π/4] In Eercises 9, set up the integral to compute the arc length of the function on the given interval Tr to compute the integral b hand, and use a CAS to compute the integral Also, use Simpson s Rule with n = 4 to approimate the arc length ) f () = on [0, ] 3) f () = 0 on [0, ] 4) f () = on [0, ] 5) f () = ln on [, e] 6) f () = on [, ] (Note: this describes the top half of a circle with radius ) 7) f () = /9 on [ 3, 3] (Note: this describes the top half of an ellipse with a major ais of length 6 and a minor ais of length ) 8) f () = on [, ] 9) f () = sec on [ π/4, π/4]

43 64 Densit, Mass, and Center of Mass Densit, Mass, and Center of Mass Motivating Questions In this section, we strive to understand the ideas generated b the following important questions: How are mass, densit, and volume related? How is the mass of an object with varing densit computed? What is the center of mass of an object, and how are definite integrals used to compute it? Introduction We have seen in several different circumstances how studing the units on the integrand and variable of integration enables us to better understand the meaning of a definite integral For instance, if v(t) is the velocit of an object moving along an ais, measured in feet per second, while t measures time in seconds, then both the definite integral and its Riemann sum approimation, b n v(t) dt v(t i ) t, a i= have their overall units given b the product of the units of v(t) and t: (feet/sec) (sec) = feet Thus, b a v(t) dt measures the total change in position (in feet) of the moving object This tpe of unit analsis will be particularl helpful to us in what follows To begin, in the following preview activit we consider two different definite integrals where the integrand is a function that measures how a particular quantit is distributed over a region and think about how the units on the integrand and the variable of integration indicate the meaning of the integral Preview Activit 64 In each of the following scenarios, we consider the distribution of a quantit along an ais (a) Suppose that the function c() = e 0 models the densit of traffic on a straight road, measured in cars per mile, where is number of miles east of a major interchange, and consider the definite integral 0 ( e 0 ) d i What are the units on the product c()? ii What are the units on the definite integral and its Riemann sum

44 464 Chapter 6 Applications of Integration approimation given b 0 c() d n c( i )? i= iii Evaluate the definite integral 0 c() d = 0 ( e 0 ) d and write one sentence to eplain the meaning of the value ou find (b) On a 6 foot long shelf filled with books, the function B models the distribution of the weight of the books, measured in pounds per inch, where is the number of inches from the left end of the bookshelf Let B() be given b the rule B() = 05 + (+) i What are the units on the product B()? ii What are the units on the definite integral and its Riemann sum approimation given b 36 B() d n B( i )? i= iii Evaluate the definite integral 7 0 B() d = 7 0 (05 + (+) ) d and write one sentence to eplain the meaning of the value ou find Densit The mass of a quantit, tpicall measured in metric units such as grams or kilograms, is a measure of the amount of a quantit In a corresponding wa, the densit of an object measures the distribution of mass per unit volume For instance, if a brick has mass 3 kg and volume 000 m 3, then the densit of the brick is 3kg kg = m3 m 3 As another eample, the mass densit of water is 000 kg/m 3 Each of these relationships demonstrate the following general principle Densit, Mass, & Volume For an object of constant densit d, with mass m and volume V, d = m, or m = d V V But what happens when the densit is not constant? If we consider the formula m = d V, it is reminiscent of two other equations that we have used frequentl in recent work: for

45 64 Densit, Mass, and Center of Mass 465 a bod moving in a fied direction, distance = rate time, and, for a rectangle, its area is given b A = l w These formulas hold when the principal quantities involved, such as the rate the bod moves and the height of the rectangle, are constant When these quantities are not constant, we have turned to the definite integral for assistance The main idea in each situation is that b working with small slices of the quantit that is varing, we can use a definite integral to add up the values of small pieces on which the quantit of interest (such as the velocit of a moving object) are approimatel constant For eample, in the setting where we have a nonnegative velocit function that is not constant, over a short time interval t we know that the distance traveled is approimatel v(t) t, since v(t) is almost constant on a small interval, and for a constant rate, distance = rate time Similarl, if we are thinking about the area under a nonnegative function f whose value is changing, on a short interval the area under the curve is approimatel the area of the rectangle whose height is f () and whose width is : f () Both of these principles are represented visuall in Figure 67 In a similar wa, if we consider the setting where the densit of some quantit is not constant, the definite integral enables us to still compute the overall mass of the quantit Throughout, we will focus on problems where the densit varies in onl one dimension, sa along a single ais, and think about how mass is distributed relative to location along the ais Let s consider a thin bar of length b that is situated so its left end is at the origin, where = 0, and assume that the bar has constant cross-sectional area of cm We let the function ρ() represent the mass densit function of the bar, measured in grams per cubic centimeter That is, given a location, ρ() tells us approimatel how much mass will be found in a onecentimeter wide slice of the bar at If we now consider a thin slice of the bar of width, as pictured in Figure 68, the volume of such a slice is the crosssectional area times Since the cross-sections each have constant area cm, it follows that the volume of the slice is cm 3 Moreover, since mass is the product of densit and volume (when densit is constant), we see that the mass of this given slice is approimatel ft/sec v(t) t = v(t) sec f() = f() Figure 67: At left, estimating a small amount of distance traveled, v(t) t, and at right, a small amount of area under the curve, f () ρ() Figure 68: A thin bar of constant cross-sectional area cm with densit function ρ() g/cm 3 mass slice ρ() g cm 3 cm3 = ρ() g

46 466 Chapter 6 Applications of Integration Hence, for the corresponding Riemann sum (and thus for the integral that it approimates), n b ρ( i ) ρ() d, i= 0 we see that these quantities measure the mass of the bar between 0 and b (The Riemann sum is an approimation, while the integral will be the eact mass) At this point, we note that we will be focused primaril on situations where mass is distributed relative to horizontal location,, for objects whose cross-sectional area is constant In that setting, it makes sense to think of the densit function ρ() with units mass per unit length, such as g/cm Thus, when we compute ρ() on a small slice, the resulting units are g/cm cm = g, which thus measures the mass of the slice The general principle follows Mass For an object of constant cross-sectional area whose mass is distributed along a single ais according to the function ρ() (whose units are units of mass per unit of length), the total mass, M of the object between = a and = b is given b M = b a ρ() d Eample A thin bar occupies the interval 0 and it has a densit in kg/m of ρ() = + Find the mass of the bar Solution The mass of the bar in kilograms is b m = ρ() d a = ( + ) d 0 ) = ( = 4 3 kg

47 64 Densit, Mass, and Center of Mass 467 Activit 64 Consider the following situations in which mass is distributed in a nonconstant manner (a) Suppose that a thin rod with constant cross-sectional area of cm has its mass distributed according to the densit function ρ() = e 0, where is the distance in cm from the left end of the rod, and the units on ρ() are g/cm If the rod is 0 cm long, determine the eact mass of the rod (b) Consider the cone that has a base of radius 4 m and a height of 5 m Picture the cone ling horizontall with the center of its base at the origin and think of the cone as a solid of revolution i Write and evaluate a definite integral whose value is the volume of the cone ii Net, suppose that the cone has uniform densit of 800 kg/m 3 What is the mass of the solid cone? iii Now suppose that the cone s densit is not uniform, but rather that the cone is most dense at its base In particular, assume that the densit of the cone is uniform across cross sections parallel to its base, but that in each such cross section that is a distance units from the origin, the densit of the cross section is given b the function ρ() = , measured in kg/m + 3 Determine and evaluate a definite integral whose value is the mass of this cone of non-uniform densit Do so b first thinking about the mass of a given slice of the cone units awa from the base; remember that in such a slice, the densit will be essentiall constant (c) Let a thin rod of constant cross-sectional area cm and length cm have its mass be distributed according to the densit function ρ() = 5 ( 5), measured in g/cm Find the eact location z at which to cut the bar so that the two pieces will each have identical mass Weighted Averages The concept of an average is a natural one, and one that we have used repeatedl as part of our understanding of the meaning of the definite integral If we have n values a, a,, a n, we know that their average is given b class grade grade points credits chemistr B calculus A histor B- 7 3 pscholog B- 7 3 Table 6: A college student s semester grades a + a + + a n, n and for a quantit being measured b a function f on an interval [a, b], the average value of the quantit on [a, b] is b f () d b a a As we continue to think about problems involving the distribution of mass, it is natural to consider the idea of a weighted

48 468 Chapter 6 Applications of Integration average, where certain quantities involved are counted more in the average A common use of weighted averages is in the computation of a student s GPA, where grades are weighted according to credit hours Let s consider the scenario in Table 6 If all of the classes were of the same weight (ie, the same number of credits), the student s GPA would simpl be calculated b taking the average = 3 But since the chemistr and calculus courses have higher weights (of 5 and 4 credits respectivel), we actuall compute the GPA according to the weighted average = 36 The weighted average reflects the fact that chemistr and calculus, as courses with higher credits, have a greater impact on the students grade point average Note particularl that in the weighted average, each grade gets multiplied b its weight, and we divide b the sum of the weights In the following activit, we eplore further how weighted averages can be used to find the balancing point of a phsical sstem Activit 64 For quantities of equal weight, such as two children on a teeter-totter, the balancing point is found b taking the average of their locations When the weights of the quantities differ, we use a weighted average of their respective locations to find the balancing point (a) Suppose that a shelf is 6 feet long, with its left end situated at = 0 If one book of weight lb is placed at = 0, and another book of weight lb is placed at = 6, what is the location of, the point at which the shelf would (theoreticall) balance on a fulcrum? (b) Now, sa that we place four books on the shelf, each weighing lb: at = 0, at =, at 3 = 4, and at 4 = 6 Find, the balancing point of the shelf (c) How does change if we change the location of the third book? Sa the locations of the -lb books are = 0, =, 3 = 3, and 4 = 6 (d) Net, suppose that we place four books on the shelf, but of varing weights: at = 0 a -lb book, at = a 3-lb book, and 3 = 4 a -lb book, and at 4 = 6 a -lb book Use a weighted average of the locations to find, the balancing point of the shelf How does the balancing point in this scenario compare to that found in (b)? (e) What happens if we change the location of one of the books? Sa

49 64 Densit, Mass, and Center of Mass 469 that we keep everthing the same in (d), ecept that 3 = 5 How does change? (f) What happens if we change the weight of one of the books? Sa that we keep everthing the same in (d), ecept that the book at 3 = 4 now weighs lbs How does change? (g) Eperiment with a couple of different scenarios of our choosing where ou move the location of one of the books to the left, or ou decrease the weight of one of the books (h) Write a couple of sentences to eplain how adjusting the location of one of the books or the weight of one of the books affects the location of the balancing point of the shelf Think carefull here about how our changes should be considered relative to the location of the balancing point of the current scenario Center of Mass In Activit 64, we saw that the balancing point of a sstem of point-masses (such as books on a shelf) is found b taking a weighted average of their respective locations In the activit, we were computing the center of mass of a sstem of masses distributed along an ais, which is the balancing point of the ais on which the masses rest In the activit, we actuall used weight rather than mass Since weight is computed b the gravitational constant times mass, the computations for the balancing point result in the same location regardless of whether we use weight or mass, since the gravitational constant is present in both the numerator and denominator of the weighted average Center of Mass For a collection of n masses m,, m n that are distributed along a single ais at the locations,, n, the center of mass is given b = m + m + n m n m + m + + m n What if we instead consider a thin bar over which densit is distributed continuousl? If the densit is constant, it is obvious that the balancing point of the bar is its midpoint But if densit is not constant, we must compute a weighted average Let s sa that the function ρ() tells us the densit distribution along the bar, measured in g/cm If we slice the bar into small sections, this enables us to think of the bar as holding a collection of adjacent point-masses For a slice of thickness at location i, note that the mass of the slice, m i, satisfies m i ρ( i ) Taking n slices of the bar, we can approimate its center of mass b ρ() Figure 69: A thin bar of constant cross-sectional area with densit function ρ() g/cm ρ( ) + ρ( ) + + n ρ( n ) ρ( ) + ρ( ) + + ρ( n )

50 470 Chapter 6 Applications of Integration Rewriting the sums in sigma notation, it follows that n i= i ρ( i ) n i= ρ( i) (6) Moreover, it is apparent that the greater the number of slices, the more accurate our estimate of the balancing point will be, and that the sums in Equation (6) can be viewed as Riemann sums Hence, in the limit as n, we find that the center of mass is given b the quotient of two integrals Center of Mass For a thin rod of densit ρ() distributed along an ais from = a to = b, the center of mass of the rod is given b = b a b a ρ() d ρ() d Note particularl that the denominator of is the mass of the bar, and that this quotient of integrals is simpl the continuous version of the weighted average of locations,, along the bar Eample A thin bar occupies the interval 0 and it has a densit in kg/m of ρ() = + Find the center of mass of the bar Solution From Eample, the mass of the bar in kilograms is 4 3 kg We just need to find b a ρ() d b a ρ() d = = = ( + ) d 0 ( + 3 ) d 0 ) = 8 The center of mass,, 00is 8 4/3 = 7 ( Activit 64 3 Consider a thin bar of length 0 cm whose densit is distributed according to the function ρ() = 4 + 0, where = 0 represents the left end

51 64 Densit, Mass, and Center of Mass 47 of the bar Assume that ρ is measured in g/cm and is measured in cm (a) Find the total mass, M, of the bar (b) Without doing an calculations, do ou epect the center of mass of the bar to be equal to 0, less than 0, or greater than 0? Wh? (c) Compute, the eact center of mass of the bar (d) What is the average densit of the bar? (e) Now consider a different densit function, given b p() = 4e 00073, also for a bar of length 0 cm whose left end is at = 0 Plot both ρ() and p() on the same aes Without doing an calculations, which bar do ou epect to have the greater center of mass? Wh? (f) Compute the eact center of mass of the bar described in (e) whose densit function is p() = 4e Check the result against the prediction ou made in (e) Summar In this section, we encountered the following important ideas: For an object of constant densit D, with volume V and mass m, we know that m = D V If an object with constant cross-sectional area (such as a thin bar) has its densit distributed along an ais according to the function ρ(), then we can find the mass of the object between = a and = b b b m = ρ() d a For a sstem of point-masses distributed along an ais, sa m,, m n at locations,, n, the center of mass,, is given b the weighted average = n i= im i i= n m i If instead we have mass continuousl distributed along an ais, such as b a densit function ρ() for a thin bar of constant cross-sectional area, the center of mass of the portion of the bar between = a and = b is given b b a ρ() d = b a ρ() d In each situation, represents the balancing point of the sstem of masses or of the portion of the bar

52 47 Chapter 6 Applications of Integration Eercises Terms and Concepts ) T/F: The integral formula for computing Arc Length was found b first approimating arc length with straight line segments ) T/F: The integral formula for computing Arc Length includes a square root, meaning the integration is probabl eas Problems 3) Let a thin rod of length a have densit distribution function ρ() = 0e 0, where is measured in cm and ρ in grams per centimeter (a) If the mass of the rod is 30 g, what is the value of a? (b) For the 30g rod, will the center of mass lie at its midpoint, to the left of the midpoint, or to the right of the midpoint? Wh? (c) For the 30g rod, find the center of mass, and compare our prediction in (b) (d) At what value of should the 30g rod be cut in order to form two pieces of equal mass? 4) Consider two thin bars of constant cross-sectional area, each of length 0 cm, with respective mass densit functions ρ() = + and p() = e 0 (a) Find the mass of each bar (b) Find the center of mass of each bar (c) Now consider a new 0 cm bar whose mass densit function is f () = ρ() + p() i Eplain how ou can easil find the mass of this new bar with little to no additional work ii Similarl, compute 0 0 f () d as simpl as possible, in light of earlier computations iii True or false: the center of mass of this new bar is the average of the centers of mass of the two earlier bars Write at least one sentence to sa wh our conclusion makes sense 5) Consider the curve given b = f () = e 5 + (30 )e 05(30 ) (a) Plot this curve in the window = 0 30, = 0 3 (with constrained scaling so the units on the and ais are equal), and use it to generate a solid of revolution about the -ais Eplain wh this curve could generate a reasonable model of a baseball bat (b) Let and be measured in inches Find the total volume of the baseball bat generated b revolving the given curve about the -ais Include units on our answer (c) Suppose that the baseball bat has constant weight densit, and that the weight densit is 06 ounces per cubic inch Find the total weight of the bat whose volume ou found in (b) (d) Because the baseball bat does not have constant cross-sectional area, we see that the amount of weight concentrated at a location along the bat is determined b the volume of a slice at location Eplain wh we can think about the function ρ() = 06π f () (where f is the function given at the start of the problem) as being the weight densit function for how the weight of the baseball bat is distributed from = 0 to = 30 (e) Compute the center of mass of the baseball bat

53 65 Phsics Applications: Work, Force, and Pressure Phsics Applications: Work, Force, and Pressure Motivating Questions In this section, we strive to understand the ideas generated b the following important questions: How do we measure the work accomplished b a varing force that moves an object a certain distance? What is the total force eerted b water against a dam? How are both of the above concepts and their corresponding use of definite integrals similar to problems we have encountered in the past involving formulas such as distance equals rate times time and mass equals densit times volume? Introduction In our work to date with the definite integral, we have seen several different circumstances where the integral enables us to measure the accumulation of a quantit that varies, provided the quantit is approimatel constant over small intervals For instance, based on the fact that the area of a rectangle is A = l w, if we wish to find the area bounded b a nonnegative curve = f () and the -ais on an interval [a, b], a representative slice of width has area A slice = f (), and thus as we let the width of the representative slice tend to zero, we find that the eact area of the region is A = b a f () d In a similar wa, if we know that the velocit of a moving object is given b the function = v(t), and we wish to know the distance the object travels on an interval [a, b] where v(t) is nonnegative, we can use a definite integral to generalize the fact that d = r t when the rate, r, is constant More specificall, on a short time interval t, v(t) is roughl constant, and hence for a small slice of time, d slice = v(t) t, and so as the width of the time interval t tends to zero, the eact distance traveled is given b the definite integral d = b a v(t) dt Finall, when we recentl learned about the mass of an object of non-constant densit, we saw that since M = D V (mass equals densit times volume, provided that densit is constant), if we can consider a small slice of an object on which the densit is approimatel constant, a definite integral ma be used to

54 474 Chapter 6 Applications of Integration determine the eact mass of the object For instance, if we have a thin rod whose cross sections have constant densit, but whose densit is distributed along the ais according to the function = ρ(), it follows that for a small slice of the rod that is thick, M slice = ρ() In the limit as 0, we then find that the total mass is given b M = b a ρ() d Note that all three of these situations are similar in that we have a basic rule (A = l w, d = r t, M = D V) where one of the two quantities being multiplied is no longer constant; in each, we consider a small interval for the other variable in the formula, calculate the approimate value of the desired quantit (area, distance, or mass) over the small interval, and then use a definite integral to sum the results as the length of the small intervals is allowed to approach zero It should be apparent that this approach will work effectivel for other situations where we have a quantit of interest that varies We net turn to the notion of work: from phsics, a basic principal is that work is the product of force and distance For eample, if a person eerts a force of 0 pounds to lift a 0- pound weight 4 feet off the ground, the total work accomplished is W = F d = 0 4 = 80 foot-pounds If force and distance are measured in English units (pounds and feet), then the units on work are foot-pounds If instead we work in metric units, where forces are measured in Newtons and distances in meters, the units on work are Newton-meters Of course, the formula W = F d onl applies when the force is constant while it is eerted over the distance d In Preview Activit 65, we eplore one wa that we can use a definite integral to compute the total work accomplished when the force eerted varies Preview Activit 65 A bucket is being lifted from the bottom of a 50-foot deep well; its weight (including the water), B, in pounds at a height h feet above the water is given b the function B(h) When the bucket leaves the water, the bucket and water together weigh B(0) = 0 pounds, and when the bucket reaches the top of the well, B(50) = pounds Assume that the bucket loses water at a constant rate (as a function of height, h) throughout its journe from the bottom to the top of the well (a) Find a formula for B(h) (b) Compute the value of the product B(5) h, where h = feet

55 65 Phsics Applications: Work, Force, and Pressure 475 Include units on our answer Eplain wh this product represents the approimate work it took to move the bucket of water from h = 5 to h = 7 (c) Is the value in (b) an over- or under-estimate of the actual amount of work it took to move the bucket from h = 5 to h = 7? Wh? (d) Compute the value of the product B() h, where h = 05 feet Include units on our answer What is the meaning of the value ou found? (e) More generall, what does the quantit W slice = B(h) h measure for a given value of h and a small positive value of h? (f) Evaluate the definite integral 50 0 B(h) dh What is the meaning of the value ou find? Wh? Work Because work is calculated b the rule W = F d, whenever the force F is constant, it follows that we can use a definite integral to compute the work accomplished b a varing force For eample, suppose that in a setting similar to the problem posed in Preview Activit 65, we have a bucket being lifted in a 50-foot well whose weight at height h is given b B(h) = + 8e 0h In contrast to the problem in the preview activit, this bucket is not leaking at a constant rate; but because the weight of the bucket and water is not constant, we have to use a definite integral to determine the total work that results from lifting the bucket Observe that at a height h above the water, the approimate work to move the bucket a small distance h is W slice = B(h) h = ( + 8e 0h ) h Hence, if we let h tend to 0 and take the sum of all of the slices of work accomplished on these small intervals, it follows that the total work is given b W = 50 0 B(h) dh = 50 0 ( + 8e 0h ) dh While is a straightforward eercise to evaluate this integral eactl using the Fundamental Theorem of Calculus, in applied settings such as this one we will tpicall use computing technolog to find accurate approimations of integrals that are of interest to us Here, it turns out that W = 50 0 ( + 8e 0h ) dh foot-pounds Our work in the most recent eample above emplos the following important general principle

56 476 Chapter 6 Applications of Integration Work For an object being moved in the positive direction along an ais,, b a force F(), the total work to move the object from a to b is given b W = b a F() d Eample How much work is performed pulling a 60 m climbing rope up a cliff face, where the rope has a mass of 66 g/m? Solution We need to create a force function F() on the interval [0, 60] To do so, we must first decide what is measuring: it is the length of the rope still hanging or is it the amount of rope pulled in? As long as we are consistent, either approach is fine We adopt for this eample the convention that is the amount of rope pulled in This seems to match intuition better; pulling up the first 0 meters of rope involves = 0 to = 0 instead of = 60 to = 50 As is the amount of rope pulled in, the amount of rope still hanging is 60 This length of rope has a mass of 66 g/m, or 0066 kg/m The the mass of the rope still hanging is 0066(60 ) kg; multipling this mass b the acceleration of gravit, 98 m/s, gives our variable force function F() = (98)(0066)(60 ) = 06468(60 ) Thus the total work performed in pulling up the rope is 60 W = 06468(60 ) d =, 644 J 0 B comparison, consider the work done in lifting the entire rope 60 meters The rope weights = N, so the work appling this force for 60 meters is =, 3848 J This is eactl twice the work calculated before (and we leave it to the reader to understand wh) Eample A bo of 00 lb of sand is being pulled up at a uniform rate a distance of 50 ft over minute The sand is leaking from the bo at a rate of lb/s The bo itself weighs 5 lb and is pulled b a rope weighing lb/ft How much work is done lifting just the rope? How much work is done lifting just the bo and sand?

57 65 Phsics Applications: Work, Force, and Pressure What is the total amount of work performed? Solution We start b forming the force function F r () for the rope (where the subscript denotes we are considering the rope) As in the previous eample, let denote the amount of rope, in feet, pulled in (This is the same as saing denotes the height of the bo) The weight of the rope with feet pulled in is F r () = 0(50 ) = 0 0 (Note that we do not have to include the acceleration of gravit here, for the weight of the rope per foot is given, not its mass per meter as before) The work performed lifting the rope is 50 W r = (0 0) d = 50 ft lb 0 The sand is leaving the bo at a rate of lb/s As the vertical trip is to take one minute, we know that 60 lb will have left when the bo reaches its final height of 50 ft Again letting represent the height of the bo, we have two points on the line that describes the weight of the sand: when = 0, the sand weight is 00 lb, producing the point (0, 00); when = 50, the sand in the bo weighs 40 lb, producing the point (50, 40) The slope of this line is =, giving the equation of the weight of the sand at height as w() = + 00 The bo itself weighs a constant 5 lb, so the total force function is F b () = + 05 Integrating from = 0 to = 50 gives the work performed in lifting bo and sand: 50 W b = ( + 05) d = 3750 ft lb 0 3 The total work is the sum of W r and W b : = 4000 ft lb We can also arrive at this via integration: 50 W = (F r () + F b ()) d 0 50 = ( ) d 0 50 = ( 4 + 5) d 0 = 4000 ft lb Eample 3 Hooke s Law states that the force required to compress or stretch a spring units from its natural length is proportional to ; that is, this force is F() = k for some constant k A force of 0 lb stretches a spring from a length of 7 inches to a length of inches How much work was performed in stretching the spring to this length? Solution In man was, we are not at all concerned with the actual length of the spring, onl with the amount of its change Hence, we do not care that 0 lb of force stretches the spring to a length of inches, but

58 478 Chapter 6 Applications of Integration rather that a force of 0 lb stretches the spring b 5 in This is illustrated in Figure??; we onl measure the change in the spring s length, not the overall length of the spring Converting the units of length to feet, we have F(5/) = 5/k = 0 lb Thus k = 48 lb/ft and F() = 48 We compute the total work performed b integrating F() from = 0 to = 5/: 5/ W = 48 d 0 = 4 5/ 0 = 5/ ft lb Activit 65 Consider the following situations in which a varing force accomplishes work (a) Suppose that a heav rope hangs over the side of a cliff The rope is 00 feet long and weighs 03 pounds per foot; initiall the rope is full etended How much work is required to haul in the entire length of the rope? (Hint: set up a function F(h) whose value is the weight of the rope remaining over the cliff after h feet have been hauled in) (b) A leak bucket is being hauled up from a 00 foot deep well When lifted from the water, the bucket and water together weigh 40 pounds As the bucket is being hauled upward at a constant rate, the bucket leaks water at a constant rate so that it is losing weight at a rate of 0 pounds per foot What function B(h) tells the weight of the bucket after the bucket has been lifted h feet? What is the total amount of work accomplished in lifting the bucket to the top of the well? (c) Now suppose that the bucket in (b) does not leak at a constant rate, but rather that its weight at a height h feet above the water is given b B(h) = 5 + 5e 005h What is the total work required to lift the bucket 00 feet? What is the average force eerted on the bucket on the interval h = 0 to h = 00? (d) From phsics, Hooke s Law for springs states that the amount of force required to hold a spring that is compressed (or etended) to a particular length is proportionate to the distance the spring is compressed (or etended) from its natural length That is, the force to compress (or etend) a spring units from its natural length is F() = k for some constant k (which is called the spring constant) For springs, we choose to measure the force in pounds and the distance the spring is compressed in feet Suppose that a force of 5 pounds etends a particular spring 4 inches (/3 foot) beond its natural length i Use the given fact that F(/3) = 5 to find the spring constant k

59 65 Phsics Applications: Work, Force, and Pressure 479 ii Find the work done to etend the spring from its natural length to foot beond its natural length iii Find the work required to etend the spring from foot beond its natural length to 5 feet beond its natural length Work: Pumping Liquid from a Tank In certain geographic locations where the water table is high, residential homes with basements have a peculiar feature: in the basement, one finds a large hole in the floor, and in the hole, there is water For eample, in Figure 630 where we see a sump crock Essentiall, a sump crock provides an outlet for water that ma build up beneath the basement floor; of course, as that water rises, it is imperative that the water not flood the basement Hence, in the crock we see the presence of a floating pump that sits on the surface of the water: this pump is activated b elevation, so when the water level reaches a particular height, the pump turns on and pumps a certain portion of the water out of the crock, hence relieving the water buildup beneath the foundation One of the questions we d like to answer is: how much work does a sump pump accomplish? To that end, let s suppose that we have a sump crock that has the shape of a frustum of a cone, as pictured in Figure 63 Assume that the crock has a diameter of 3 feet at its surface, a diameter of 5 feet at its base, and a depth of 4 feet In addition, suppose that the sump pump is set up so that it pumps the water verticall up a pipe to a drain that is located at ground level just outside a basement window To accomplish this, the pump must send the water to a location 9 feet above the surface of the sump crock It turns out to be advantageous to think of the depth below the surface of the crock as being the independent variable, so, in problems such as this one we tpicall let the positive -ais point down, and the positive -ais to the right, as pictured in the figure As we think about the work that the pump does, we first realize that the pump sits on the surface of the water, so it makes sense to think about the pump moving the water one slice at a time, where it takes a thin slice from the surface, pumps it out of the tank, and then proceeds to pump the net slice below For the sump crock described in this eample, each slice of water is clindrical in shape We see that the radius of each approimatel clindrical slice varies according to the linear function = f () that passes through the points (0, 5) and (4, 075), where is the depth of the particular slice in the tank; it is a straightforward eercise to find that f () = Figure 630: A sump crock Image credit to (0, 5) = f() (4, 075) Figure 63: A sump crock with approimatel clindrical cross-sections that is 4 feet deep, 5 feet in diameter at its base, and 3 feet in diameter at its top