Chapter 6 Notes, Stewart 8e
|
|
- Randell Calvin Griffith
- 6 years ago
- Views:
Transcription
1 Contents 6. Area between curves Area between the curve and the -ais Overview of Area of a Region Between Two Curves Area of a Region Between Two Curves Volume Volume B Slicing Volumes of Solids of Revolution - Disks & Washers Volumes b Clindrical Shells Clindrical shells formula Shell Method Work Work Done b a Constant Force Work Done b a Variable Force Hooke s Law (Robert Hooke ): Pumping Problems Average Value of a Function Average Value of a Function Mean Value Theorem for Integral Calculus
2 6. Area between curves 6.. Area between the curve and the -ais Definition 6.. Let f() be continuous on [a,b]. The area of the region between the graph of f() and the -ais is A = b a f()d Let f() be continuous on [a,b]. The area of the region between the graph of f() and the -ais is A = b a f()d Eample 6... Set up the integral(s) needed to find the area of the region bounded b ) = 3, the -ais and = and = = ) = and the -ais. =
3 3) = and the -ais on [,]. = Overview of Area of a Region Between Two Curves With a few modifications the area under a curve represented b a definite integral can be etended to find the area between to curves. Observe the following graphs of f() and g(). Since both graphs lie above the -ais, we can geometricall interpret the area of the region between the graphs as the area of the region under f() minus the area of the region under g(). Area of the region between f() and g() f() g() 3 Area of the region under g() Area of the region under f()
4 Although we have just shown one case, i.e. when f and g are both positive, this condition is not necessar. All we need in order to evaluate the integrand [f() g()] is that f() and g() both be continuous and g() f() on the interval [a,b] Area of a Region Between Two Curves Finding the Area Between the Curves Using Vertical Rectangles; Integrating with respect to. If f() and g() are continuous on [a,b] and g() f() for all in [a,b], then the area of the region bounded b the graphs of f() and g() and the vertical lines = a and = b is A = b a [f() g()]d where f() is the upper curve and g() is the lower curve. In the net picture, ou can see that we simpl partition the interval [a,b] and draw our representative rectangles. Let s drawn a few rectangles to illustrate the concept. Steps to Find the Area Between Two Curves integrating with restpect to. Graph the curves and draw a representative rectangle. This reveals which curve is f() (upper curve) and which is g()(lower curve). It also helps find the limits of integration if ou do not alread know them.. Find the limits of integration; ou ma need to find the pts of intersection. 3. Write a formula for f() g(). Simplif it if ou can. 4. Integrate [f() g()] from a to b. The number ou get is the area. NOTE: If ou have ver complicated functions ou can use a graphing calculator, Matlab or another computer algebra sstem to draw the functions, this will greatl speed up our work. The main reason for doing step is to see how man times the graphs cross (if ever) in the specified domain. It also is to determine which function is dominant in which intervals. If ou don t have a quick means to graph the functions, ou can set the two functions equal to each other and tr to solve for where the are equal. This will give ou the intersection points of the two graphs. Once ou know these, ou can determine which function is dominant in which subintervals. Then just proceed as before. 4
5 Eample 6... Find the indicated areas ) Find the area of the region bounded b = and = ) Find the area of the region bounded b =, = ( ) and =. =, = ( ), = 5
6 3) Find the area of the region bounded b = 3, =, = and = = 3 3 = 6
7 Finding the Area Between the Curves Using Horizontal Rectangles; Integrating with respect to. The procedure of adding rectangles works no matter how we draw the rectangles. We can just of easil draw our rectangles horizontall and integrate with respect to, (our representative rectangles will now be parallel to the - ais instead of the - ais). Consider the following graphs, we will draw horizontal rectangles in one and vertical in the other: Horizontal rectangles Vertical rectangles 3 3 In the first picture, we onl need to integrate one time. In the second picture, we would have to integrate twice using different limits of integration and different functions. Finding the Area Between Two Curves integrating with respect to If h() and k() are continuous on [c,d] and k() h() for all in [c,d], then the area of the region bounded b the graphs of h() and k() and the horizontal lines = c and = d is A = d where h() is the right curve and k() is the left curve. c [h() k()]d; Steps to Finding the Area Between Two Curves integrating with restpect to. Graph the curves and draw a representative rectangle. This reveals which curve is h() (right curve) and which is k() (left curve). It also helps find the limits of integration if ou do not alread know them.. Find the limits of integration; ou ma need to find the points of intersection. 3. Write a formula for h() k(). Simplif it if ou can. 4. Integrate [h() k()] from c to d. The number ou get is the area. 7
8 Eample Find the indicated areas ) Set up the integrals to find the area of the region bounded b + = 4 and =. 3 4 ) Set up the integrals to find the area of the region bounded b =, = ( ) and =. =, = ( ), = 8
9 6. Volume 6.. Volume B Slicing We will be tring to find the volume of a solid shaped using the sum of cross section areas times a width. We will be driving toward developing a Riemann Sum so that we can transform them into integrals. Definition 6.. The volume of a solid of known integrable cross-section area A() from = a to = b is the integral of A() from a to b, V = b Steps to Find Volumes b the Method of Slicing. Sketch the solid and a tpical cross section. a A() d.. Find a formula for A(), the area of a tpical cross section. 3. Find the limits of integration. 4. Integrate A() to find the volume, V = b a A() d. Eample 6... Find the volumes of the solids if the solid lies between planes perpendicular to the -ais at = and = 4. The cross sections perpendicular to the -ais between these planes run from the parabola = to the parabola =. The cross sections are semi-circular disks with diameters in the -plane. Note: the area of a semicircular disk is given b A = πr. = 3 4 = Eample 6... Find the volume of the solid that lies between planes perpendicular to the -ais at = and =. In each case, the cross sections perpendicular to the -ais, between these planes, run from the semicircle = to the semicircle =. ) Assume the cross sections are squares with bases in the -plane. Y X 9
10 ) The cross sections are squares with diagonals in the -plane. (The length of a square s diagonal is times the length of its sides) Y X Eample The base of the solid is the region between the curve = sin() and the interval [,π] on the -ais. The cross sections perpendicular to the -ais are vertical equilateral triangles with bases running from the -ais to the curve. Note: the area of a triangle is given b A = bh or A = 3 4 s = sin() π Y X Eample The solid lies between the planes perpendicular to the -ais at = π 3 and = π 3. The cross sections perpendicular to the -ais are circular disks with diameters running from the curve = tan to the curve = sec.
11 6.. Volumes of Solids of Revolution - Disks & Washers Introduction In this part of the section, we begin with a known planar shape and rotate that shape about a line. The resulting three dimensional shape is known as a solid of revolution. The line about which we rotate the plane shape, is called the ais of rotation. Certain solids of revolution that can be generated like a clinder, a cone, or a sphere, we can find their volumes using formulas from geometr. However, when the solid or revolution takes on a non-regular shape, like a spool, a bullet, or a blimp, etc... there are often times no eas geometric volume formulas. So we fall back on integral calculus to compute the volume. A. Disks Method: Rotation about the - ais:. If we revolve a continuous function = R(), on [a,b], and the -ais, about the ais as our ais of revolution, then we have a solid formed (with no hole); the cross section perpendicular to the ais of revolution will be a disk of radius (R() ) which will have area A() = π(radius) = π[r()].. The volume of the solid generated b revolving about the -ais the region between the -ais and the graph of the continuous function = R(), a b, is V = b 3. Steps to find the volume of the solid of revolution a π[r()] d a. Graph the region. b. Determine the ais of revolution. c. Determine radius R(). d. Determine the limits of integration (Determined b the region). e. Integrate. Eample Find the volume of the solid generated b revolving the region defined b =, = about the -ais. =, =..5.5.
12 B. Disks Method: Rotation about the -ais: (a) Similarl, we can find the volume of the solid when the region is rotated about the -ais. (b) The volume of the solid generated b revolving about the -ais the region between the -ais and the graph of the continuous function = R(), c d, is V = b a π[r()] d Eample Find the volume of the solid generated b revolving the region defined b = 3, = 8 and = about the -ais. 8 = 3, = C. Disks Method: Rotation about a line other than one of the aes: If ou are revolving the planar object about one of the lines that bound the function, then ou proceed in prett much the same wa, since ou still have a solid with no hole. In this tpe of problem, the radius of the disk is still the distance between the curve and the ais of revolution (i.e., subtract radius R from the ais of revolution or vise versa, whichever gives a positive length). Eample Find the volume of the solid generated b revolving the region defined b = and = about =. = =
13 D. Washer Method: (a) If the region we revolve to generate the solid does not border on or cross the ais of revolution, the solid will have a hole in it. The cross section perpendicular to the ais of revolution are washers instead of disk. (b) The dimensions of a tpical washer are given: Outer Radius: R() Inner Radius: r() (c) Thus the area of the washer will be given as: A() = π(outer radius) π(inner radius) = π[r()] π[r()] A() = π ( [R()] [r()] ) (d) The volume of such a solid of revolution would be given as: V() = b a π ( [R()] [r()] ) d (e) This formula can be adjusted to be integrated with respect to. (f) The disk formula for finding volume is just the washer formula with r() =. Eample Find the volume of the solid generated b revolving the region defined b = and = about the -ais. = = 3
14 Eample Find the volume of the solid generated b revolving the region defined b =, = and = 4 about the -ais. =, = 4 4 Eample 6... Find the volume of the solid generated b revolving the region defined b = +, = +, = and = about = 3. = +, = + 3 4
15 6.3 Volumes b Clindrical Shells 6.3. Clindrical shells formula. Consider a representative rectangle that has the following characteristics: w = width of the rectangle h = height of the rectangle p = distance between ais of revolution and center of the rectangle. When this rectangle is revolved about the ais of revolution it forms a clindrical shell or tube of thickness w. In order to find the volume of this shell, we ll consider two clinders, one that has radius equal to R() = p+ w the other has radius r() = p w, since p is the average radius of the shell. Recall that the volume of a clinder is V = π(radius) (height) = πr h. So the volume of the shell is: V = πr h πr h = π(p+w/) h π(p w/) h = π [ (p +pw+w /4) (p pw +w /4) ] h = π(pw)h = π(average radius)(width)(height). An alternative wa to remember formula is to consider a clindrical shell with outer radius r and inner radius r. Now imagine cutting and unrolling this clindrical shell to get a nearl flat rectangular solid. Using the fact that the volume of a rectangular solid is V =(length)(width)(height); the volume of this rectangular solid would be V = π(r)(r r )h = (πr)h r. If we now take a solid of revolution and want to approimate its volume b adding the volumes of the shells swept out b n rectangles and take the limit of this sum as r the volume of the solid is: V = lim r k= n (πr)h r = b a πrh dr. 5
16 6.3. Shell Method A. Rotation about the -ais. The volume of the solid generated b revolving the region between the -ais and the graph of the continuous function = f() >, a < < b about the -ais, is V = π(shell radius)(shell height) d V = b. Steps to find the volume of the solid of revolution a πf()d a. Graph the region. b. Determine the ais of revolution. c. Determine shell radius. d. Determine shell height. e. Determine the limits of integration (Determined b the region). f. Integrate. Eample Find the volume of the solid generated b revolving the region defined b = + 4, = from = to = about the -ais. = + 4 6
17 Eample Find the volume of the solid generated b revolving the region defined b =, = 4 from about the -ais. =, = B. Rotation about the -ais The volume of the solid generated b revolving the region between the -ais and the graph of the continuous function = g(), c d about the -ais, is V = π(shell radius)(shell height) d V = d c πg() d Eample Find the volume of the solid generated b revolving the region in the first quadrant bounded b = 3 and = about the -ais. = 3.5 7
18 C. Rotation about a line other than one of the aes: Eample Find the volume of the solid generated b revolving the region in the first quadrant bounded b = 3, = 9 and = about: a. the line = 9 b. the line = = 3, =
19 Eample Find the volume of the solid generated b revolving the region bounded b = and = + about the line = 3. 4 =, = + 3 Eample Find the volume of the solid generated b revolving the region in the first quadrant bounded b =, the -ais and the line = about the line =. =, = 3 9
20 6.4 Work 6.4. Work Done b a Constant Force Definition 6.3. If an object is moved a distance D in the direction of an applied constant force F, then the work W done b the force is defined as W = FD.. Notice that b this definition, if the object ou are tring to move, does not move, then no work has been done, even though ou have eerted work in our efforts to move it.. There are man tpes of forces - centrifugal, electromotive, and gravitational, etc. We tpicall measure forces according to their abilit to accelerate a unit of mass over time. Units Sstem of Measurement Units for Force Units for Distance Units for Work U.S. pound (lb) foot (ft) foot-pounds (ftlb) inch (in) inch-pounds (in-lb) S.I. dne centimeters (cm) erg newton (N) meters (m) Newton-meter (N-m) Joule (J) NOTE: joule = ( newton)( meter) = 7 ergs Conversions pound = newtons slug = 4.59 kilograms newton =.48 pounds kilogram =.6854 slugs dne =.48 pounds gram =.6854 slugs Eample A 5 lb man is climbing a ft pole. Calculate the work done b the man Work Done b a Variable Force Definition 6.4. If an object is moved along a straight line b a continuousl varing force F(), then the work W done b the force as the object is moved from = a to = b is given b: W = lim n n f( i) = i= b a f()d
21 Eample A mountain climber is about to haul up a 5 m length of hanging rope. How much work will it take if the rope weighs.634 N/m? Eample A 5 lb bucket is lifted from the ground into the air b pulling in ft of rope, weighing.6 lbs, over a constant speed. The bucket contains gallons of H (6lbs). How much work was spent lifting the H, the bucket and the rope? Eample A 75 ft cable weighing 6 lb/ft is connected to an elevator weighing 5 lb. Find the work done in lifting the elevator to a height of 5 ft.
22 6.4.3 Hooke s Law (Robert Hooke ): Definition 6.5. The force F required to compress or to stretch a spring (within its elastic limits) is proportional to the distance d that the spring is compressed or stretched from its original length (natural length). Mathematicall speaking: F = k where the constant of proportionalit k (also known as the force constant or spring constant) depends on the specific nature of the spring and is the number of length units the spring is compressed/stretched from its natural length. Eample A spring of natural length is stretched to 5 under a weight of lb. a) Find the work done to stretch it from to 3. b) Find the work done to stretch it from 5 to 8. Eample It takes in-lb of work to stretch a spring from its natural length of to. a) Determine the force constant. b) Determine the work done to compress the spring from to 8.
23 6.4.4 Pumping Problems A. The strateg: When finding work done to pump liquids, we need to think of thin slabs of water being removed and then summing those slabs up. Eventuall, taking the limit of Riemann Sums to come up with an integral. B. The Weight of Water: Because of variations in the earth s gravitational field, the weight of a cubic foot of water at sea level can var from about 6.6 lbs at the equator to as much as 6.59 lbs near the poles, a variation of about.5%. We will use a tpical densit estimate of δ = 6.5 lbs ft 3 for our calculations. Using the S.I. units of measurement the densit of a cubic meter of water is kg m 3. Therefore, the estimated weight of the water is δ = 9.8 m s kg m 3 = 98 N m 3. C. Placement of Tank: Whether we place the tank above or below ground does not affect the amount of work done to pump the liquid out of the tank. We will sometimes place a tank underground in order to make the algebra easier. D. Steps in Finding Work Done During Pumping:. Draw a figure with a coordinate sstem.. Find the volume V of a thin horizontal slab/slice of liquid. 3. Find the weight/force F of a thin horizontal slab/slice of liquid. F = δ( V) 4. Find the distance d that the liquid has to travel. 5. Find the work W it takes to lift the slab to its destination. W = Fd = δ( V)d 6. Find the limits of integration. (defined b the region) 7. Integrate the work epression from the base to the surface of the liquid. Eample Constant Shaped Tank. An aquarium with dimensions 3 ft is full of H O (weighing 6.5 lbs/ft 3 ). Find the work in lifting and pushing the H O over the top. 3
24 . What if the tank is full of H O and all the H O is pumped out 8 feet above the top. (Although the water is moved up a spigot above the top of the tank, for simplicit we disregard the adjustment needed for this change in the receptacle and assume that the entire slab is still being moved up the additional distance.) Eample Variable Shaped Tank. Find the work in pumping kerosene, weighing 5. lbs/ft 3, over the rim of a trough which is ft long and has a semicircular end of diameter 6 ft if the tank is filled to /3 of its height. 4
25 . A tank in the form of a right circular cone, verte down, is filled with H O to a depth of / its height. If the height of the tank is ft and its diameter is 8 ft, find the work done pumping all the H O 5 ft above the top of the tank. 3. To design the interior of a huge stainless steel tank, ou revolve the curve =, 4, about the -ais. The container, with dimensions in meters, is to be filled with seawater which weighs, N/m 3. How much work will it take to empt the tank b pumping the seawater to the tanks top? 5
26 6.5 Average Value of a Function 6.5. Average Value of a Function Definition 6.6. If f() is integrable on [a,b], its average(mean) value on [a,b] is fave = b a b a f()d Eample Find the average value of f() = on [,4] Mean Value Theorem for Integral Calculus Definition 6.7. The Mean Value Theorem for Definite Integrals (MVT): If f() is continuous on [a,b], then at some point c in [a,b], f(c) = b a Eample Appl MVT to the eample above. b a f()d Geometric Interpretation of the MVT For positive functions f(), there is a number c such that the rectangle with base [a,b] and height f(c) has the same area as the region under the graph of f() from a to b
27 Eample Find the average value of = 9+ on [,4]. Appl the MVT to this problem. Eample In a certain cit the temperature (in F) t hours after 9 a.m. was approimated b the function T(t) = 5+4sin ( πt ). Find the average temperature during the period from 9 a.m. to 9 p.m. 7
foot (ft) inch (in) foot (ft) centimeters (cm) meters (m) Joule (J)
Math 176 Calculus Sec. 6.4: Work I. Work Done by a Constant Force A. Def n : If an object is moved a distance D in the direction of an applied constant force F, then the work W done by the force is defined
More informationU.S. pound (lb) foot (ft) foot-pounds (ft-lb) pound (lb) inch (in) inch-pounds (in-lb) tons foot (ft) foot-tons (ft-ton)
Math 1206 Calculus Sec. 6.4: Work I. Work Done by a Constant Force A. Def n : If an object is moved a distance D in the direction of an applied constant force F, then the work W done by the force is defined
More informationMAC 2311 Chapter 5 Review Materials Topics Include Areas Between Curves, Volumes (disks, washers, and shells), Work, and Average Value of a Function
MAC Chapter Review Materials Topics Include Areas Between Curves, Volumes (disks, washers, and shells), Work, and Average Value of a Function MULTIPLE CHOICE. Choose the one alternative that best completes
More informationChapter 7 Applications of Integration
Chapter 7 Applications of Integration 7.1 Area of a Region Between Two Curves 7.2 Volume: The Disk Method 7.3 Volume: The Shell Method 7.5 Work 7.6 Moments, Centers of Mass, and Centroids 7.7 Fluid Pressure
More informationChapter 6: Applications of Integration
Chapter 6: Applications of Integration Section 6.3 Volumes by Cylindrical Shells Sec. 6.3: Volumes: Cylindrical Shell Method Cylindrical Shell Method dv = 2πrh thickness V = න a b 2πrh thickness Thickness
More informationMath 75B Practice Midterm III Solutions Chapter 6 (Stewart) Multiple Choice. Circle the letter of the best answer.
Math 75B Practice Midterm III Solutions Chapter 6 Stewart) English system formulas: Metric system formulas: ft. = in. F = m a 58 ft. = mi. g = 9.8 m/s 6 oz. = lb. cm = m Weight of water: ω = 6.5 lb./ft.
More information(a) Use washer cross sections: a washer has
Section 8 V / (sec tan ) / / / / [ tan sec tan ] / / (sec sec tan + tan ) + tan sec / + A( ) s w (sec tan ), and / V (sec tan ), which b / same method as in part (a) equals A cross section has width w
More informationChapter 6: Applications of Integration
Chapter 6: Applications of Integration Section 6.4 Work Definition of Work Situation There is an object whose motion is restricted to a straight line (1-dimensional motion) There is a force applied to
More information5.6 Work. Common Units Force Distance Work newton (N) meter (m) joule (J) pound (lb) foot (ft) Conversion Factors
5.6 Work Page 1 of 7 Definition of Work (Constant Force) If a constant force of magnitude is applied in the direction of motion of an object, and if that object moves a distance, then we define the work
More informationMath 122 Fall Solutions to Homework #5. ( ) 2 " ln x. y = x 2
Math 1 Fall 8 Solutions to Homework #5 Problems from Pages 383-38 (Section 7.) 6. The curve in this problem is defined b the equation: = ( ) " ln and we are interested in the part of the curve between
More informationRectangular box of sizes (dimensions) w,l,h wlh Right cylinder of radius r and height h r 2 h
Volumes: Slicing Method, Method of Disks and Washers -.,.. Volumes of Some Regular Solids: Solid Volume Rectangular bo of sizes (dimensions) w,l,h wlh Right clinder of radius r and height h r h Right cone
More information6.5 Work and Fluid Forces
6.5 Work and Fluid Forces Work Work=Force Distance Work Work=Force Distance Units Force Distance Work Newton meter Joule (J) pound foot foot-pound (ft lb) Work Work=Force Distance Units Force Distance
More informationAPPLICATIONS OF INTEGRATION
6 APPLICATIONS OF INTEGRATION APPLICATIONS OF INTEGRATION 6.4 Work In this section, we will learn about: Applying integration to calculate the amount of work done in performing a certain physical task.
More informationCHAPTER 6 Applications of Integration
PART II CHAPTER Applications of Integration Section. Area of a Region Between Two Curves.......... Section. Volume: The Disk Method................. 7 Section. Volume: The Shell Method................
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
--review Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the area of the shaded region. ) f() = + - ) 0 0 (, 8) 0 (0, 0) - - - - - - -0
More informationChapter 6 Some Applications of the Integral
Chapter 6 Some Applications of the Integral Section 6.1 More on Area a. Representative Rectangle b. Vertical Separation c. Example d. Integration with Respect to y e. Example Section 6.2 Volume by Parallel
More informationTriple Integrals. y x
Triple Integrals. (a) If is an solid (in space), what does the triple integral dv represent? Wh? (b) Suppose the shape of a solid object is described b the solid, and f(,, ) gives the densit of the object
More informationFluid force problems
Roberto s Notes on Integral Calculus Chapter 5: Basic applications of integration Section 1 Fluid force problems What ou need to know alread: How to use the four step process to set up an integral. What
More informationPractice Exam 1 Solutions
Practice Exam 1 Solutions 1a. Let S be the region bounded by y = x 3, y = 1, and x. Find the area of S. What is the volume of the solid obtained by rotating S about the line y = 1? Area A = Volume 1 1
More informationf x and the x axis on an interval from x a and
Unit 6: Chapter 8 Areas and Volumes & Density Functions Part 1: Areas To find the area bounded by a function bwe use the integral: f d b a b 0 f d f d. a b a f and the ais on an interval from a and. This
More informationMath 262 Exam 1 - Practice Problems. 1. Find the area between the given curves:
Mat 6 Exam - Practice Problems. Find te area between te given curves: (a) = x + and = x First notice tat tese curves intersect wen x + = x, or wen x x+ =. Tat is, wen (x )(x ) =, or wen x = and x =. Next,
More informationlim x c) lim 7. Using the guidelines discussed in class (domain, intercepts, symmetry, asymptotes, and sign analysis to
Math 7 REVIEW Part I: Problems Using the precise definition of the it, show that [Find the that works for any arbitrarily chosen positive and show that it works] Determine the that will most likely work
More informationwe make slices perpendicular to the x-axis. If the slices are thin enough, they resemble x cylinders or discs. The formula for the x
Math Learning Centre Solids of Revolution When we rotate a curve around a defined ais, the -D shape created is called a solid of revolution. In the same wa that we can find the area under a curve calculating
More informationApplications of Integration to Physics and Engineering
Applications of Integration to Physics and Engineering MATH 211, Calculus II J Robert Buchanan Department of Mathematics Spring 2018 Mass and Weight mass: quantity of matter (units: kg or g (metric) or
More informationMATH141: Calculus II Exam #1 review 6/8/2017 Page 1
MATH: Calculus II Eam # review /8/7 Page No review sheet can cover everything that is potentially fair game for an eam, but I tried to hit on all of the topics with these questions, as well as show you
More informationMath 76 Practice Problems for Midterm II Solutions
Math 76 Practice Problems for Midterm II Solutions 6.4-8. DISCLAIMER. This collection of practice problems is not guaranteed to be identical, in length or content, to the actual exam. You may expect to
More informationFor the intersections: cos x = 0 or sin x = 1 2
Chapter 6 Set-up examples The purpose of this document is to demonstrate the work that will be required if you are asked to set-up integrals on an exam and/or quiz.. Areas () Set up, do not evaluate, any
More informationIntegrals. D. DeTurck. January 1, University of Pennsylvania. D. DeTurck Math A: Integrals 1 / 61
Integrals D. DeTurck University of Pennsylvania January 1, 2018 D. DeTurck Math 104 002 2018A: Integrals 1 / 61 Integrals Start with dx this means a little bit of x or a little change in x If we add up
More informationEvaluate the following limit without using l Hopital s Rule. x x. = lim = (1)(1) = lim. = lim. = lim = (3 1) =
5.4 1 Looking ahead. Example 1. Indeterminate Limits Evaluate the following limit without using l Hopital s Rule. Now try this one. lim x 0 sin3x tan4x lim x 3x x 2 +1 sin3x 4x = lim x 0 3x tan4x ( ) 3
More information11.1 Double Riemann Sums and Double Integrals over Rectangles
Chapter 11 Multiple Integrals 11.1 ouble Riemann Sums and ouble Integrals over Rectangles Motivating Questions In this section, we strive to understand the ideas generated b the following important questions:
More informationQuiz 6 Practice Problems
Quiz 6 Practice Problems Practice problems are similar, both in difficulty and in scope, to the type of problems you will see on the quiz. Problems marked with a are for your entertainment and are not
More information5.5 Volumes: Tubes. The Tube Method. = (2π [radius]) (height) ( x k ) = (2πc k ) f (c k ) x k. 5.5 volumes: tubes 435
5.5 volumes: tubes 45 5.5 Volumes: Tubes In Section 5., we devised the disk method to find the volume swept out when a region is revolved about a line. To find the volume swept out when revolving a region
More informationChapter 8. Applications of Definite Integrals. π π. 1 and 2. The graph is. x 2x + 3 = 0 has no real solutions, since
Chapter 8 Applications of Definite Integrals Section 8 Accumulation and Net Change (pp 8 96) Eploration Revisiting Eample t 8 st () 8 + + t + 8 s () 8+ + 8 s () 8+ + 6 and The graph is + + Section 8 7
More informationDepartment of Mathematical x 1 x 2 1
Contents Limits. Basic Factoring Eample....................................... One-Sided Limit........................................... 3.3 Squeeze Theorem.......................................... 4.4
More informationCHAPTER 1 MEASUREMENTS AND VECTORS
CHPTER 1 MESUREMENTS ND VECTORS 1 CHPTER 1 MESUREMENTS ND VECTORS 1.1 UNITS ND STNDRDS n phsical quantit must have, besides its numerical value, a standard unit. It will be meaningless to sa that the distance
More informationApplications of Definite Integrals
58_Ch7_pp378-433.qd /3/6 :3 PM Page 378 Chapter 7 Applications of Definite Integrals T he art of potter developed independentl in man ancient civilizations and still eists in modern times. The desired
More informationNOT FOR SALE 6 APPLICATIONS OF INTEGRATION. 6.1 Areas Between Curves. Cengage Learning. All Rights Reserved. ( ) = 4 ln 1 = 45.
6 APPLICATIONS OF INTEGRATION 6. Areas Between Curves... 4. 8 ( ) 8 8 4 4 ln ( ln 8) 4 ln 45 4 ln 8 ( ) ( ) ( ) + + + + + ( ) ( 4) ( +6) + ( 8 + 7) 9 5. ( ) + ( +) ( + ) + 4 6. ( sin ) +cos 8 8 + 7. The
More informationMathematics 309 Conic sections and their applicationsn. Chapter 2. Quadric figures. ai,j x i x j + b i x i + c =0. 1. Coordinate changes
Mathematics 309 Conic sections and their applicationsn Chapter 2. Quadric figures In this chapter want to outline quickl how to decide what figure associated in 2D and 3D to quadratic equations look like.
More informationPre-Algebra Homework 10 Geometry: Solutions
Pre-Algebra Homework 10 Geometry: Solutions Use π = 3 for your calculations 1. I am building a ramp. The length of the ramp along the ground is 12 meters and it s 5 meters high and 5 meters wide. I want
More information1 x 3 3x. x 1 2x 1. 2x 1 2x 1. 2x 1 2x 1. x 2 +4x 1 j. lim. x3 +2x x 5. x2 9
MATHEMATICS 57 Final Eamination Review Problems. Let f 5. Find each of the following. a. fa+b b. f f. Find the domain of each function. a. f b. g +. The graph of f + is the same as the graph of g ecept
More informationApplications of Integration
8 Applications of Integration º½ Ö ØÛ Ò ÙÖÚ We have seen how integration can be used to find an area between a curve and the x-axis With very little change we can find some areas between curves; indeed,
More informationHigher. Integration 1
Higher Mathematics Contents Indefinite Integrals RC Preparing to Integrate RC Differential Equations A Definite Integrals RC 7 Geometric Interpretation of A 8 Areas between Curves A 7 Integrating along
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. C) 2
Cal II- Final Review Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Epress the following logarithm as specified. ) ln 4. in terms of ln and
More informationOne of the most common applications of Calculus involves determining maximum or minimum values.
8 LESSON 5- MAX/MIN APPLICATIONS (OPTIMIZATION) One of the most common applications of Calculus involves determining maimum or minimum values. Procedure:. Choose variables and/or draw a labeled figure..
More informationWork. 98N We must exert a force of 98N to raise the object. 98N 15m 1470Nm. One Newton- meter is called a Joule and the
ork Suppose an object is moving in one dimension either horizontally or vertically. Suppose a Force which is constant in magnitude and in the same direction as the object's motion acts on that object.
More information5.6 RATIOnAl FUnCTIOnS. Using Arrow notation. learning ObjeCTIveS
CHAPTER PolNomiAl ANd rational functions learning ObjeCTIveS In this section, ou will: Use arrow notation. Solve applied problems involving rational functions. Find the domains of rational functions. Identif
More information1985 AP Calculus AB: Section I
985 AP Calculus AB: Section I 9 Minutes No Calculator Notes: () In this eamination, ln denotes the natural logarithm of (that is, logarithm to the base e). () Unless otherwise specified, the domain of
More informationApplications of Integration
Math 112 Spring 2019 Lab 3 Name: Section: Score: Applications of Integration In Lab 2 we explored one application of integration, that of finding the volume of a solid. Here, we explore a few more of the
More informationFind the following limits. For each one, if it does not exist, tell why not. Show all necessary work.
Calculus I Eam File Spring 008 Test #1 Find the following its. For each one, if it does not eist, tell why not. Show all necessary work. 1.) 4.) + 4 0 1.) 0 tan 5.) 1 1 1 1 cos 0 sin 3.) 4 16 3 1 6.) For
More informationMathematics Placement Examination (MPE)
Practice Problems for Mathematics Placement Eamination (MPE) Revised June, 011 When ou come to New Meico State Universit, ou ma be asked to take the Mathematics Placement Eamination (MPE) Your inital placement
More information18. Two different ways to find the volume of a cone
. Two different was to find the volume of a cone Suppose ou didn t alread know that the volume of a solid cone of radius R and height H is 3 πr H. How would ou find out? One wa would be to chop up the
More informationDepartment of Mathematical 1 Limits. 1.1 Basic Factoring Example. x 1 x 2 1. lim
Contents 1 Limits 2 1.1 Basic Factoring Example...................................... 2 1.2 One-Sided Limit........................................... 3 1.3 Squeeze Theorem..........................................
More informationd` = 1+( dy , which is part of the cone.
7.5 Surface area When we did areas, the basic slices were rectangles, with A = h x or h y. When we did volumes of revolution, the basic slices came from revolving rectangles around an axis. Depending on
More informationGrade 11 Mathematics Practice Test
Grade Mathematics Practice Test Nebraska Department of Education 00 Directions: On the following pages are multiple-choice questions for the Grade Practice Test, a practice opportunit for the Nebraska
More informationSection 1.5 Formal definitions of limits
Section.5 Formal definitions of limits (3/908) Overview: The definitions of the various tpes of limits in previous sections involve phrases such as arbitraril close, sufficientl close, arbitraril large,
More informationMATHEMATICS LEVEL 2 TEST FORM B Continued
Mathematics Level Test Form B For each of the following problems, decide which is the BEST of the choices given. If the eact numerical value is not one of the choices, select the choice that best approimates
More information8.2 APPLICATIONS TO GEOMETRY
8.2 APPLICATIONS TO GEOMETRY In Section 8.1, we calculated volumes using slicing and definite integrals. In this section, we use the same method to calculate the volumes of more complicated regions as
More informationMATHEMATICS LEVEL 2. MATHEMATICS LEVEL 2 Continued GO ON TO THE NEXT PAGE USE THIS SPACE FOR SCRATCHWORK. 1. If xy 0 and 3x = 0.
MATHEMATICS LEVEL For each of the following problems, decide which is the BEST of the choices given. If the eact numerical value is not one of the choices, select the choice that best approimates this
More information17. Find the moments of inertia I x, I y, I 0 for the lamina of. 4. D x, y 0 x a, 0 y b ; CAS. 20. D is enclosed by the cardioid r 1 cos ; x, y 3
SCTION 2.5 TRIPL INTGRALS 69 2.4 XRCISS. lectric charge is distributed over the rectangle, 2 so that the charge densit at, is, 2 2 (measured in coulombs per square meter). Find the total charge on the
More informationSecond-Order Linear Differential Equations C 2
C8 APPENDIX C Additional Topics in Differential Equations APPENDIX C. Second-Order Homogeneous Linear Equations Second-Order Linear Differential Equations Higher-Order Linear Differential Equations Application
More information4 The Cartesian Coordinate System- Pictures of Equations
The Cartesian Coordinate Sstem- Pictures of Equations Concepts: The Cartesian Coordinate Sstem Graphs of Equations in Two Variables -intercepts and -intercepts Distance in Two Dimensions and the Pthagorean
More informationc) domain {x R, x 3}, range {y R}
Answers Chapter 1 Functions 1.1 Functions, Domain, and Range 1. a) Yes, no vertical line will pass through more than one point. b) No, an vertical line between = 6 and = 6 will pass through two points..
More informationArkansas Tech University MATH 2924: Calculus II Dr. Marcel B. Finan. Solutions to Assignment 7.6. sin. sin
Arkansas Tech University MATH 94: Calculus II Dr. Marcel B. Finan Solutions to Assignment 7.6 Exercise We have [ 5x dx = 5 ] = 4.5 ft lb x Exercise We have ( π cos x dx = [ ( π ] sin π x = J. From x =
More informationLab 5 Forces Part 1. Physics 211 Lab. You will be using Newton s 2 nd Law to help you examine the nature of these forces.
b Lab 5 Forces Part 1 Phsics 211 Lab Introduction This is the first week of a two part lab that deals with forces and related concepts. A force is a push or a pull on an object that can be caused b a variet
More informationMath 323 Exam 2 - Practice Problem Solutions. 2. Given the vectors a = 1,2,0, b = 1,0,2, and c = 0,1,1, compute the following:
Math 323 Eam 2 - Practice Problem Solutions 1. Given the vectors a = 2,, 1, b = 3, 2,4, and c = 1, 4,, compute the following: (a) A unit vector in the direction of c. u = c c = 1, 4, 1 4 =,, 1+16+ 17 17
More informationMath 116 First Midterm October 9, 2017
On m honor, as a student, I have neither given nor received unauthorized aid on this academic work. Initials: Do not write in this area Your Initials Onl: Math 116 First Midterm October 9, 217 Your U-M
More information0.1 Work. W net = T = T f T i,
.1 Work Contrary to everyday usage, the term work has a very specific meaning in physics. In physics, work is related to the transfer of energy by forces. There are essentially two complementary ways to
More information5. Find the intercepts of the following equations. Also determine whether the equations are symmetric with respect to the y-axis or the origin.
MATHEMATICS 1571 Final Examination Review Problems 1. For the function f defined by f(x) = 2x 2 5x find the following: a) f(a + b) b) f(2x) 2f(x) 2. Find the domain of g if a) g(x) = x 2 3x 4 b) g(x) =
More informationANOTHER FIVE QUESTIONS:
No peaking!!!!! See if you can do the following: f 5 tan 6 sin 7 cos 8 sin 9 cos 5 e e ln ln @ @ Epress sin Power Series Epansion: d as a Power Series: Estimate sin Estimate MACLAURIN SERIES ANOTHER FIVE
More informationSpace Coordinates and Vectors in Space. Coordinates in Space
0_110.qd 11//0 : PM Page 77 SECTION 11. Space Coordinates and Vectors in Space 77 -plane Section 11. -plane -plane The three-dimensional coordinate sstem Figure 11.1 Space Coordinates and Vectors in Space
More information6 = 1 2. The right endpoints of the subintervals are then 2 5, 3, 7 2, 4, 2 9, 5, while the left endpoints are 2, 5 2, 3, 7 2, 4, 9 2.
5 THE ITEGRAL 5. Approimating and Computing Area Preliminar Questions. What are the right and left endpoints if [, 5] is divided into si subintervals? If the interval [, 5] is divided into si subintervals,
More informationIntegration to Compute Volumes, Work. Goals: Volumes by Slicing Volumes by Cylindrical Shells Work
Week #8: Integration to Compute Volumes, Work Goals: Volumes by Slicing Volumes by Cylindrical Shells Work 1 Volumes by Slicing - 1 Volumes by Slicing In the integration problems considered in this section
More informationFunctions of Several Variables
Chapter 1 Functions of Several Variables 1.1 Introduction A real valued function of n variables is a function f : R, where the domain is a subset of R n. So: for each ( 1,,..., n ) in, the value of f is
More informationChapter 5: Quadratic Equations and Functions 5.1 Modeling Data With Quadratic Functions Quadratic Functions and Their Graphs
Ch 5 Alg Note Sheet Ke Chapter 5: Quadratic Equations and Functions 5.1 Modeling Data With Quadratic Functions Quadratic Functions and Their Graphs Definition: Standard Form of a Quadratic Function The
More informationy=1/4 x x=4y y=x 3 x=y 1/3 Example: 3.1 (1/2, 1/8) (1/2, 1/8) Find the area in the positive quadrant bounded by y = 1 x and y = x3
Eample: 3.1 Find the area in the positive quadrant bounded b 1 and 3 4 First find the points of intersection of the two curves: clearl the curves intersect at (, ) and at 1 4 3 1, 1 8 Select a strip at
More informationSummary and Vocabulary
Chapter 2 Chapter 2 Summar and Vocabular The functions studied in this chapter are all based on direct and inverse variation. When k and n >, formulas of the form = k n define direct-variation functions,
More information221 MATH REFRESHER session 3 SAT2015_P06.indd 221 4/23/14 11:39 AM
Math Refresher Session 3 1 Area, Perimeter, and Volume Problems Area, Perimeter, and Volume 301. Formula Problems. Here, you are given certain data about one or more geometric figures, and you are asked
More information2.2 Equations of Lines
660_ch0pp07668.qd 10/16/08 4:1 PM Page 96 96 CHAPTER Linear Functions and Equations. Equations of Lines Write the point-slope and slope-intercept forms Find the intercepts of a line Write equations for
More informationAP Calculus Free-Response Questions 1969-present AB
AP Calculus Free-Response Questions 1969-present AB 1969 1. Consider the following functions defined for all x: f 1 (x) = x, f (x) = xcos x, f 3 (x) = 3e x, f 4 (x) = x - x. Answer the following questions
More informationAP Calculus AB/BC ilearnmath.net
CALCULUS AB AP CHAPTER 1 TEST Don t write on the test materials. Put all answers on a separate sheet of paper. Numbers 1-8: Calculator, 5 minutes. Choose the letter that best completes the statement or
More informationMathematics. Mathematics 2. hsn.uk.net. Higher HSN22000
Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still Notes. For
More information8.4 Density and 8.5 Work Group Work Target Practice
8.4 Density and 8.5 Work Group Work Target Practice 1. The density of oil in a circular oil slick on the surface of the ocean at a distance meters from the center of the slick is given by δ(r) = 5 1+r
More information2007 AP Calculus AB Free-Response Questions Section II, Part A (45 minutes) # of questions: 3 A graphing calculator may be used for this part
2007 AP Calculus AB Free-Response Questions Section II, Part A (45 minutes) # of questions: 3 A graphing calculator may be used for this part 1. Let R be the region in the first and second quadrants bounded
More informationTrigonometric Functions
Trigonometric Functions This section reviews radian measure and the basic trigonometric functions. C ' θ r s ' ngles ngles are measured in degrees or radians. The number of radians in the central angle
More informationMathematics Open Textbooks. Follow this and additional works at: Part of the Mathematics Commons
GALILEO, Universit Sstem of Georgia GALILEO Open Learning Materials Mathematics Open Tetbooks Mathematics Spring 05 Armstrong Calculus Michael Tiemeer Armstrong State Universit, MichaelTiemeer@armstrongedu
More information6.1 Area Between Curves. Example 1: Calculate the area of the region between the parabola y = 1 x 2 and the line y = 1 x
AP Calculus 6.1 Area Between Curves Name: Goal: Calculate the Area between curves Keys to Success: Top Curve Bottom Curve (integrate w/respect to x or dx) Right Curve Left Curve (integrate w/respect to
More informationConic Sections CHAPTER OUTLINE. The Circle Ellipses and Hyperbolas Second-Degree Inequalities and Nonlinear Systems FIGURE 1
088_0_p676-7 /7/0 :5 PM Page 676 (FPG International / Telegraph Colour Librar) Conic Sections CHAPTER OUTLINE. The Circle. Ellipses and Hperbolas.3 Second-Degree Inequalities and Nonlinear Sstems O ne
More informationFinal Exam Review / AP Calculus AB
Chapter : Final Eam Review / AP Calculus AB Use the graph to find each limit. 1) lim f(), lim f(), and lim π - π + π f 5 4 1 y - - -1 - - -4-5 ) lim f(), - lim f(), and + lim f 8 6 4 y -4 - - -1-1 4 5-4
More information10.2 The Unit Circle: Cosine and Sine
0. The Unit Circle: Cosine and Sine 77 0. The Unit Circle: Cosine and Sine In Section 0.., we introduced circular motion and derived a formula which describes the linear velocit of an object moving on
More informationENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 3 Solutions [Multiple Integration; Lines of Force]
ENGI 44 Advanced Calculus for Engineering Facult of Engineering and Applied Science Problem Set Solutions [Multiple Integration; Lines of Force]. Evaluate D da over the triangular region D that is bounded
More informationCHAPTER 2: Partial Derivatives. 2.2 Increments and Differential
CHAPTER : Partial Derivatives.1 Definition of a Partial Derivative. Increments and Differential.3 Chain Rules.4 Local Etrema.5 Absolute Etrema 1 Chapter : Partial Derivatives.1 Definition of a Partial
More informationMathematics. Mathematics 2. hsn.uk.net. Higher HSN22000
hsn.uk.net Higher Mathematics UNIT Mathematics HSN000 This document was produced speciall for the HSN.uk.net website, and we require that an copies or derivative works attribute the work to Higher Still
More information1. Find A and B so that f x Axe Bx. has a local minimum of 6 when. x 2.
. Find A and B so that f Ae B has a local minimum of 6 when.. The graph below is the graph of f, the derivative of f; The domain of the derivative is 5 6. Note there is a cusp when =, a horizontal tangent
More informationSections 8.1 & 8.2 Areas and Volumes
Sections 8.1 & 8.2 Areas and Volumes Goal. Find the area between the functions y = f(x) and y = g(x) on the interval a x b. y=f(x) y=g(x) x = a x x x b=x 0 1 2 3 n y=f(x) y=g(x) a b Example 1. Find the
More information( afa, ( )) [ 12, ]. Math 226 Notes Section 7.4 ARC LENGTH AND SURFACES OF REVOLUTION
Math 6 Notes Section 7.4 ARC LENGTH AND SURFACES OF REVOLUTION A curve is rectifiable if it has a finite arc length. It is sufficient that f be continuous on [ab, ] in order for f to be rectifiable between
More informationChapter 8: Radical Functions
Chapter 8: Radical Functions Chapter 8 Overview: Types and Traits of Radical Functions Vocabulary:. Radical (Irrational) Function an epression whose general equation contains a root of a variable and possibly
More informationHOMEWORK SOLUTIONS MATH 1910 Sections 6.1, 6.2, 6.3 Fall 2016
HOMEWORK SOLUTIONS MATH 191 Sections.1,.,. Fall 1 Problem.1.19 Find the area of the shaded region. SOLUTION. The equation of the line passing through ( π, is given by y 1() = π, and the equation of the
More informationAnalytic Geometry in Three Dimensions
Analtic Geometr in Three Dimensions. The Three-Dimensional Coordinate Sstem. Vectors in Space. The Cross Product of Two Vectors. Lines and Planes in Space The three-dimensional coordinate sstem is used
More informationHigher. Integration 89
hsn.uk.net Higher Mathematics UNIT UTCME Integration Contents Integration 89 Indefinite Integrals 89 Preparing to Integrate 9 Differential Equations 9 Definite Integrals 9 Geometric Interpretation of Integration
More informationVectors and the Geometry of Space
Chapter 12 Vectors and the Geometr of Space Comments. What does multivariable mean in the name Multivariable Calculus? It means we stud functions that involve more than one variable in either the input
More information