5.2 Volumes: Disks and Washers


 Jason Riley
 2 years ago
 Views:
Transcription
1 4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of crosssection or slice. In this section, we restrict our ttention to specil cse in which the solid is generted by rotting region in the yplne bout horizontl or verticl line. We cll solid formed in this wy solid of revolution nd we cll the line n is of rottion. If the is of rottion coincides with boundry of the region (s in the mrgin figure) then the crosssections of the region perpendiculr to the is of rottion will be disks, mking it reltively esy to find formul for the re of crosssection: A() = re of disk (rdius) The rdius is often function of, the loction of the crosssection. Emple. Find the volume of the solid (shown in the mrgin) formed by rotting the region in the first qudrnt bounded by the curve y = nd the line = 4 bout the is. Solution. Any slice perpendiculr to the is (nd to the yplne) will yield circulr crosssection with rdius equl to the distnce between the curve y = nd the is, so the volume of the region is given by: π d = π 4 d 8 4 = π or bout 6.8 cubic inches. Sometimes the boundry curve intersects the is of rottion. Emple. The region between the grph of f () = nd the horizontl line y = for is revolved bout the horizontl line y = to form solid (see mrgin). Compute the volume of the solid. Solution. The mrgin figure shows crosssections for severl vlues of, ll of them disks. If, then the rdius of the disk is r() = ; if, then r() =. We could split up the volume computtion into two seprte integrls, using A() r() = π for nd A() r() for, but: π ( )
2 5. volumes: disks nd wshers 4 for ll so we cn insted compute the volume with single integrl: π d 4 + d = 46π 5 or bout Prctice. Find the volume of the solid formed by revolving the region between f () = 3 nd the horizontl line y = bout the line y = for 3 (see mrgin). Volumes of Revolved Regions ( Disk Method ) If then the region constrined by the grph of y = f (), the horizontl line y = L nd the intervl, b is revolved bout the horizontl line y = L then the volume of the resulting solid is: A() d = π (rdius) d = π f () L d We often refer to this technique s the disk method becuse revolving thin rectngulr slice of the region (tht we might use in Riemnn sum to pproimte the re of the region) results in disk. If the region between the grph of f nd the is (L = ) is revolved bout the is, then the previous formul reduces to: π f () d Emple 3. Find the volume generted when the region between one rch of the sine curve (for π) nd () the is is revolved bout the is nd (b) the line y = is revolved bout the line y =. Solution. () The rdius of ech circulr slice (see mrgin) is just the height of the function y = sin(): π sin() d 4 sin() π sin () d cos() d π π 4.93
3 4 pplictions of definite integrls (b) Here we use the more generl disk formul with L = : V sin() d sin () sin() + d 4 cos() sin() + d π sin() + cos() 4 3π 4 π + = 3π 4 π or pproimtely.. Prctice. Find the volume generted when () the region between the prbol y = (for ) nd the is is revolved bout the is nd (b) the region between the prbol y = (for ) nd the line y = is revolved bout the line y =. Emple 4. Given tht f () d = 4 nd f () d = 7, represent the volume of ech solid shown in the mrgin s definite integrl, nd evlute those integrls. Solution. () Here the is of rottion is y = so: π (rdius) d = (b) Here the is of rottion is y = so: π (rdius) d = π f () d f () d = 7π f () + d π f () ( ) d ( f ()) d + f () d (5 ) = 9π ( f ()) + f () + d d (c) This is not solid of revolution, even though the crosssections re disks. Ech disk hs dimeter equl to the function height, so the rdius of ech disk is hlf tht height, nd the volume is: The lst one is left for you. f () π d f () d = 7π 4 Prctice 3. Set up nd evlute n integrl to compute the volume of the lst solid shown in the mrgin.
4 5. volumes: disks nd wshers 43 Solids with Holes Some solids hve holes : for emple, we might drill cylindricl hole through sphericl solid (such s bll bering) to crete prt for n engine. One pproch involves using n integrl (or using geometry) to compute the volume of the outer solid, then use nother integrl (or geometry) to compute the volume of the hole cut out of the originl solid, nd finlly subtrcting the second result from the first. You should be ble to use this pproch in the net problem. Prctice 4. Compute the volume of the solid shown in the mrgin. A specil cse of solid with hole results from rotting region bounded by two curves round n is tht does not intersect the region. Emple 5. Compute the volume of the solid shown in the mrgin. Solution. The fce for slice mde t hs re: A() = re of BIG circle re of smll circle BIG rdius π smll rdius Here the BIG rdius is the distnce from the line y = + to the is, or R() = ( + ) = + ; similrly, the smll rdius is the distnce from the curve y = to the is, or r() = =, hence the crosssectionl re is: A() + π + + The curves intersect where: = + = + = = ± 4( ) = ± 5 Clerly we need > for this region, so the left endpoint of integrtion must be = + 5 while the right endpoint is =, so the volume of the solid is: V + + d which simplifies to π ( π ) 3 ( )
5 44 pplictions of definite integrls The previous Emple etends the disk method to more generl technique often clled the wsher method becuse big disk with smller disk cut out of the middle resembles wsher ( smll flt ring used with nuts nd bolts). Volumes of Revolved Regions ( Wsher Method ) If then the region constrined by the grphs of y = f () nd y = g() nd the intervl, b is revolved bout horizontl line the volume of the resulting solid is: π (R()) π (r()) d where R() represents the distnce from the is of rottion to the frthest curve from tht is, nd r() represents the distnce from the is to the closest curve. If r() =, the wsher method becomes the disk method. When pplying the wsher method, you should: grph the region drw representtive rectngulr slice of tht region check tht revolving the slice bout the is of rottion results in wsher locte the limits of integrtion set up n integrl evlute the integrl If you re unble to find n ntiderivtive for the integrnd of your integrl, you cn consult n integrl tble or use numericl methods to pproimte the volume of the solid. You might lso need to use numericl methods to locte where the boundry curves of the region intersect. Emple 6. Find the volume of the solid generted by rotting the region between the curves y = nd y = bout the () is (b) yis (c) the line = (d) the line y = 5. Solution. () The curves intersect where = = ( ) =, so the limits of integrtion should involve = nd =. Revolving verticl slice of the region with width bout the is yields wsher with big rdius R() = = (the line y = is frthest from the is) nd smll rdius r() =
6 5. volumes: disks nd wshers 45 = (the prbol is closest to the is when ). So the volume of the solid is: V () π( ) d 4 4 d π = 64π 5 (b) A verticl slice revolved round the yis does not result in wsher so insted we try slicing horizontlly. A horizontl slice of thickness y revolved round the yis does result in wsher. The big rdius is the distnce from the prbol (where = y) to the yis (where = ) so R(y) = y. Similrly, the smll rdius is the distnce from the line (where = y ) to the yis (where = ), so r(y) = y. Becuse the vrible of integrtion is now y, we need yvlues for the limits of integrtion. At the lower intersection point of the two curves, = y = ; t the upper intersection point, = y = = = 4. So the volume of the solid is: y=4 V ( ( y) y ) π dy y 4 y dy y= y y = 8π 3 3 (c) This solid resembles the one from prt (b), ecept now the rdii re both bigger becuse the region (nd the curves tht form the boundry of the region) re frther wy from the is of rottion: R() = y ( ) = y + nd r() = y ( ) = y + : y=4 V ( ( y + ) y ) π + dy y= (y + ( y + ) y 4 y dy ) 4 y + y + dy 4 3 y 3 4 y = 6π 3 3 (d) For this solid, slicing the region verticlly s in prt () results in wshers, but here the ner nd fr roles of the curves re reversed: the prbol is frthest wy from y = 5 while the line is closest. The rdii re R() = 5 nd r() = 5 : = V (5 ) (5 ) d = 36π 5 = The detils of evluting this definite integrl re left to you. Prctice 5. Find the volume of the solid generted by rotting the region between the curves y = nd y = bout the () the line = 5 (b) the line y = 5.
7 46 pplictions of definite integrls 5. Problems In Problems, find the volume of the solid generted when the region in the first qudrnt bounded by the given curves is rotted bout the is.. y =, = 5. y = sin(), 3. y = cos(), 3 4. y = 3 5. y = 7 6. y = y = 5 8. = y 9 3. Use clculus to determine the volume of sphere of rdius r. (Revolve the region bounded by the is nd the top hlf of the circle + y = r bout the is.) 33. Compute the volume swept out when the top hlf of the ellipticl region bounded by 5 + y 3 = is revolved round the is (see figure below). 9. = y. + y = y = y = 5 In Problems 3 3, compute the volume of the solid formed when the region between the given curves is rotted bout the specified is. 3. y =, y = 4 bout the is 4. y =, y = 4 bout the yis 5. y =, y = 4 bout the yis 6. y =, y = 4 bout the is 7. y =, y = 3 bout the is 8. y = sec(), y = cos(), 3 bout the is 9. y = sec(), y = cos(), 3 bout the is. y =, y = 4 bout y = 3. y =, y = 4 bout y = 4. y =, y = 4 bout = 4 3. y =, y = 4 bout = 3 4. y =, y = 4 bout = 5. y = sin(), y =, = bout y = 3 6. y = sin(), y =, bout y = 7. y =, y = 3, bout = 8. y =, y = 3, bout = 4 9. y =, y = 3, bout y = 3. y =, y = 3, bout y = 3 3. Use clculus to compute the volume of sphere of rdius. (A sphere is formed when the region bounded by the is nd the top hlf of the circle + y = is revolved bout the is.) 34. Compute the volume swept out when the top hlf of the ellipticl region bounded by + y b = is revolved round the is. 35. Compute the volume of the region shown below. 36. Compute the volume of sphere of rdius 5 with hole of rdius 3 drilled through its center.
8 5. volumes: disks nd wshers Compute the volume of the region shown in the mrgin. 38. Determine the volume of the doughnut (clled torus, see lower mrgin figure) generted by rotting disk of rdius r with center R units wy from the is bout the is. 39. () Find the re between f () = nd the is for, nd M. Wht is the limit of the re for M when M? (b) Find the volume swept out when the region in prt () is revolved bout the is for, nd M. Wht is the limit of the volume for M when M? 5. Prctice Answers. 3 3 π (3 ) ) d ( ) d. () Slicing the region verticlly nd rotting the slice bout the is results in disks, so the volume of the solid is: 3. π d 3 4 d 5 5 = 3π 5 (b) Here the slices etend from y = to y = so the rdius of ech disk is nd the volume is: π d + d = 3π d = 56π 5 π 3 f () d 9 6 f () + ( f ()) d 5 9 d 6 f () d + f () d = 9π 4. The volume we wnt cn be obtined by subtrcting the volume of the bo from the volume of the truncted cone generted by the rotted line segment. The volume of the truncted cone is: π + d d = 56π 3 while the volume of the bo is () = 4 so the volume of the solid shown in the grph is 56π
9 48 pplictions of definite integrls 5. () Slicing the region verticlly nd rotting the slice bout the line = 5 results in something other thn wsher, so we insted slice the region horizontlly. The slice etends from = y (frthest from the is of rottion) to = y (closest), so the volume of the solid is: ( π 5 y ) π (5 y) dy = 3π 3 (b) Slicing the region verticlly nd rotting the slice bout the line y = 5 results in wshers, so the volume is: ( π 5 ) π (5 ) d = 36π 5
5 Applications of Definite Integrals
5 Applictions of Definite Integrls The previous chpter introduced the concepts of definite integrl s n re nd s limit of Riemnn sums, demonstrted some of the properties of integrls, introduced some methods
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationAPPLICATIONS OF DEFINITE INTEGRALS
Chpter 6 APPICATIONS OF DEFINITE INTEGRAS OVERVIEW In Chpter 5 we discovered the connection etween Riemnn sums ssocited with prtition P of the finite closed intervl [, ] nd the process of integrtion. We
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationNot for reproduction
AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationThe problems that follow illustrate the methods covered in class. They are typical of the types of problems that will be on the tests.
ADVANCED CALCULUS PRACTICE PROBLEMS JAMES KEESLING The problems tht follow illustrte the methods covered in clss. They re typicl of the types of problems tht will be on the tests. 1. Riemnn Integrtion
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More information7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?
7.1 Integrl s Net Chnge Clculus 7.1 INTEGRAL AS NET CHANGE Distnce versus Displcement We hve lredy seen how the position of n oject cn e found y finding the integrl of the velocity function. The chnge
More informationIndefinite Integral. Chapter Integration  reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationMath 0230 Calculus 2 Lectures
Mth Clculus Lectures Chpter 7 Applictions of Integrtion Numertion of sections corresponds to the text Jmes Stewrt, Essentil Clculus, Erly Trnscendentls, Second edition. Section 7. Ares Between Curves Two
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors PreChpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationMath 113 Exam 1Review
Mth 113 Exm 1Review September 26, 2016 Exm 1 covers 6.17.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More informationMultiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution
Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge
More informationMath 131. Numerical Integration Larson Section 4.6
Mth. Numericl Integrtion Lrson Section. This section looks t couple of methods for pproimting definite integrls numericlly. The gol is to get good pproimtion of the definite integrl in problems where n
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationx = a To determine the volume of the solid, we use a definite integral to sum the volumes of the slices as we let!x " 0 :
Clculus II MAT 146 Integrtion Applictions: Volumes of 3D Solids Our gol is to determine volumes of vrious shpes. Some of the shpes re the result of rotting curve out n xis nd other shpes re simply given
More information5.2 Exponent Properties Involving Quotients
5. Eponent Properties Involving Quotients Lerning Objectives Use the quotient of powers property. Use the power of quotient property. Simplify epressions involving quotient properties of eponents. Use
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationCh AP Problems
Ch. 7.7. AP Prolems. Willy nd his friends decided to rce ech other one fternoon. Willy volunteered to rce first. His position is descried y the function f(t). Joe, his friend from school, rced ginst him,
More informationSpace Curves. Recall the parametric equations of a curve in xyplane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xyplne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More informationMTH 122 Fall 2008 Essex County College Division of Mathematics Handout Version 10 1 October 14, 2008
MTH 22 Fll 28 Essex County College Division of Mthemtics Hndout Version October 4, 28 Arc Length Everyone should be fmilir with the distnce formul tht ws introduced in elementry lgebr. It is bsic formul
More informationragsdale (zdr82) HW2 ditmire (58335) 1
rgsdle (zdr82) HW2 ditmire (58335) This printout should hve 22 questions. Multiplechoice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc
More informationTest , 8.2, 8.4 (density only), 8.5 (work only), 9.1, 9.2 and 9.3 related test 1 material and material from prior classes
Test 2 8., 8.2, 8.4 (density only), 8.5 (work only), 9., 9.2 nd 9.3 relted test mteril nd mteril from prior clsses Locl to Globl Perspectives Anlyze smll pieces to understnd the big picture. Exmples: numericl
More informationChapter 7: Applications of Integrals
Chpter 7: Applictions of Integrls 78 Chpter 7 Overview: Applictions of Integrls Clculus, like most mthemticl fields, egn with tring to solve everd prolems. The theor nd opertions were formlized lter. As
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More informationThammasat University Department of Common and Graduate Studies
Sirindhorn Interntionl Institute of Technology Thmmst University Deprtment of Common nd Grdute Studies Semester: 3/2008 Instructors: Dr. Prpun Suksompong MAS 6: Lecture Notes 7 6 Applictions of the Definite
More informationAPPM 1360 Exam 2 Spring 2016
APPM 6 Em Spring 6. 8 pts, 7 pts ech For ech of the following prts, let f + nd g 4. For prts, b, nd c, set up, but do not evlute, the integrl needed to find the requested informtion. The volume of the
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More information50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS
68 CHAPTE MULTIPLE INTEGALS 46. e da, 49. Evlute tn 3 4 da, where,. [Hint: Eploit the fct tht is the disk with center the origin nd rdius is smmetric with respect to both es.] 5. Use smmetr to evlute 3
More informationCalculus  Activity 1 Rate of change of a function at a point.
Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus
More information[ ( ) ( )] Section 6.1 Area of Regions between two Curves. Goals: 1. To find the area between two curves
Gols: 1. To find the re etween two curves Section 6.1 Are of Regions etween two Curves I. Are of Region Between Two Curves A. Grphicl Represention = _ B. Integrl Represention [ ( ) ( )] f x g x dx = C.
More information4.5 THE FUNDAMENTAL THEOREM OF CALCULUS
4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere
More informationP 1 (x 1, y 1 ) is given by,.
MA00 Clculus nd Bsic Liner Alger I Chpter Coordinte Geometr nd Conic Sections Review In the rectngulr/crtesin coordintes sstem, we descrie the loction of points using coordintes. P (, ) P(, ) O The distnce
More informationThe base of each cylinder is called a crosssection.
6. Volume y Slicing Gol: To find the volume of olid uing econd emeter clculu Volume y CroSection Volume y Dik Volume y Wher Volume y Slicing Volume y Shell 6. Volume y Slicing 6. Volume y Slicing Gol:
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationMath 113 Exam 2 Practice
Mth 3 Exm Prctice Februry 8, 03 Exm will cover 7.4, 7.5, 7.7, 7.8, 8.3 nd 8.5. Plese note tht integrtion skills lerned in erlier sections will still be needed for the mteril in 7.5, 7.8 nd chpter 8. This
More informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
More informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More information14.4. Lengths of curves and surfaces of revolution. Introduction. Prerequisites. Learning Outcomes
Lengths of curves nd surfces of revolution 4.4 Introduction Integrtion cn be used to find the length of curve nd the re of the surfce generted when curve is rotted round n xis. In this section we stte
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationl 2 p2 n 4n 2, the total surface area of the
Week 6 Lectures Sections 7.5, 7.6 Section 7.5: Surfce re of Revolution Surfce re of Cone: Let C be circle of rdius r. Let P n be n nsided regulr polygon of perimeter p n with vertices on C. Form cone
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationChapter 8.2: The Integral
Chpter 8.: The Integrl You cn think of Clculus s doulewide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in
More informationNUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.
NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationRiemann Integrals and the Fundamental Theorem of Calculus
Riemnn Integrls nd the Fundmentl Theorem of Clculus Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University September 16, 2013 Outline Grphing Riemnn Sums
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationSECTION 94 Translation of Axes
94 Trnsltion of Aes 639 Rdiotelescope For the receiving ntenn shown in the figure, the common focus F is locted 120 feet bove the verte of the prbol, nd focus F (for the hperbol) is 20 feet bove the verte.
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING
6 Courtes NASA APPLICATIONS OF THE DEFINITE INTEGRAL IN GEOMETRY, SCIENCE, AND ENGINEERING Clculus is essentil for the computtions required to lnd n stronut on the Moon. In the lst chpter we introduced
More informationcos 3 (x) sin(x) dx 3y + 4 dy Math 1206 Calculus Sec. 5.6: Substitution and Area Between Curves
Mth 126 Clculus Sec. 5.6: Substitution nd Are Between Curves I. USubstitution for Definite Integrls A. Th m 6Substitution in Definite Integrls: If g (x) is continuous on [,b] nd f is continuous on the
More informationProf. Anchordoqui. Problems set # 4 Physics 169 March 3, 2015
Prof. Anchordoui Problems set # 4 Physics 169 Mrch 3, 15 1. (i) Eight eul chrges re locted t corners of cube of side s, s shown in Fig. 1. Find electric potentil t one corner, tking zero potentil to be
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationReversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b
Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite
More information2 b. , a. area is S= 2π xds. Again, understand where these formulas came from (pages ).
AP Clculus BC Review Chpter 8 Prt nd Chpter 9 Things to Know nd Be Ale to Do Know everything from the first prt of Chpter 8 Given n integrnd figure out how to ntidifferentite it using ny of the following
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2019
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS MATH00030 SEMESTER 208/209 DR. ANTHONY BROWN 7.. Introduction to Integrtion. 7. Integrl Clculus As ws the cse with the chpter on differentil
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationNotes on Calculus II Integral Calculus. Miguel A. Lerma
Notes on Clculus II Integrl Clculus Miguel A. Lerm November 22, 22 Contents Introduction 5 Chpter. Integrls 6.. Ares nd Distnces. The Definite Integrl 6.2. The Evlution Theorem.3. The Fundmentl Theorem
More informationI. Equations of a Circle a. At the origin center= r= b. Standard from: center= r=
11.: Circle & Ellipse I cn Write the eqution of circle given specific informtion Grph circle in coordinte plne. Grph n ellipse nd determine ll criticl informtion. Write the eqution of n ellipse from rel
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationMathematics Extension 2
S Y D N E Y B O Y S H I G H S C H O O L M O O R E P A R K, S U R R Y H I L L S 005 HIGHER SCHOOL CERTIFICATE TRIAL PAPER Mthemtics Extension Generl Instructions Totl Mrks 0 Reding Time 5 Minutes Attempt
More informationApplications of Definite Integral
Chpter 5 Applitions of Definite Integrl 5.1 Are Between Two Curves In this setion we use integrls to find res of regions tht lie between the grphs of two funtions. Consider the region tht lies between
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationReview Problem for Midterm #1
Review Problem for Midterm # Midterm I:  :5.m Fridy (Sep. ), Topics: 5.8 nd 6.6. Office Hours before the midterm I: W  pm (Sep 8), Th 5 pm (Sep 9) t UH 8B Solution to quiz cn be found t http://mth.utoledo.edu/
More informationImproper Integrals. Introduction. Type 1: Improper Integrals on Infinite Intervals. When we defined the definite integral.
Improper Integrls Introduction When we defined the definite integrl f d we ssumed tht f ws continuous on [, ] where [, ] ws finite, closed intervl There re t lest two wys this definition cn fil to e stisfied:
More informationObjectives. Materials
Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the
More informationThe practical version
Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht
More informationCHAPTER 4 MULTIPLE INTEGRALS
CHAPTE 4 MULTIPLE INTEGAL The objects of this chpter re fivefold. They re: (1 Discuss when sclrvlued functions f cn be integrted over closed rectngulr boxes in n ; simply put, f is integrble over iff
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More information