Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution
|
|
- Gwenda Casey
- 5 years ago
- Views:
Transcription
1 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 Plnr Are In the limit s d the totl number of pnels A = b d = b f()d Volume of Solid of evolution ) Disk Method : rotte = f() bout the is to form solid. =f() b rotte round -is The disk hs volume of V = π. The totl volume between nd b cn be determined s: V = b π d Note: The vlue of = f() is substituted into the formultion for re nd the resulting eqution is integrted between nd b. 1
2 b) Shell Method: Find ring defined with ring re: π. The volume of the ring is given b V = (π ) The volume of the solid is determined b solving the integrl =f() V = 0 π d Either method cn be used, which ever is most convenient. b rotte round -is Surfce Are of Solid of evolution the rc length cn be defined using Eq. 7.15: rc length s =f() b rotte round -is s = = ( ) d 1 + d d + d d rotte bout the is, where the surfce re is defined s A surf = π s the totl surfce re is given s A surfce = b ( ) d π 1 + d d Emple: 3.1 Find the re in the positive qudrnt bounded b = 1 nd = 3 4
3 Emple: 3. Find the volume of cone with bse rdius nd height h, rotted bout the is using the disk method. h Emple: 3.3 Find the volume of cone with bse rdius nd height h, rotted bout the is using the shell method. ring Emple: 3.4 Find the surfce re of cone with bse rdius nd height h, rotted bout the is. =- h s s A s 3
4 Numericl Integrtion eding Trim 8.7 Numericl Integrtion Assignment web pge To find: I = b f() d First, subdivide the intervl b into n pnels or strips of finite width, where or h = (b )/n Trpeoidl ule The objective of the trpeoidl rule is to find the re under the curve using geometric pnels tht pproimte the re. We know tht ect vlue of the integrl is I ect = b f() d tht cn lso be equted to the pproimte re determined b the trpeoidl rule plus n error term I ect = I trp + ɛ + I trp + C h + D h 4 + E h }{{} error terms (unknown) I trp = h [ ] n 1 f() + f(b) + f i i=1 where the pnel width is h. The lrgest term in the error is ɛ h indicting tht the trpeoidl method is nd order method, i.e. if we hlf the pnel sie we should see decrese in the error of fctor of 4. 4
5 Simpson s ule Insted of using simple liner pproimtion of the curve, s in the Trpeoidl rule, Simpson s rule uses prbolic representtion of the curve. The re under the curve is now given b Are = i i f()d (C 1 + C + C 3 )d i 1 i 1 Prbols re used for ech pnel pir, therefore we require n, even number of pnels with (n + 1) odd number of points. I simp = h 3 f() + f(b) + 4 n 1 i=1,,3,... f i + n f i i=,4,6,... nd the error term is given b ɛ = C 1 h 4 + D 1 h 6 + E 1 h 8 + The lrgest error term is ɛ h 4 indicting 4th order method. Emple: 3.5 Find the vlue for I = 0 4 d using Trpeoidl rule nd Simpson s rule 5
6 omberg Integrtion omberg integrtion is bsed on procedure tht elimintes the error term from the trpeoidl rule clcultions. In essence it provides procedure for getting n ccurte nswer without the need for ver smll h. It is efficient nd minimies the potentil for round off errors. The ect vlue of the integrl cn be written s I ect = I trp + C h + D h 4 + E h }{{} error terms (unknown) (1) If we rewrite the trpeoidl rule with pnel width of h/ I trp = h/ f() + f(b) + (n 1) i= f( i ) The ect vlue cn then be written s ( ) h ( ) h 4 ( ) h 6 I ect = I trp + C + D + E +... }{{} error terms (unknown) () If we multipl Eq. () b 4 nd subtrct Eq. (1), we get 3 I ect = (4I trp I trp1 ) + ( ) h 4 + ( ) h I ect = (4I trp I trp1 ) 3 + ( ) h4 3 + ( ) h The leding term gives better estimte of I ect becuse the error terms re proportiontel smller. Emple: 3.5b Find the vlue for I = 0 4 d using omberg integrtion 6
7 Double Integrls eding Trim 13.1 Double Integrls nd Double Iterted Integrls 13. Evl. of Double Integrls b Double Iterted Integrls 13.7 Double Iterted Integrls in Polr Coordintes Assignment web pge ssignment #7 Crtesin Coordintes Find the re in the + ve qudrnt bounded b = 1 4 nd = 3. The bsic re element in D is 1/8 region - = 3 = 1/4 A t rbitrr (, ) A = We cn build this re up into strip b summing over, keeping fied. 1/ A strip = 1/4 = 3 fied Sum up ll strips to get the totl re A = 1/ =0 1/4 = 3 In the limit s d nd d we get double integrl s follows A = 1/ =0 ( ) 1/4 d d = 3 7
8 Polr Coordintes In Crtesin coordintes our re element ws A =, which in differentil form gve us A = dd A = r r r We cn chnge the principl coordintes into polr coordintes b trnsforming nd into r nd θ. r = + r r θ = tn 1 (/) The Polr coordinte re element becomes when integrted becomes A = r r θ A = rdrdθ Emple: 3.6 Find the re in the + ve qudrnt bounded b circles 1 re A to be found = 1 ( 1) + = 1 8
9 Surfce Ares from Double Integrls eding Trim 13.3 Ares nd Volumes of Solids of evolution 13.6 Surfce Are Assignment web pge ssignment #7 surfce re element S surfce = f(,) S ^ n projection in (, ) plne re element A = in ^ k A How is S relted to A? Imgine shining light verticll down through S to get A. 1. the surfce is defined s = f(, ). redefine s F = f(, ) where the surfce is given s F = 0 - F > 0 nd F < 0 will the regions bove nd below the surfce, respectivel 3. the grdient of the function F is given s F = ( f ), f, 1 F is the perpendiculr to the surfce nd the perpendiculr to the tngent plnes n = F 9
10 4. get the unit norml vector s follows ˆn = F F = î f ĵ f + ˆk ( ) f ( ) f find the component of the S surfce projected onto ˆk from Trim 1.5 we know tht A = cos θ S Note, when θ = 0 A = S (this is the surfce prllel to the plne. In generl, k n k n A S A = cos }{{ θ} S ˆn ˆk ˆn ˆk = ˆn ˆk cos θ = cos θ A = S(ˆn ˆk) = S 1 F since ˆn ˆk produces numertor of ˆk ˆk = 1 nd denomintor of F errnging the bove eqution, we cn solve for S. In the limit ( ) f ( ) f ds = }{{} F dd }{{} da Given the surfce = f(, ), the surfce re is S = ( ) f ( ) f dd where is the projection of the f(, ) surfce down onto the (, ) plne. While this is the most common form of the eqution, we could lso find S b projecting onto nother coordinte plne. Sometimes it is more convenient to do it this w. See Trim 14.6 for pplicble equtions. 10
11 Emple: 3.7 Find the surfce re in the + ve octnt for = f(, ) = 4. 4 plne 4 = -/ Emple: 3.8 Given the sphere, + + =, derive the formul for surfce re. Emple: 3.9 Find the volume formed in the + ve octnt between the coordinte plnes nd the surfce = f(, ) = 4 4 plne 4 = -/ 11
12 Emple: 3.10 Find the men vlue of = f() = sin in the domin = 0 to = π. Emple: 3.10b Find the men vlue of temperture for T = f(, ) = 4. T of plte t (,) = f(,) 4 plte =0, =0, =-/ Emple: 3.10c Derive the formul for the volume of revolution. for the following sphere: + + =. surfce = f(,) =0 =0 region + = 1
13 Triple Integrls eding Trim 13.8 Triple Integrls nd Triple Iterted Integrls 13.9 Volumes Assignment web pge ssignment #8 Volume Clcultions in Crtesin Coordintes The triple integrl cn be identified s V d d d }{{} dv volume element or V f d d d dd up the dv elements in,, directions, i.e. triple sum. Consider the solid defined b + = 4 in the positive octnt. Find the volume of this solid between the coordinte plnes nd the plne + = 6. 6 volume to be found + = 4 6 clinder + = 4 plne + = 6 13
14 Strt with volume element t rbitrr (,, ) in spce inside dv Build up slice - sum columns over, keeping, fied. dv = d d d Build up column - sum over keeping, constnt. =0 sum over d column volume = = 6- ( 6 =0 Evlution of the integrl gives ) d dd sum column over d 4 slice volume = =0 Finll sum the slices over V = ( 6 =0 ) d ddd d d V = = (6 )dd = 4 d ( 4 ) d (4 )d = 6π use tbles if necessr Emple: 3.11 Find the volume of the prboloid, = + for 0 4. Consider onl the + ve octnt, i.e. 1/4 of the volume. 4 = + surfce sum projection onto (,) plne + = 4 14
15 Volume Clcultions in Clindricl nd Sphericl Coordintes eding Trim Triple Iterted Integrls in Clindricl Coordintes 13.1 Triple Iterted Integrls in Sphericl Coordintes Assignment web pge ssignment #8 Clindricl Coordintes point: volume element: P (r, θ, ) i.e. polr in, plne plus dv = r dr dθ d bsed on links to Crtesin coordintes r = + θ = tn 1 (/) = or = r cos θ = r sin θ = where 0 r, nd 0 θ π Tpicll we build up column, wedge slice nd then the totl volume, given s rdrdθd The mth opertions re esier when we hve i-smmetric sstems, i.e. clinders nd cones Sphericl Coordintes r sin f d r sinf r da rdf dr point: P (r, θ, φ) volume element: dv = (r sin φdφ) rdrdθ }{{}}{{} re height bsed on links to Crtesin coordintes dv = da dr = r sin f dr d f d 15
16 r = + + θ = tn 1 (/) ( ) φ = cos or = r sin φ cos θ = r sin φ sin θ = r cos φ where 0 r ; 0 θ π; 0 φ π. Note: for 0 φ π the sinφ is lws + ve for dv + ve. The solution procedure involves building up columns, slices s before to obtin the totl volume, given s r sin φ dr dθ dφ Emple: 3.1 Find the volume bounded b clinder, + = nd prboloid, = + + = = + Sphericl Coordinte Emple Emple: 3.13 Derive formul for the volume of sphere with rdius, + + = 16
17 Moments of Are/ Mss / Volume eding Trim 13.5 Centres of Mss nd Moments of Inerti Centres of Mss nd Moments of Inerti Assignment web pge ssignment #9 Centroids, Centers of Mss etc. -D cse: thin plte of constnt thickness Sometimes, single integrls work, s in -D cse, where the thickness is given s t nd is constnt or function of position s t(, ). The mteril densit is given s ρ (kg/m 3 ), gin constnt or function of position s ρ(, ). We sometimes use the mss per unit re of the plte, ρ = ρ t (kg/m ). re mss bsic element da = d d dm = ρ t d d or ρ d d totl re A = d d M = dm = ρ t d d first moment of re first moment of mss (weight b distnce from is) bout is da = d d dm = ρ t d d totl F = da dm bout is F = da dm centroid coordintes = da A = da A center of mss coordintes c = c = dm M dm M second moments da da dm dm 17
18 3-D cse: We use the sme bsic ides but the bsic element is now V = ddd -D Objects 3-D Objects plte of thickness t region volumev d V t rbitrr (,,) re element da volume is t da mss is (,) t da Quntities of interest in pplictions such s dnmics. Are: A = da (V olume = ta) Mss: M = ρ(, )tda where ρ(, ) = densit of mteril in (kg/m 3 ) t point (, ) Centroid = geometricl center of object = da 1st moment of re bout is A = da 1st moment of re bout is A Center of Mss: useful in dnmics problems c = dm M = ρ(, )tda M c = dm M = ρ(, )tda M Note: tht if the object densit is uniform, then the centroid nd center of mss re the sme. nd Moments of Are nd Mss: Moments of Inerti nd moment of re bout: is I = da nd moment of mss bout: is I = ρ(, )tda (similr formuls for I bout the is) projection onto (,) plne defines mss is (,,) dv Quntities of interest in pplictions such s dnmics. Volume: V = V dv Mss: M = ρ(,, )dv V where ρ(,, ) = densit of mteril in (kg/m 3 ) t point (,, ) Centroid = geometricl center of object V = dv 1st moment of volume bout plne V V = dv 1st moment of volume bout plne V V = dv 1st moment of volume bout plne V Center of Mss: useful in dnmics problems ρ(,, )dv V c = M similr formuls for c nd c nd Moments of Are nd Mss: Polr Moments of Inerti volume moment bout: is J = V ( + )dv mss moment bout: is J = V ( + )ρ(,, )dv (similr formuls for J bout the is) nd (similr formuls for J bout the is) 18
19 Emple: 3.14 Find the centroid, center of mss nd the 1st moment of mss for qurter circle of rdius with n inner circle of rdius / mde of led with densit of ρ 1 = 11, 000 kg/m 3 nd n outer circle of rdius mde luminum with densit of ρ =, 500 kg/m 3. The thickness is uniform throughout t t = 10 mm. + = / ρ 1 = 11 g/cm = 110 kg/m 1 ρ =.5 g/cm = 5 kg/m / Emple: 3.15 Find the re of the prboloid = + below the plne = 1 Emple: 3.16 Find the moment of inerti bout the is of the re enclosed b the crdioid r = (1 cos θ) Emple: 3.17 Find the center of grvit of homogeneous solid hemisphere of rdius 19
Multiple 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: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge
More information(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence
Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll
More informationSolutions to Problems Integration in IR 2 and IR 3
Solutions to Problems Integrtion in I nd I. For ec of te following, evlute te given double integrl witout using itertion. Insted, interpret te integrl s, for emple, n re or n verge vlue. ) dd were is te
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 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 informationIn Mathematics for Construction, we learnt that
III DOUBLE INTEGATION THE ANTIDEIVATIVE OF FUNCTIONS OF VAIABLES In Mthemtics or Construction, we lernt tht the indeinite integrl is the ntiderivtive o ( d ( Double Integrtion Pge Hence d d ( d ( The ntiderivtive
More information5.2 Volumes: Disks and Washers
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 cross-section or slice. In this section, we restrict
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 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 informationReference. Vector Analysis Chapter 2
Reference Vector nlsis Chpter Sttic Electric Fields (3 Weeks) Chpter 3.3 Coulomb s Lw Chpter 3.4 Guss s Lw nd pplictions Chpter 3.5 Electric Potentil Chpter 3.6 Mteril Medi in Sttic Electric Field Chpter
More informationJUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 8 (First moments of a volume) A.J.Hobson
JUST THE MATHS UNIT NUMBER 3.8 INTEGRATIN APPLICATINS 8 (First moments of volume) b A.J.Hobson 3.8. Introduction 3.8. First moment of volume of revolution bout plne through the origin, perpendiculr to
More information(6.5) Length and area in polar coordinates
86 Chpter 6 SLICING TECHNIQUES FURTHER APPLICATIONS Totl mss 6 x ρ(x)dx + x 6 x dx + 9 kg dx + 6 x dx oment bout origin 6 xρ(x)dx x x dx + x + x + ln x ( ) + ln 6 kg m x dx + 6 6 x x dx Centre of mss +
More informationChapter 1 VECTOR ALGEBRA
Chpter 1 VECTOR LGEBR INTRODUCTION: Electromgnetics (EM) m be regrded s the stud of the interctions between electric chrges t rest nd in motion. Electromgnetics is brnch of phsics or electricl engineering
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 informationTotal Score Maximum
Lst Nme: Mth 8: Honours Clculus II Dr. J. Bowmn 9: : April 5, 7 Finl Em First Nme: Student ID: Question 4 5 6 7 Totl Score Mimum 6 4 8 9 4 No clcultors or formul sheets. Check tht you hve 6 pges.. Find
More informationElectromagnetics P5-1. 1) Physical quantities in EM could be scalar (charge, current, energy) or vector (EM fields).
Electromgnetics 5- Lesson 5 Vector nlsis Introduction ) hsicl quntities in EM could be sclr (chrge current energ) or ector (EM fields) ) Specifing ector in -D spce requires three numbers depending on the
More informationragsdale (zdr82) HW2 ditmire (58335) 1
rgsdle (zdr82) HW2 ditmire (58335) This print-out should hve 22 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. 00 0.0 points A chrge of 8. µc
More informationSome Methods in the Calculus of Variations
CHAPTER 6 Some Methods in the Clculus of Vritions 6-. If we use the vried function ( α, ) α sin( ) + () Then d α cos ( ) () d Thus, the totl length of the pth is d S + d d α cos ( ) + α cos ( ) d Setting
More informationEigen Values and Eigen Vectors of a given matrix
Engineering Mthemtics 0 SUBJECT NAME SUBJECT CODE MATERIAL NAME MATERIAL CODE : Engineering Mthemtics I : 80/MA : Prolem Mteril : JM08AM00 (Scn the ove QR code for the direct downlod of this mteril) Nme
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More informationMATH 253 WORKSHEET 24 MORE INTEGRATION IN POLAR COORDINATES. r dr = = 4 = Here we used: (1) The half-angle formula cos 2 θ = 1 2
MATH 53 WORKSHEET MORE INTEGRATION IN POLAR COORDINATES ) Find the volume of the solid lying bove the xy-plne, below the prboloid x + y nd inside the cylinder x ) + y. ) We found lst time the set of points
More informationPhysics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016
Physics 333, Fll 16 Problem Set 7 due Oct 14, 16 Reding: Griffiths 4.1 through 4.4.1 1. Electric dipole An electric dipole with p = p ẑ is locted t the origin nd is sitting in n otherwise uniform electric
More information10.5. ; 43. The points of intersection of the cardioid r 1 sin and. ; Graph the curve and find its length. CONIC SECTIONS
654 CHAPTER 1 PARAETRIC EQUATIONS AND POLAR COORDINATES ; 43. The points of intersection of the crdioid r 1 sin nd the spirl loop r,, cn t be found ectl. Use grphing device to find the pproimte vlues of
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 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 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 informationMath 113 Exam 1-Review
Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 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 informationFinal Exam Solutions, MAC 3474 Calculus 3 Honors, Fall 2018
Finl xm olutions, MA 3474 lculus 3 Honors, Fll 28. Find the re of the prt of the sddle surfce z xy/ tht lies inside the cylinder x 2 + y 2 2 in the first positive) octnt; is positive constnt. olution:
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 cross-section in plne pssing through
More informationEngg. Math. I (Unit-II)
Dr. Stish Shukl of 7 Engg. Mth. I Unit-II) Integrl Clculus iemnn Integrl) The ide. Suppose, f be continuous function defined on [, b nd we wnt to clculte the re bounded by this function with the -is from
More informationChapter 9. Arc Length and Surface Area
Chpter 9. Arc Length nd Surfce Are In which We ppl integrtion to stud the lengths of curves nd the re of surfces. 9. Arc Length (Tet 547 553) P n P 2 P P 2 n b P i ( i, f( i )) P i ( i, f( i )) distnce
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 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 n-sided regulr polygon of perimeter p n with vertices on C. Form cone
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 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 informationPROPERTIES OF AREAS In general, and for an irregular shape, the definition of the centroid at position ( x, y) is given by
PROPERTES OF RES Centroid The concept of the centroid is prol lred fmilir to ou For plne shpe with n ovious geometric centre, (rectngle, circle) the centroid is t the centre f n re hs n is of smmetr, the
More informationConducting Ellipsoid and Circular Disk
1 Problem Conducting Ellipsoid nd Circulr Disk Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 (September 1, 00) Show tht the surfce chrge density σ on conducting ellipsoid,
More informationHung problem # 3 April 10, 2011 () [4 pts.] The electric field points rdilly inwrd [1 pt.]. Since the chrge distribution is cylindriclly symmetric, we pick cylinder of rdius r for our Gussin surfce S.
More informationExercise Qu. 12. a2 y 2 da = Qu. 18 The domain of integration: from y = x to y = x 1 3 from x = 0 to x = 1. = 1 y4 da.
MAH MAH Eercise. Eercise. Qu. 7 B smmetr ( + 5)dA + + + 5 (re of disk with rdius ) 5. he first two terms of the integrl equl to becuse is odd function in nd is odd function in. (see lso pge 5) Qu. b da
More informationMULTIPLE INTEGRALS. A double integral of a positive function is a volume, which is the limit of sums of volumes of rectangular columns.
5 MULTIPL INTGALS A double integrl of positive function is volume, which is the limit of sums of volumes of rectngulr columns. In this chpter we etend the ide of definite integrl to double nd triple integrls
More informationThomas Whitham Sixth Form
Thoms Whithm Sith Form Pure Mthemtics Unit C Alger Trigonometry Geometry Clculus Vectors Trigonometry Compound ngle formule sin sin cos cos Pge A B sin Acos B cos Asin B A B sin Acos B cos Asin B A B cos
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 informationPartial Differential Equations
Prtil Differentil Equtions Notes by Robert Piché, Tmpere University of Technology reen s Functions. reen s Function for One-Dimensionl Eqution The reen s function provides complete solution to boundry
More informationSpace Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne 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 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 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 information2A1A Vector Algebra and Calculus I
Vector Algebr nd Clculus I (23) 2AA 2AA Vector Algebr nd Clculus I Bugs/queries to sjrob@robots.ox.c.uk Michelms 23. The tetrhedron in the figure hs vertices A, B, C, D t positions, b, c, d, respectively.
More informationLecture 13 - Linking E, ϕ, and ρ
Lecture 13 - Linking E, ϕ, nd ρ A Puzzle... Inner-Surfce Chrge Density A positive point chrge q is locted off-center inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
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 informationGeometric and Mechanical Applications of Integrals
5 Geometric nd Mechnicl Applictions of Integrls 5.1 Computing Are 5.1.1 Using Crtesin Coordintes Suppose curve is given by n eqution y = f(x), x b, where f : [, b] R is continuous function such tht f(x)
More informationStudent Handbook for MATH 3300
Student Hndbook for MATH 3300 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 0.5 0 0.5 0.5 0 0.5 If people do not believe tht mthemtics is simple, it is only becuse they do not relize how complicted life is. John Louis
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 informationHomework Assignment 3 Solution Set
Homework Assignment 3 Solution Set PHYCS 44 6 Ferury, 4 Prolem 1 (Griffiths.5(c The potentil due to ny continuous chrge distriution is the sum of the contriutions from ech infinitesiml chrge in the distriution.
More informationExam 1 Solutions (1) C, D, A, B (2) C, A, D, B (3) C, B, D, A (4) A, C, D, B (5) D, C, A, B
PHY 249, Fll 216 Exm 1 Solutions nswer 1 is correct for ll problems. 1. Two uniformly chrged spheres, nd B, re plced t lrge distnce from ech other, with their centers on the x xis. The chrge on sphere
More informationProblems for HW X. C. Gwinn. November 30, 2009
Problems for HW X C. Gwinn November 30, 2009 These problems will not be grded. 1 HWX Problem 1 Suppose thn n object is composed of liner dielectric mteril, with constnt reltive permittivity ɛ r. The object
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 informationMA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations
LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll
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 informationYear 12 Mathematics Extension 2 HSC Trial Examination 2014
Yer Mthemtics Etension HSC Tril Emintion 04 Generl Instructions. Reding time 5 minutes Working time hours Write using blck or blue pen. Blck pen is preferred. Bord-pproved clcultors my be used A tble of
More information4. F = i + sin xj. (page 811) 1. F = xi + xj. 5. F = e x i + e x j. 2. F = xi + yj. 3. F = yi + xj. 6. F = (x 2 y) = 2xi j.
ETION 5 PAGE 8 R A ADAM: ALULU HAPTER 5 ection 5 pge 8 VETOR FIELD Vector nd clr Fields F i + j The field lines stisf d d, ie, d d The field lines re +, stright lines prllel to 4 F i + sin j The field
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 informationTheoretische Physik 2: Elektrodynamik (Prof. A.-S. Smith) Home assignment 4
WiSe 1 8.1.1 Prof. Dr. A.-S. Smith Dipl.-Phys. Ellen Fischermeier Dipl.-Phys. Mtthis Sb m Lehrstuhl für Theoretische Physik I Deprtment für Physik Friedrich-Alexnder-Universität Erlngen-Nürnberg Theoretische
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 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 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 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 informationJUST THE MATHS SLIDES NUMBER INTEGRATION APPLICATIONS 11 (Second moments of an area (A)) A.J.Hobson
JUST THE MATHS SLIDES NUMBER. INTEGRATIN APPLICATINS (Second moments of n re (A)) b A.J.Hobson.. Introduction..2 The second moment of n re bout the -xis.. The second moment of n re bout the x-xis UNIT.
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must
More information13.3. The Area Bounded by a Curve. Introduction. Prerequisites. Learning Outcomes
The Are Bounded b Curve 3.3 Introduction One of the importnt pplictions of integrtion is to find the re bounded b curve. Often such n re cn hve phsicl significnce like the work done b motor, or the distnce
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 informationCalculus 2: Integration. Differentiation. Integration
Clculus 2: Integrtion The reverse process to differentition is known s integrtion. Differentition f() f () Integrtion As it is the opposite of finding the derivtive, the function obtined b integrtion is
More informationMATH 13 FINAL STUDY GUIDE, WINTER 2012
MATH 13 FINAL TUY GUI, WINTR 2012 This is ment to be quick reference guide for the topics you might wnt to know for the finl. It probbly isn t comprehensive, but should cover most of wht we studied in
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. Limits - L Hopital s Rule. x c. x c. f x. g x. x c 0 6 = 1 6. D. -1 E. nonexistent. ln ( x 1 ) 1 x 2 1. ( x 2 1) 2. 2x x 1.
A. Limits - L Hopitl s Rule Wht you re finding: L Hopitl s Rule is used to find limits of the form f ( ) lim where lim f or lim f limg. c g = c limg( ) = c = c = c How to find it: Try nd find limits by
More information7.6 The Use of Definite Integrals in Physics and Engineering
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 7.6 The Use of Definite Integrls in Physics nd Engineering It hs been shown how clculus cn be pplied to find solutions to geometric problems
More informationCHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6. VOLUMES USING CROSS-SECTIONS. A() ;, ; (digonl) ˆ Ȉ È V A() d d c d 6 (dimeter) c d c d c ˆ 6. A() ;, ; V A() d d. A() (edge) È Š È Š È ;, ; V A() d d 8
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 informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationTime : 3 hours 03 - Mathematics - March 2007 Marks : 100 Pg - 1 S E CT I O N - A
Time : hours 0 - Mthemtics - Mrch 007 Mrks : 100 Pg - 1 Instructions : 1. Answer ll questions.. Write your nswers ccording to the instructions given below with the questions.. Begin ech section on new
More informationA-Level Mathematics Transition Task (compulsory for all maths students and all further maths student)
A-Level Mthemtics Trnsition Tsk (compulsory for ll mths students nd ll further mths student) Due: st Lesson of the yer. Length: - hours work (depending on prior knowledge) This trnsition tsk provides revision
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 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 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 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 information1.2. Linear Variable Coefficient Equations. y + b "! = a y + b " Remark: The case b = 0 and a non-constant can be solved with the same idea as above.
1 12 Liner Vrible Coefficient Equtions Section Objective(s): Review: Constnt Coefficient Equtions Solving Vrible Coefficient Equtions The Integrting Fctor Method The Bernoulli Eqution 121 Review: Constnt
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 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 information12 TRANSFORMING BIVARIATE DENSITY FUNCTIONS
1 TRANSFORMING BIVARIATE DENSITY FUNCTIONS Hving seen how to trnsform the probbility density functions ssocited with single rndom vrible, the next logicl step is to see how to trnsform bivrite probbility
More informationDistributed Forces: Centroids and Centers of Gravity
Distriuted Forces: Centroids nd Centers of Grvit Introduction Center of Grvit of D Bod Centroids nd First Moments of Ares nd Lines Centroids of Common Shpes of Ares Centroids of Common Shpes of Lines Composite
More information200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes
PHYSICS 132 Smple Finl 200 points 5 Problems on 4 Pges nd 20 Multiple Choice/Short Answer Questions on 5 pges 1 hour, 48 minutes Student Nme: Recittion Instructor (circle one): nme1 nme2 nme3 nme4 Write
More informationUniversity of Washington Department of Chemistry Chemistry 453 Winter Quarter 2010 Homework Assignment 4; Due at 5p.m. on 2/01/10
University of Wshington Deprtment of Chemistry Chemistry 45 Winter Qurter Homework Assignment 4; Due t 5p.m. on // We lerned tht the Hmiltonin for the quntized hrmonic oscilltor is ˆ d κ H. You cn obtin
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
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 informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationPlane curvilinear motion is the motion of a particle along a curved path which lies in a single plane.
Plne curiliner motion is the motion of prticle long cured pth which lies in single plne. Before the description of plne curiliner motion in n specific set of coordintes, we will use ector nlsis to describe
More informationCONIC SECTIONS. Chapter 11
CONIC SECTIONS Chpter. Overview.. Sections of cone Let l e fied verticl line nd m e nother line intersecting it t fied point V nd inclined to it t n ngle α (Fig..). Fig.. Suppose we rotte the line m round
More informationElectromagnetism Answers to Problem Set 10 Spring 2006
Electromgnetism 76 Answers to Problem Set 1 Spring 6 1. Jckson Prob. 5.15: Shielded Bifilr Circuit: Two wires crrying oppositely directed currents re surrounded by cylindricl shell of inner rdius, outer
More informationFinal Review, Math 1860 Thomas Calculus Early Transcendentals, 12 ed
Finl Review, Mth 860 Thoms Clculus Erly Trnscendentls, 2 ed 6. Applictions of Integrtion: 5.6 (Review Section 5.6) Are between curves y = f(x) nd y = g(x), x b is f(x) g(x) dx nd similrly for x = f(y)
More informationEquations of Motion. Figure 1.1.1: a differential element under the action of surface and body forces
Equtions of Motion In Prt I, lnce of forces nd moments cting on n component ws enforced in order to ensure tht the component ws in equilirium. Here, llownce is mde for stresses which vr continuousl throughout
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 information