50. Use symmetry to evaluate xx D is the region bounded by the square with vertices 5, Prove Property 11. y y CAS
|
|
- Whitney Reed
- 5 years ago
- Views:
Transcription
1 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 4 da, where is the region bounded b the squre with vertices 5, 47. Prove Propert. nd, In evluting double integrl over region, sum of 5. Compute s da, where is the disk iterted integrls ws obtined s follows:, b first identifing the integrl s the volume of solid. f, da f, d d 3 3 f, d d Sketch the region nd epress the double integrl s n iterted integrl with reversed order of integrtion. CAS 5. Grph the solid bounded b the plne z nd the prboloid z 4 nd find its ect volume. (Use our CAS to do the grphing, to find the equtions of the boundr curves of the region of integrtion, nd to evlute the double integrl.).3 UBLE INTEGALS IN PLA CINATES Polr coordintes were introduced in Section 9.3. Suppose tht we wnt to evlute double integrl f, da, where is one of the regions shown in Figure. In either cse the description of in terms of rectngulr coordintes is rther complicted but is esil described using polr coordintes. + = + =4 + = FIGUE () =s(r, ) r, πd (b) =s(r, ) r, πd P(r, )=P(, ) ecll from Figure tht the polr coordintes r, of point re relted to the rectngulr coordintes, b the equtions r r r cos r sin FIGUE The regions in Figure re specil cses of polr rectngle r, r b, which is shown in Figure 3. In order to compute the double integrl f, da, where is polr rectngle, we divide the intervl, b into m subintervls r i, r i with lengths r i r i r i nd we divide the intervl, into n subintervls j, j with lengths j j j. Then the circles r r i nd the rs j divide the polr rectngle into the smll polr rectngles shown in Figure 4.
2 SECTIN.3 UBLE INTEGALS IN PLA CINATES 683 = j = j_ = r=b ij (r i *, j*) Î j r= å =å r=r i r=r i_ FIGUE 3 Polr rectngle FIGUE 4 ividing into polr subrectngles The center of the polr subrectngle hs polr coordintes ij r, r i r r i, j j r i * r i r i j* j j We compute the re of ij using the fct tht the re of sector of circle with rdius r nd centrl ngle is r. Subtrcting the res of two such sectors, ech of which hs centrl ngle, we find tht the re of is j ij A ij r i j r i j r i r i j r i r i r i r i j r i * r i j Although we hve defined the double integrl f, da in terms of ordinr rectngles, it cn be shown tht, for continuous functions f, we lws obtin the sme nswer using polr rectngles. The rectngulr coordintes of the center of ij re r i * cos j*, r i * sin j*, so tpicl iemnn sum is m n i j f r i * cos j*, r i * sin j* A ij m n i j f r i * cos j*, r i * sin j* r i * r i j If we write written s tr, rfr cos, r sin m n i j, then the iemnn sum in Eqution cn be tr i *, j* r i j which is iemnn sum for the double integrl b tr, dr d
3 684 CHAPTE MULTIPLE INTEGALS Therefore, we hve f, da lim m ri, j l m n i j lim m ri, j l m n i j f r i * cos j*, r i * sin j* A ij b f r cos, r sin rdrd tr i *, j* r i j b tr, dr d CHANGE T PLA CINATES IN A UBLE INTEGAL If f is continuous on polr rectngle given b r b,, where, then f, da b f r cos, r sin rdrd d da dr The formul in () ss tht we convert from rectngulr to polr coordintes in double integrl b writing r cos nd r sin, using the pproprite limits of integrtion for r nd, nd replcing da b rdrd. Be creful not to forget the dditionl fctor r on the right side of Formul. A clssicl method for remembering this is shown in Figure 5, where the infinitesiml polr rectngle cn be thought of s n ordinr rectngle with dimensions rd nd dr nd therefore hs re da rdrd. FIGUE 5 r r d EXAMPLE Evlute 3 4 da, where is the region in the upper hlfplne bounded b the circles nd 4. SLUTIN The region cn be described s,, 4 It is the hlf-ring shown in Figure (b), nd in polr coordintes it is given b r,. Therefore, b Formul, 3 4 da 3r cos 4r sin rdrd 3r cos 4r 3 sin dr d Here we use the trigonometric identit sin cos s discussed in Section 6.. [r 3 cos r 4 sin ] r r d 7 cos 5 sin d [7 cos 5 cos ] d 7 sin sin 5
4 SECTIN.3 UBLE INTEGALS IN PLA CINATES 685 z (,, ) V EXAMPLE Find the volume of the solid bounded b the plne z nd the prboloid z. SLUTIN If we put z in the eqution of the prboloid, we get. This mens tht the plne intersects the prboloid in the circle, so the solid lies under the prboloid nd bove the circulr disk given b [see Figures 6 nd ()]. In polr coordintes is given b r,. Since r, the volume is FIGUE 6 V da d r r 3 dr r r rdrd r 4 4 If we hd used rectngulr coordintes insted of polr coordintes, then we would hve obtined V da s s d d which is not es to evlute becuse it involves finding 3 d. = r=h ( ) Wht we hve done so fr cn be etended to the more complicted tpe of region shown in Figure 7. It s similr to the tpe II rectngulr regions considered in Section.. In fct, b combining Formul in this section with Formul..5, we obtin the following formul. å =å r=h ( ) 3 then If f is continuous on polr region of the form r, f, da FIGUE 7 =s(r, ) å, h ( ) r h ( )d In prticulr, tking f,, h, nd h h in this formul, we see tht the re of the region bounded b,, nd r h is A, h r h h f r cos, r sin rdrd h da h r h rdrd d h d nd this grees with Formul
5 686 CHAPTE MULTIPLE INTEGALS FIGUE 8 (-)@+ = (or r= cos ) z V EXAMPLE 3 Find the volume of the solid tht lies under the prboloid z, bove the -plne, nd inside the clinder. SLUTIN The solid lies bove the disk whose boundr circle hs eqution or, fter completing the squre, (See Figures 8 nd 9.) In polr coordintes we hve r nd r cos, so the boundr circle becomes r r cos, or r cos. Thus the disk is given b r,, r cos nd, b Formul 3, we hve V da 4 cos4 d 8 cos r rdrd cos 4 d 8 r 4 cos 4 cos d d FIGUE 9 cos cos 4 d 3 [ 3 sin 8 sin 4] 3.3 EXECISES 4 A region is shown. ecide whether to use polr coordintes or rectngulr coordintes nd write f, da s n iterted integrl, where f is n rbitrr continuous function on Sketch the region whose re is given b the integrl nd evlute the integrl rdrd 6. 4 cos rdrd 7 Evlute the given integrl b chnging to polr coordintes. 7. da, where is the disk with center the origin nd rdius 3 8. da, where is the region tht lies to the left of the -is between the circles nd 4 9. cos da, where is the region tht lies bove the -is within the circle 9. s4 da, where, 4,. rctn da, where, 4,. e da, where is the region in the first qudrnt enclosed b the circle 5
6 SECTIN.3 UBLE INTEGALS IN PLA CINATES Use polr coordintes to find the volume of the given solid. 3. Under the cone z s nd bove the disk 4 4. Below the prboloid z 8 nd bove the -plne 5. A sphere of rdius 6. Inside the sphere z 6 nd outside the clinder 4 7. Above the cone z s nd below the sphere z 8. Bounded b the prboloid z nd the plne z 7 in the first octnt 9. Inside both the clinder 4 nd the ellipsoid 4 4 z 64. () A clindricl drill with rdius r is used to bore hole through the center of sphere of rdius r. Find the volume of the ring-shped solid tht remins. (b) Epress the volume in prt () in terms of the height h of the ring. Notice tht the volume depends onl on h, not on or. Use double integrl to find the re of the region.. ne loop of the rose r cos 3. The region enclosed b the curve r 4 3 cos 3 6 Evlute the iterted integrl b converting to polr coordintes r r 3 3 s9 sin d d s dd s d d s s d d 7. A swimming pool is circulr with 4-ft dimeter. The depth is constnt long est-west lines nd increses linerl from ft t the south end to 7 ft t the north end. Find the volume of wter in the pool. 8. An griculturl sprinkler distributes wter in circulr pttern of rdius ft. It supplies wter to depth of e r feet per hour t distnce of r feet from the sprinkler. () Wht is the totl mount of wter supplied per hour to the region inside the circle of rdius centered t the sprinkler? (b) etermine n epression for the verge mount of wter per hour per squre foot supplied to the region inside the circle of rdius. 9. Use polr coordintes to combine the sum s s d d s into one double integrl. Then evlute the double integrl. 3. () We define the improper integrl (over the entire plne where is the disk with rdius nd center the origin. Show tht (b) An equivlent definition of the improper integrl in prt () is where S is the squre with vertices,. Use this to show tht (c) educe tht (d) B mking the chnge of vrible t s, show tht (This is fundmentl result for probbilit nd sttistics.) 3. Use the result of Eercise 3 prt (c) to evlute the following integrls. () I lim l e d e da e da e da e da lim l S e d e d e d s e d s d d s s4 d d e d d e da (b) s e d
MULTIPLE 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 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 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 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: olumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge
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 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 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 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 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 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 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 informationSECTION 9-4 Translation of Axes
9-4 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 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 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 informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: How to identify the leding coefficients nd degrees of polynomils How to dd nd subtrct polynomils How to multiply polynomils
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 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 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 informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
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 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 cross-section in plne pssing through
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 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 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 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 informationForm 5 HKCEE 1990 Mathematics II (a 2n ) 3 = A. f(1) B. f(n) A. a 6n B. a 8n C. D. E. 2 D. 1 E. n. 1 in. If 2 = 10 p, 3 = 10 q, express log 6
Form HK 9 Mthemtics II.. ( n ) =. 6n. 8n. n 6n 8n... +. 6.. f(). f(n). n n If = 0 p, = 0 q, epress log 6 in terms of p nd q.. p q. pq. p q pq p + q Let > b > 0. If nd b re respectivel the st nd nd terms
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 informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewise-liner function f, for 4, is shown below.
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 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 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 informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More informationsec x over the interval (, ). x ) dx dx x 14. Use a graphing utility to generate some representative integral curves of the function Curve on 5
Curve on Clcultor eperience Fin n ownlo (or type in) progrm on your clcultor tht will fin the re uner curve using given number of rectngles. Mke sure tht the progrm fins LRAM, RRAM, n MRAM. (You nee to
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 informationCalculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.
Clculus Module C Ares Integrtion Copright This puliction The Northern Alert Institute of Technolog 7. All Rights Reserved. LAST REVISED Mrch, 9 Introduction to Ares Integrtion Sttement of Prerequisite
More information10.2 The Ellipse and the Hyperbola
CHAPTER 0 Conic Sections Solve. 97. Two surveors need to find the distnce cross lke. The plce reference pole t point A in the digrm. Point B is meters est nd meter north of the reference point A. Point
More informationA LEVEL TOPIC REVIEW. factor and remainder theorems
A LEVEL TOPIC REVIEW unit C fctor nd reminder theorems. Use the Fctor Theorem to show tht: ) ( ) is fctor of +. ( mrks) ( + ) is fctor of ( ) is fctor of + 7+. ( mrks) +. ( mrks). Use lgebric division
More informationMAC 1105 Final Exam Review
1. Find the distnce between the pir of points. Give n ect, simplest form nswer nd deciml pproimtion to three plces., nd, MAC 110 Finl Em Review, nd,0. The points (, -) nd (, ) re endpoints of the dimeter
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 ) dx dx x sec x over the interval (, ).
Curve on 6 For -, () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor
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 informationSample Problems for the Final of Math 121, Fall, 2005
Smple Problems for the Finl of Mth, Fll, 5 The following is collection of vrious types of smple problems covering sections.8,.,.5, nd.8 6.5 of the text which constitute only prt of the common Mth Finl.
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 information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
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 informationSection - 2 MORE PROPERTIES
LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when
More informationr 0 ( ) cos( ) r( )sin( ). 1. Last time, we calculated that for the cardioid r( ) =1+sin( ),
Wrm up Recll from lst time, given polr curve r = r( ),, dx dy dx = dy d = (r( )sin( )) d (r( ) cos( )) = r0 ( )sin( )+r( ) cos( ) r 0 ( ) cos( ) r( )sin( ).. Lst time, we clculted tht for crdioid r( )
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
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 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 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 informationPractice Final. Name: Problem 1. Show all of your work, label your answers clearly, and do not use a calculator.
Nme: MATH 2250 Clculus Eric Perkerson Dte: December 11, 2015 Prctice Finl Show ll of your work, lbel your nswers clerly, nd do not use clcultor. Problem 1 Compute the following limits, showing pproprite
More informationR(3, 8) P( 3, 0) Q( 2, 2) S(5, 3) Q(2, 32) P(0, 8) Higher Mathematics Objective Test Practice Book. 1 The diagram shows a sketch of part of
Higher Mthemtics Ojective Test Prctice ook The digrm shows sketch of prt of the grph of f ( ). The digrm shows sketch of the cuic f ( ). R(, 8) f ( ) f ( ) P(, ) Q(, ) S(, ) Wht re the domin nd rnge of
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 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 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 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 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 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
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 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 informationJackson 2.26 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson 2.26 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: The two-dimensionl region, ρ, φ β, is bounded by conducting surfces t φ =, ρ =, nd φ = β held t zero
More informationChapter 1: Logarithmic functions and indices
Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4
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 informationHOMEWORK SOLUTIONS MATH 1910 Sections 7.9, 8.1 Fall 2016
HOMEWORK SOLUTIONS MATH 9 Sections 7.9, 8. Fll 6 Problem 7.9.33 Show tht for ny constnts M,, nd, the function yt) = )) t ) M + tnh stisfies the logistic eqution: y SOLUTION. Let Then nd Finlly, y = y M
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 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 informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
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 informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More information8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS
8 FURTHER APPLICATIONS OF THE INTEGRAL AND TAYLOR POLYNOMIALS 8 Arc Length nd Surfce Are Preliminr Questions Which integrl represents the length of the curve cos between nd π? π π + cos d, + sin d Let
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 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 informationy b y y sx 2 y 2 z CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
ECION.8 CHANGE OF VAIABLE IN MULIPLE INEGAL 73 CA tive -is psses throgh the point where the prime meridin (the meridin throgh Greenwich, Englnd) intersects the eqtor. hen the ltitde of P is nd the longitde
More informationNUMERICAL INTEGRATION
NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls
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 informationThe Trapezoidal Rule
_.qd // : PM Pge 9 SECTION. Numericl Integrtion 9 f Section. The re of the region cn e pproimted using four trpezoids. Figure. = f( ) f( ) n The re of the first trpezoid is f f n. Figure. = Numericl Integrtion
More informationr = cos θ + 1. dt ) dt. (1)
MTHE 7 Proble Set 5 Solutions (A Crdioid). Let C be the closed curve in R whose polr coordintes (r, θ) stisfy () Sketch the curve C. r = cos θ +. (b) Find pretriztion t (r(t), θ(t)), t [, b], of C in polr
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationFurther applications of integration UNCORRECTED PAGE PROOFS
. Kick off with CAS. Integrtion recognition. Solids of revolution. Volumes Further pplictions of integrtion. Arc length, numericl integrtion nd grphs of ntiderivtives.6 Wter flow.7 Review . Kick off with
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 informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationMath 1132 Worksheet 6.4 Name: Discussion Section: 6.4 Work
Mth 1132 Worksheet 6.4 Nme: Discussion Section: 6.4 Work Force formul for springs. By Hooke s Lw, the force required to mintin spring stretched x units beyond its nturl length is f(x) = kx where k is positive
More informationfractions Let s Learn to
5 simple lgebric frctions corne lens pupil retin Norml vision light focused on the retin concve lens Shortsightedness (myopi) light focused in front of the retin Corrected myopi light focused on the retin
More information1. Find the derivative of the following functions. a) f(x) = 2 + 3x b) f(x) = (5 2x) 8 c) f(x) = e2x
I. Dierentition. ) Rules. *product rule, quotient rule, chin rule MATH 34B FINAL REVIEW. Find the derivtive of the following functions. ) f(x) = 2 + 3x x 3 b) f(x) = (5 2x) 8 c) f(x) = e2x 4x 7 +x+2 d)
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 informationJUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 12 (Second moments of an area (B)) A.J.Hobson
JUST THE MATHS UNIT NUMBE 13.1 INTEGATION APPLICATIONS 1 (Second moments of n re (B)) b A.J.Hobson 13.1.1 The prllel xis theorem 13.1. The perpendiculr xis theorem 13.1.3 The rdius of grtion of n re 13.1.4
More informationZ b. f(x)dx. Yet in the above two cases we know what f(x) is. Sometimes, engineers want to calculate an area by computing I, but...
Chpter 7 Numericl Methods 7. Introduction In mny cses the integrl f(x)dx cn be found by finding function F (x) such tht F 0 (x) =f(x), nd using f(x)dx = F (b) F () which is known s the nlyticl (exct) solution.
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 informationAnti-derivatives/Indefinite Integrals of Basic Functions
Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationLoudoun Valley High School Calculus Summertime Fun Packet
Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember!
More informationMATH , Calculus 2, Fall 2018
MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
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 informationThe Fundamental Theorem of Calculus, Particle Motion, and Average Value
The Fundmentl Theorem of Clculus, Prticle Motion, nd Averge Vlue b Three Things to Alwys Keep In Mind: (1) v( dt p( b) p( ), where v( represents the velocity nd p( represents the position. b (2) v ( dt
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 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 informationJackson 2.7 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell
Jckson.7 Homework Problem Solution Dr. Christopher S. Bird University of Msschusetts Lowell PROBLEM: Consider potentil problem in the hlf-spce defined by, with Dirichlet boundry conditions on the plne
More information