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by: Reyes Glover


Marketplace > University of Texas at Austin > Mathematics (M) > M 427L > ADV CALCULUS FOR APPLICATNS II
Reyes Glover
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This 20 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 427L at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 55 views. For similar materials see /class/181493/m-427l-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15
M427L HANDOUT APPLICATIONS OF THE CHANGE OF VARIABLES FORMULA amp IMPROPER INTEGRALS SALMAN BUTT July 19 2007 Moments of Inertia If we want to study the dynamics of a rotating rigid loody7 the moment of inertia tells us about how inertia acts in each of the m y and z axes If we have a solid W with uniform density 6 then Essentially7 the moment of inertia tells about an objects resistance to rotation and depends on the shape of the object Exercise Find the moment of inertia around the y axis for the ball z2 y2 l 22 g R2 if the mass density is a constant 6 Gravitational Fields of Solid Objects Suppose we have some object that occupies a domain W with mass density 6z7y72 This is our attracting object think of W as a planet The gravitational potential V for W acting on a mass m at 1 34121 is Improper Integrals Recall the situation of improper integrals 7 ie integrals of unbounded functions on intervals or integrals of functions on unbounded intervals 7 from single variable calculus These integrals were evaluated simply by using a limiting process We will do a similar process in higher dimensions For the sake of simplicity we will only consider functions of two variables and we will assume our function is non negative Consider a functions f D 7 R that is continuous except for points on the boundary of D This is the rst improper integral we will consider Suppose the z range for D is 071 and the y range for D is l gt1ltgt7 gt2lt90gtl3 We take some 776 small enough so that a 7 lt b 7 7 and gt1z 6 lt gt2z 7 6 for all a g x g b and consider the region D7 a 7 g x g b 7 67 1z 6 g y gt2z 7 6 C D We know 1 is continuous and bounded on Dmg so the integral ffDn 6 1 1A exists We can then consider what happens as we push D7 out to D7 ie what happens as 7 6 7 07 0 If lim fdA 71gt5H0gt0 DW5 exists7 we say that the integral of 1 over D is convergent or that f is integrable over D We then de ne ffD 1 1A to be this limit Exercise Evaluate dz dy where D is bounded by z 17 z y and z 2y Unfortunately it is usually not possible to evaluate such limits so directly and easily Most in tegrals one encounters in practice are usually more complicated and produce difficulties in applying the above technique But if we develop some of our theory we can get a work around to handle some of these more complicated integrals Essentially the idea is to exploit the iterated integral not just the double integral For those interested I am talking to you math majorsl read the discussion in the text from the bottom of page 409 to page 411 But we want to collect the main result known as Fubini s theorem Theorem 1 Fubini s Theorem So what is this theorem saying All it says is that if f and D are nice as long as one of the integrals above exists then all the others exist and are all equal to each other Exercise Show that the integral fol faxVa 7 y dy dz exists and compute its value We next consider a different and slightly easier situation suppose our non negative function f is unde ned at some isolated point in the domain D Let mo yo be the isolated point at which 1 is not de ned let D5 D5z0 yo be the disk of radius 6 centered at 0 yo If f is continuous at all other points of D then DDNA exists We then say that ffD 1 1A is convergent or that f is integrable over D if lim fdA no DDltS exists In this case we de ne ffD 1 1A to be this limit Exercise Evaluate ffDz2 yz 23 dA where D is the unit disk in R2 One can generalize this situation to de ne the integral of a non negative function 1 that is continuous except at a nite number of points in D One can also de ne the integral for a function that has a nite number of discontinuities on D or at points on the boundary of D If you want 1 to take on positive and negative values a little more theory is required This is the beginning of Lebesgue integration theory and is well beyond the scope of this course But it is helpful to know that Lebesgue integration is the next step77 in analysis after a course such as this Finally7 if we want to consider how to integrate a function over an unbounded region7 we simply do what we do in the single variable case we simply take some parameter or parameters to in nity Exercise Let D be the unbounded region de ned as the set 73472 with z2 y2 22 2 1 Evaluate the improper integral dm dy dz D m2 y2 22 as A Brief Calang dr Lha Duadnc swims Prupectluns 835 Is Jlx y 7 l7 v x yl F mm 7 1n n z 3 A BRIEF CATALOG OF THE DUADFHC SURFACES PROJECTIONS w rcprcscnlcd rundiidns crude variable by curves in me Mala1e We will repre senl functions uriwd variables by surfaces in Maespace lions dr Ich am AX2 Ey1sz D vExzEVzHXIvJz K0i Such surfaces are called mm sinum By sunable lransla uns and rotalinnsoflhc Coordinate axes we can simpliry such aqualiuns and thereby show um die nondegenerate r quadn39cs all into nine disnnu types l mu ellipsoid 2 I11 byperhuluid drone sheet 3 he hyperbolold dnwa sheds 4 the quadric mad 5 the elliptic parahclold illc hypzrbolic paxaboloid 7 quotK uarabolic cylinder 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mice in Ill I D Flgure 1522 hivelbnlnld M an ma FVGLIRE la 2 a plum pllul 51 K 0 5 u ellipse 1 1L 0 m l b 0 The Inc in he 11 a is me hymrhola L ml m um in me plan 3m 0 E um hypcrboln 2 1 J A sections parallel to m szplanc ml plane are hyperbolas lru 1 than llll smiuns pnmllcl m hu xyplnn as circles and v have a Ill39ptlrbnllud HfrPVUHv quotan a Fun nuns 56 Several Vanumes a The Hyperbulold of Twn Sheets Figure 1623 r l The surface inwrscclh mu coordinnw axes only at hc two veniccs 10 0 in The srf sf39um39om ncz consisl for which 2 c S 7c 39 unbounded m the 139plam wllh z Sccduns I1c mile pmus m cuordinme planes arc hypubnlns ro have rcxnmple for y 3 WC quotminner nl m sums suns vaza The sur c is symmcll39lc aboux 2 mm anurdinn planes and is centered al lhc ori D 4 The Duadmc Cane m Figure I6 4 The su ac Inlclsccls Ihc mammal axes only ax m cream The surface is m bounded Oncu mm 1ch is symmmy abnul um um cuordinnh plants 11 um in um plunc is n mm of inmrwcling lim A 39u The Irnc in m V J me origin Seuions parallnl uIhcxy planearecllipscs 1m s I hcscscuionsare Th unplr and lowcr nnninns ul39 lhe can arr milled Impm EJ vs 2 A anal Cacamg ur the Duann Sullauesl Prulecuuns 839 mm M FlGuRE la 24 and lquot hm hm insumd of 5 The Elllptic Farabaluid lFlgum 1615 mum pambelmu mung l6 2 5 4 y 39 39 39 I39h arigln is called the re ux Seamus palulch lo me xyplnnc nr ellipses secliuns parallel m lhe ulher mordiunlc plans are pamholas Hum me mm quotclliplic pnmbclloidquot Th surface u symmzlrir 3 ul me xzplanl and ubuul IIquot Vzplnnc 39 39 mo 1 i I39m ll39l rullulullml D D Functmns 1 Several Vannbles s The Hyperbolic Parammm Figure 1525 yppxane 1 alas Hem mu ILnn 39 39 quot 39 39 39 39 the mm in Ihc x Innc bu a maximum paint for It Irma in IL39 Drig39n is caund a mininuu or uidly paim of u surface I Ilymlboh narabolmd mung 15 2 5 Th 55 or he qundric surfaces are Eylimleu The mm deserves de ni nn of39 fun a surface Such a surrm is called acyMer we muner with use mmquot C The perpendicular line are nuwn as meganmmrs on c cylin 1 quotthe base curve lies in me A leanc or in a plan parullel m It 39plane um I I m an on all Values 7 Tha Parabullc Cylinder 4039 Figure 617 This surfacc is formed by an lines mm pass through m parabola r 4c39 m m perpendicular m mu Jhplanc u l a The amp Cylinder 752 A Emef Catamg nf the Duanrm Surfaces Frnjecmuns B41 me mm FIGURE V527 FlsuRE 1525 The surface is formed by an Lines Lilal pass Lhrough the ellipse x2 y F F V n a h cylinder 1 hynerhuHc Cylinder I URE 52 a 94 The Hyperbunc Cylinder Figuxe 6l9 M427K Handout PDEls Fourier Series Salman Butt April 27 2006 FOURIER SERIES We will be constructing solutions later in class involving in nite sums of sine and cosine terms These series are called Fourier series We want to determine which functions can be represented by Fourier series and how to compute the coefficients in the series corresponding to a given function Note that we begin the series with lo2 as opposed to simply rig to simplify the coefficient formula we will derive below Peiiodicitg of the Sine and Cosine Functions We need to know about the periodic nature of the terms in our Fourier series Recall that a function is said to be periodic with period T gt 0 if fz l T x for all z x l T in the domain of f The smallest value of T for which 1 is periodic is called the fundamental period If g are two periodic functions with common period T then the product fg and an arbitrary linear combination 01f l 02g are also periodic with period T So the individual terms in our series cosm7rmL and sinm7rzL are periodic with fundamental period T 2Lm Looking at every term in our series we see that every term in our series has common period 2L Orthogonalitg of the Sine and Cosine Functions Recall what it means for two vectors 7317 to be orthogonal 73 17 0 where is the dot product We adapt the notion of the dot product to functions and call this product the inner product g of two real valued functions g on the interval 04 g x g B And we then say that two functions g are orthogonal on 04 g x g B if g O A set of functions is said to be mutually orthogonal if each distinct pair of functions in the set is orthogonal The central fact is that the set sinm7rmLcosm7rmL m 1 2 is mutually orthog onal The proof of this fact comes directly from computing the integrals Some of the necessary computations are done in the book and I will not bother discussing these details The reason we care about these functions being orthogonal is because we will exploit this fact to develop a formula for the coefficients in our Fourier series The Euler Fourier Formulas Let us suppose that our Fourier series converges at some point z and let s call it That is we have M427L HANDOUT GREEN S THEOREM STOKES THEOREM amp CONSERVATIVE FIELDS SALMAN BUTT August 27 2007 We now come to the payoff of this class where we tie together differential and integral calculus for multivariable functions This will allow us to generalize the fundamental theorem of calculus to multiple variables and provide extremely powerful tools to compute integrals Green7s Theorem Green s theorem lets us relate the line integral along a closed curve C in R2 to a double integral over the region enclosed by C In order to understand the theorem7 we need to understand the orientation of the boundary of a region Simply put7 we orient the boundary of a region in R2 counterclockwise That is7 the positive orientation for the boundary of a region D can be realized as the orientation taken by someone walking along the boundary that keeps the region to the left Theorem 1 Let D be a simple region and let C be its boundary Suppose P D gt R and Q D HR are of class Cl Then In fact this theorem applies to much more regions than just simple regions The theorem applies to any nice region in R We can use Green s theorem to quickly compute the region enclosed by a simple closed curve C If C 8D for some nice region D7 then This is easily done by taking Pz7 y 7y and c2z7 y z and applying Green s theorem Exercise Find the area bounded by the z axis and one arc of the cycloid z a0 7 sint97 y 117 cos0 where a gt 0 and 9 E 07 27139 We can restate Green s theorem using the language of vector calculus Theorem 2 Curl form of Green s theorem Let D C R2 be a nice region and let 8D be its positively oriented boundary Let F Pi be a C1 vector eld on D Then Theorem 3 Divergence form of Green s theorem Let D C R2 be a nice region and let 8D be its positively oriented boundary Let ii denote the outward normal to 3D and let 3t Mt7 be an orientation preserving parametrization of 8D Then iit ez 2 ast lytllz Let F Pl be a C1 vector eld on D Then a 77 Exercise Verify the divergence theorem for D z y and the D the unit disk 2 y2 g 1 Stokes7 Theorem Stokes7 theorem relates the line integral of a vector eld around a simple closed curve C in R3 to a integral over a surface S for which C is the boundary We will discuss two cases of Stokes7 theorem Stokes Theorem for Graphs Let S be the surface that is the graph ofa function m y D a R so we have a natural parametrization aw do fuv We induce an orientation on 85 from the orientation on 8D if E a7b a R2 8t ztyt is an orientation perserving parametrization of 8D7 then the 85 is an oriented simple closed curve that is the image of 15 t H mtytfmtyt with orientation induced by 13 Geometrically7 this means that the orientation on SS is determined by walking along the boundary with your head pointing in the direction of the normal of S and the surface being to your left With this in hand7 we can state the following Theorem 4 Let S be the oriented surface de ned by a 02 function 2 fzy where my 6 D and D is a region to which Green s theorem applies Let F be a C1 uector eld on S Then Stokes Theorem for Parametrized Surfaces Let f D a R3 be a parametrization of a surface S and let 8t utut be a parametrization of 8D Suppose I is oneto one on D7 so ltIgt8D 8S That is7 SS is the oriented simple closed curve that is the image of the mapping 13 L t H ltIgtut7 ut with orientation induced by 13 Theorem 5 LetS be an oriented surface de ned by a one to one parametrization l D C R2 gt S where D is a region to which Green s theorem applies Let 8S be the oriented boundary ofS and let F be a C1 uector eld on S Then If 8S 0 then the integral on the left is zero Exercise Evaluate ffSV x lg where S is the surface 2 y2 322 1 2 g 0 and F y ix 33422 oriented outward Conservative Fields We previously note that if F Vf for some vector eld F and some real valued function f then ng d f8b 7 f a where E a b a R But when is a vector eld a gradient eld Theorem 6 Let F be a C1 vector eld de ned on R3 except possibly for a nite number ofpoints Then the following are equivalent 1 For any oriented simple closed curve C f0 F 1 0 2 For any two oriented simple curves Cl and 02 that have the same endpoints Fd Fd i 01 02 5 F Vf for some real valued function 4 V x F 6 A vector eld satisfying one and hence all of the conditions above is called a conservative vector eld In the case of a vector eld F on R2 we require that F is smooth everywhere and write F P Q Then the condition that V x F 6 reduces to As a result we nd that if F P Q is a C1 vector eld on R2 such that 8P8y eggom then FVf for some Now we also know that V V x 0 for any CZ vector eld ls there a converse to this statement lndeed Theorem 7 If F is a C1 vector eld on all of R3 ie it is smooth everywhere with F 0 then there exists a C1 vector eld G such that F V X G Exercise Let Fz 342 em sing5m cos y 22 Evaluate fg d where 8t Wits exp for t 6 01


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