Algebr Algebr is the foundtion of lulus The bsi ide behind lgebr is rewriting equtions nd simplifying expressions; this inludes suh things s ftoring, FOILing (ie, (+b)(+d) = +d+b+bd), dding frtions (remember to get ommon denomintor, b + d + b = ) nd multiplying by d bd the onjugte (the onjugte of + b is b nd multiplying gives 2 b 2 ) Below re some bsi fts tht frequently ome up ( + b)( b) = 2 b 2 ( + b) 2 = 2 + 2b + b 2 ( b) 2 = 2 2b + b 2 Another useful ft is the qudrti formul (lso known s the qudrti eqution) x 2 + bx + = 0 x = b ± b 2 4 2 ules for exponents nd logrithms re useful: x x b = x +b x, x b = x b, (x ) b = x b, ln(b) = ln + ln b, ln ( ) = ln ln b, b nd ln( b ) = b ln Of ourse it is lso useful to remember wht we nnot do For exmple, x 2 + 2 = x + is not true, even when tking test! Often ssoited with lgebr re funtions nd grphing The domin of funtion re numbers tht we n put into the funtion nd get something out while the rnge re the possible vlues tht we n get out We work with points in the plne by desribing them reltive to how fr wy from the origin we hve moved in different diretions, ie, (x, y) The distne between the two points (x 0, y 0 ) nd (x 1, y 1 ) in the plne is distne = (x 1 x 0 ) 2 + (y 1 y 0 ) 2 We hve lines (y = mx + b where m is the slope (found by rise/run) nd b the y-interept), irles ((x x 0 ) 2 + (y y 0 ) 2 = r 2 where (x 0, y 0 ) is the enter of the irle nd r the rdius), prbols (y = x 2 +bx+), ellipses (x 2 +by 2 = 1) nd hyperbols (x 2 by 2 = 1) We n desribe points using polr oordintes r nd θ where we hve x = r os θ nd y = r sin θ ometimes shpes re esier to desribe in polr oordintes, suh s irles (r = 1) or rdiods The re of retngle is (length) (width), the re of irle is πr 2 (where r is the rdius) nd the re of tringle is 1 2 (bse) (height) The volume of box is (length) (width) (height) nd the volume of sphere is 4 3 πr3 (gin r is the rdius) Trigonometry Trigonometry is bsed off the two ides tht (i) tringles re rigid nd (ii) when we sle tringle the rtios of the sides do not hnge Using these we n ssoite vlues with the ngles of tringles whih n be used to solve vrious problems relted to tringles The bsis of mny importnt fts bout tringles omes from the Pythgoren Theorem whih sys tht for right tringle with sides, b, ( being the hypotenuse) then 2 + b 2 = 2 Wht this mens is tht often when deling with tringles we look for right tringles Let θ be n ngle of right tringle, opp the length of the side opposite θ, dj the length of the side djent to θ nd hyp the length of the hypotenuse; then we hve sin θ = opp hyp, tn θ = opp dj, dj os θ = hyp, hyp se θ = dj (We either mesure ngles in degrees (360 is full revolution) or rdins (2π is full revolution) In lulus we lmost lwys will use rdins) Note tht the funtions sin nd os re periodi nd lso re lwys bounded between 1 nd 1 ( very useful ft) From the definitions hve tn θ = sin θ os θ nd se θ = 1 os θ And by using the Pythgoren Theorem we n onlude sin 2 θ + os 2 θ = 1 nd tn 2 θ + 1 = se 2 θ Other rules whih frequently ome up re the double ngle formuls sin(2θ) = 2 sin θ os θ os(2θ) = os 2 θ sin 2 θ, this n be used to redue powers of sine nd osine os 2 θ = 1 + os 2θ 2 nd sin 2 θ = We lso hve the even/odd er identities 1 os 2θ 2 sin( θ) = sin θ, nd os( θ) = os θ There re of ourse mny more formuls nd identities tht rise in trigonometry, but for our purposes
these will usully suffie (we will mke use, for exmple, of the lw of osines but in generl these re the most importnt identities) ingle vrible lulus The ide behind lulus is to study things whih re not flt by things whih re flt For exmple, urves will look like lines when we zoom in lose nd res n be broke into strips tht look like retngles; sine we know bout lines nd retngles then we n sy something bout the urves nd the re The underlying ide behind this re limits whih tell us wht should hppen bsed on wht is hppening nerby The first hlf of lulus is differentil lulus nd is bsed on the derivtive, whih n be used to find the slopes of tngent lines to urve Given funtion f(x) we sy f(x) = f f(x + h) f(x) (x) = lim h 0 h Of ourse, using the limit definition is tedious, so insted we ome up with simple rules The following re the rules for tking derivtives of the bsi building bloks of funtions = 0, x k = kx k 1 d (, ) e x = e x, sin x = os x, os x = sin x, tn x = se 2 x, se x = se x tn x 1 ln x = x, 1 rtn x = 1 + x 2 The other thing we need to be ble to do is to tke the derivtive of funtions tht hve been ombined together, by multiplition, ddition, sling, division or omposition These re omplished by the following rules f(x) + g(x) kf(x) f(x)g(x) ( ) d f(x) g(x) f(g(x)) = f (x) + g (x) = kf (x) = f (x)g(x) + f(x)g (x) = f (x)g(x) f(x)g (x) ( g(x) ) 2 = f (g(x))g (x) With the derivtive we n tell where the funtion is inresing nd deresing nd so lso n tell when it is t pek or vlley whih ours t ritil points where the derivtive is zero or undefined This helps to find min or mx so we n do optimiztion We n lso find the tngent lines whih n be used for lol pproximtion More generlly we n use Tylor polynomils to find better pproximtions The seond hlf of lulus is integrl lulus where we try to find totls (ie, res, lengths, volumes of revolution, surfe re, work, moments, enter of mss), the bsi ide to be to split things into smll piees nd dd the piees up For exmple b f(x) = (signed re between nd b) The best wy to do integrtion is by the Fundmentl Theorem of lulus whih sttes tht if F(x) is n ntiderivtive of f(x) (ie, F (x) = f(x)) then b f(x) = F(b) F() The other prt of the Fundmentl Theorem of lulus sttes ( d x ) f(t) dt = f(x) The bsi ntiderivtives re s follows: x k = 1 1 k + 1 xk+1 +, = ln x + x sin x = os x +, os x = sin x + se 2 x = tn x +, se x tn x = se x + e x = e x +, 1 = rtn x + 1 + x2 While differentition is mtter of pplying the rules, integrtion is n rt form The min tehnique is to use lgebr nd trigonometri identities to simplify the expression to one of the bove simple funtions tht we n tully find the ntiderivtive of Another importnt rule to help in simplifying is the substitution rule whih llows us to simplify integrls (this omes from the hin rule so the ide is to look for funtions inside of funtions): f ( ) g(x) g (x) = }{{}}{{} =u =du For definite integrls this is b f ( g(x) ) g (x) = g(b) g() f(u) du f(u) du, where u = g(x) ymmetry is sometimes useful for doing integrls Another importnt rule is integrtion by prts u dv = uv v du
Bsis of spe nd vetors oordinte systems rtesin oordintes desribe positions by how fr you move in eh diretion (x, y) (x units in the diretion of the x-xis nd y units in the diretion of the y-xis), or in three dimensions (x, y, z) The distne between points (x 1, y 1, z 1 ) nd (x 2, y 2, z 2 ) is given by (x 2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 A sphere (nd similrly) irle represent the set of points equidistnt (rdius r) from enter point (h, k, l); using the distne formul this beomes (x h) 2 + (y k) 2 + (z l) 2 = r 2 Polr oordintes desribe position by first inditing whih diretion to move, θ, nd then indite how fr to move, r We hve x = r os θ y = r sin θ Note x 2 + y 2 = r 2 There re two generliztions to three dimensions ylindril oordintes work by using polr oordintes in the xy plne nd then using z to desribe height o points re desribed by r, θ nd z This reltes to rtesin oordintes s follows: x = r os θ y = r sin θ z = z pheril oordintes work by first desribing diretions using two ngles φ nd θ (θ plys the sme role s before; φ mesures the ngle off of the positive z-xis nd rnges from 0 to π), finlly we indite the distne to move ρ This reltes to rtesin oordintes s follows: x = ρ sin φ os θ y = ρ sin φ sin θ z = ρ os φ Note x 2 + y 2 + z 2 = ρ 2 In generl we re most omfortble working with rtesin oordintes We use Polr/ylindril when we del with regions in the plne tht re desribed in terms of polr oordintes, ie, rdiods or irles We lmost only ever use pheril oordintes for problems involving sphere or one Vetors A vetor is n objet tht hs both diretion nd mgnitude We n del with vetors geometrilly (usully s line segments with rrows inditing orienttion) but more often we will del with vetors lgebrilly Nmely we represent them in terms of how they move in eh diretion (or in omponent form), b, = i + bj + k, ie, this vetor moves in the x-diretion, b in the y- diretion nd in the z-diretion The vetors i, j nd k re the stndrd unit vetors, ie, i = 1, 0, 0, j = 0, 1, 0 nd k = 0, 0, 1 To find vetor between two points we simply tke the differene of the ending oordintes minus the strting oordintes Addition of vetors nd sling by onstnts is done omponent wise We nnot multiply vetors together The mgnitude of vetor in omponent form is given by, b, = 2 + b 2 + 2 A vetor whih hs length 1 is lled unit vetor We n sle ny vetor by its mgnitude to get unit vetor pointing in the sme diretion Two vetors re prllel if they differ by sling ftor Dot nd ross produts The dot produt tkes two vetors nd produes number This works by multiplying the orresponding entries together nd summing, ie,, b, d, e, f = d + be + f This stisfies simple properties nmely u v = v u nd u u = u 2 The importnt thing is tht the dot produt n be interpreted geometrilly, ie, u v = u v os θ nd os θ = u v u v where θ is the ngle between the two vetors (if the tils emnte from the sme point) This gives n esy wy to find ngles between vetors A speil ngle is 90, suh vetors re lled orthogonl nd we see tht two vetors re orthogonl when u v = 0 Dot produts re useful for doing projetion, ie, ( ) u v proj v u = v v v Also work n be expressed s F D where F is fore nd D is diretion The ross produt tkes two vetors nd produes vetor This only works in dimension three!, b, d, e, f = = i b e f j d i j k b d e f f + k d b e = bf e, d f, e bd The importnt ft bout ross produts is tht u v is perpendiulr to both u nd v, so this is useful when we wnt to onstrut vetor tht is perpendiulr to two given vetors
Another osionlly useful ft bout ross produts is tht the mgnitude of the ross produt is the re of the prllelogrm with sides formed by the vetors o to find the re of tringle we tke 1 2 u v where u nd v re two sides of the tringle in vetor formt Lines nd plnes A line n be desribed by point (x 0, y 0, z 0 ) nd diretion, b, In vetor formt line is given by x, y, z = x 0, y 0, z 0 + t, b, If we solve for eh omponent we get the prmetri form x = x 0 + t y = y 0 + bt z = z 0 + t If we solve eh of these for t we get the symmetri form (lest used!) x x 0 = y y 0 b = z z 0 To find line we lwys need to find point nd diretion The diretion n ome from the vetor between two points on line, ross produt (if line needs to be perpendiulr to two vetors), or given geometrilly (suh s being perpendiulr to plne) A plne n be desribed by point nd norml vetor, b, whih is perpendiulr to eh vetor in the plne This implies 0 = x x 0, y y 0, z z 0, b, = (x x 0 ) + b(y y 0 ) + (z z 0 ) To find plne we will lwys need to find point nd norml vetor The norml vetor n be given to us or we might need to reover it geometrilly, this is often done with the id of the ross produt Vetor vlued funtions A vetor vlued funtion is funtion whih hs s output vetor The importnt point bout vetor vlued funtions is tht we n work with them pieewise, ie, to find the limit/derivtive/integrl of vetor vlued funtion we do it omponent by omponent The rules for differentition of dot produt, ross produt nd sling work just s we would expet We n use vetor vlued funtions to indite prmeteriztion of urves r(t) = x(t), y(t), z(t) ; we hve r (t) = v(t) = x (t), y (t), z (t) indites the veloity nd mrks the diretion in whih prtile long the urve is trveling The tngent vetor to urve is thus given by r (t), osionlly we will use the unit tngent vetor T(t) = r (t)/ r (t) nd the unit norml vetor N(t) = T (t)/ T (t) The tngent line for r(t) t t = t 0 is r(t 0 ) + tr (t 0 ) The elertion of r(t) is given by r (t) = (t) = x (t), y (t), z (t) To find the distne trveled by prmetri urve from t = to t = b we hve b b r (t) dt = (x (t)) 2 + (y (t)) 2 + (z (t)) 2 dt Aelertion n be deomposed into the prt tht pulls in the diretion of motion nd prt whih pulls perpendiulr to motion, ie, T = r r r = T T + N N nd urvture N = r r r The urvture is mesurement of how bent urve is For prmetri urve in three spe we hve κ = r r r 3 In the speil se when the urve is in the plne, ie, (x(t), y(t)) we hve κ = x y y x ( (x ) 2 + (y ) 2) 3/2 Qudri surfes Qudri surfes re surfes whose tres re oni setions There re six possible qudri surfes we might enounter (when in doubt sketh how the surfe intersets the xy-, xz-, nd yz-plne): x 2 Ellipsoids: 2 + y2 b 2 + z2 2 = 1 ross setions: ellipses, empty Hyperboloid of one sheet: 2 + y2 b 2 z2 2 = 1 ross setions: ellipses, hyperbols Hyperboloid of two sheets: 2 y2 b 2 z2 2 = 1 ross setions: ellipses, hyperbols, empty Ellipti prboloid: z = x2 2 + y2 b 2 ross setions: ellipses, prbols, empty x 2 x 2 Hyperboli prboloid: z = x2 2 y2 b 2 ross setions: prbols, hyperbols Ellipti one: z 2 = x2 2 + y2 b 2 ross setions: ellipses, hyperbols
Multivrible differentition Funtions of severl vribles A funtion is rule for mpping inputs to outputs A funtion of severl vribles indites one where there re severl inputs The grph of the funtion is found by plotting ll points of the form (x, y, f(x, y)) The level urves (lines in the ontour mp) re found by setting f(x, y) = k nd plotting the orresponding urves For funtions of three vribles we hve F(x, y, z) = k orresponds to level surfes The domin is the set of inputs for whih we n get n output In generl there re three problems to wth out for: division by zero squre root of negtive log of nonpositive For our lss these will be the only problems tht we will hve to del with Prtil derivtives We wnt to understnd how the funtion is hnging (ie, derivtives whih orrespond to rtes of hnge) One simple wy to do this is prtil differentition, ie, tret ll but one vrible s onstnt f x f (x, y) = lim h 0 f(x + h, y) f(x, y) (x, y) = lim h 0 h f(x, y + h) f(x, y) h The morl is tht we use the ext sme rules s before; the only thing is to remember wht is the vrible we re tking the derivtive with respet to The nottion f/ x mens the derivtive of f with respet to x or the rte of hnge of f s x hnges An lterntive nottion is f x We n lso tke higher order derivtives (inluding mixed derivtives), the nottion needs to indite whih derivtives we re tking For exmple 3 f x 2 = f yxx indites tht we tke three derivtives, with respet to x twie nd with respet to y one For the funtions tht we will enounter in our lss we hve mixed prtils re equl, ie, 2 f x = 2 f x Limits Limits re used to understnd wht vlue should be by exmining points lose by The funtion f is ontinuous funtion if lim f(x, y) = f(, b), (x,y) (,b) ie, wht we expet is wht we get Polynomils re ontinuous nd we n ombine ontinuous funtions to get ontinuous funtions (ie, ddition, subtrtion, multiplition, omposition, nd when denomintor is not zero we n use division) ome possibile pprohes for limits: We plug in the vlue nd get number, ie, funtion is ontinuous nd we re done We n simplify the funtion, ie, rewrite nd gin plug in nd be done We n show tht when pprohing in two different wys we get different vlues nd the limit does not exist (There re other possibilities tht ould hppen, but we will not enounter them in our lss) Differentibility nd tngent plne A funtion will be differentible if lolly it is flt (ie, hs tngent plne) For us differentibility boils down to sying tht f/ x nd f/ re ontinuous funtions For ontinuous funtions we hve tht for funtion f(x, y) with (x, y) ner (x 0, y 0 ) f(x, y) f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f (x 0, y 0 )(y y 0 ) }{{} = tngent plne t (x 0, y 0 ) where the pproximtion gets better s (x, y) gets loser to (x 0, y 0 ) Or more suintly, z f f x + x y This llows us to use derivtives to give estimtes for hnge of n output for smll hnges in the input We n lso use Tylor polynomils to get better pproximtions, ie, z = f + f x (x x 0 ) + f y (y y 0 ) + 1 2 f xx(x x 0 ) 2 + f xy (x x 0 )(y y 0 ) + 1 2 f yy(y y 0 ) 2 Grdient nd its properties The grdient of funtion f, denoted f, is the vetor omposed of the prtil derivtives f = f x, f y or g = g x, g y, g z A diretionl derivtive mesures the rte of hnge of funtion f s we move in the diretion u (where u is unit vetor) We hve D u f = f u For diretionl derivtives we lwys work to nil down the diretion, ie, interpret the problem to give vetor nd mke it unit vetor by sling The reltionship between diretionl derivtes nd grdients give some importnt properties of grdients We hve the following:
f(p) is mximum rte of inrese t p f(p) points in the diretion whih gives the mximum rte of inrese t p f(p) is mximum rte of derese t p f(p) points in the diretion whih gives the mximum rte of derese t p Grdient vetors re perpendiulr to level urves nd level surfes The lst property will be useful in finding tngent plnes to impliitly defined surfes, ie, surfes of the form F(x, y, z) = k We n interpret suh surfe s level surfe nd then note tht F will be perpendiulr so beomes the norml vetor for the tngent plne hin rule The hin rule dels with the problem of wht to do when we hve funtion inside of funtion For exmple if we hve z = f(x, y) nd x = x(t) nd y = y(t), then z is relly funtion of t o we hve dz dt = x dt + dy dt, ie, we need to determine the effet tht t hs through the vrible x, nd through the vrible y nd then we sum up to get the totl effet imilrly we hve tht if x = x(s, t) nd y = y(s, t) then s = x ds + dy ds, t = x dt + dy dt In generl to pply the hin rule grph the reltionship between the vribles s tree nd use the tree to show how to ombine ll the prtil derivtives We n use the hin rule in the se of impliit funtions For exmple when F(x, y) = k then while if F(x, y, z) = k then dy = F x F y x = F x F z nd = F y F z Mxim nd minim A mximum is point where the funtion is lrger either thn points nerby when lol; or for ll points when globl imilrly for minimum When our funtion is ontinuous on losed nd bounded set it must hieve mximum nd minimum At mximum (similrly minimum) we nnot get bigger, so the grdient should not be nonzero o we hve tht mximums nd minimum will our t ritil points, these inlude: Where f = 0 (ie, ritil points) Where f is undefined At boundry For funtion we find ritil points by looking t f nd where it is 0, equivlently where the prtil derivtives re 0 One we hve suh ritil points we n sometimes determine wht type of ritil point it is by use of the eond Prtils Test This is done by omputing D = f xx f yy (f xy ) 2 We then evlute this funtion t the ritil point(s); we hve the following: If D < 0 then it is sddle, neither mx or min If D > 0 nd f xx < 0 (or f yy < 0) then it is mximum If D > 0 nd f xx > 0 (or f yy > 0) then it is minimum In the se when we re finding mximum/minimum over losed nd bounded set we then do the following: 1 Find ll ritil points in the interior using f 2 Find ll ritil points on boundry (redue dimensions down) 3 Plug in list of ritil points into funtion 4 Lrgest number on list is mx; smllest is min Lgrnge multipliers The tehnique of Lgrnge multipliers is used to solve optimiztion problems with onstrint These re usully esy to identify, ie, there will be two funtions one tht is being mximized nd the other tht is onstrint on the vribles (ie, given tht or suh tht ) If mximizing/minimizing f(x, y) given the onstrint g(x, y) = k this boils down to solving the system of equtions f(x, y) = λ g(x, y) nd g(x, y) = k This leds to lrge system of nonliner equtions When in doubt, solve for λ nd set the vrious terms equl to lmbd equl to one nother This gives nother reltionship between x nd y tht n be used with g(x, y) = k to solve for ll points One we hve these points we plug into the funtion, the lrgest vlue is the mximum nd the smllest vlue is the minimum
Multivrible integrtion Multivrible integrtion To integrte we split our domin into smll prts, ompute the ontribution of eh prt, nd then dd it bk up to get totl o we hve f(x, y) da = lim f(x, y) A In prtie we use the tools we lerned from single vrible lulus to ompute these integrls However, we n use this when our tools re insuffiient to get n pproximtion by subdividing our domin into smll number of piees nd omputing the bove sum Integrls re liner nd so we hve (f(x, y) + bg(x, y)) da = f(x, y) da + b g(x, y) da imilrly if we n brek our region into disjoint prts nd T, ie, = T then we n brek the integrl into two prts, ie, f(x, y) da = f(x, y) da + f(x, y) da T Our method to ompute will be to trnsform the integrl into n iterted integrl This will llow us to integrte one vrible t time treting the other vribles s onstnts Integrtion with rtesin oordintes When integrting with respet to rtesin oordintes we hve da = dy or da = dy The importnt prt is to set up the bounds for integrtion In the idel se when we hve retngle, ie, x b nd y d then we hve b d f(x, y) dy When our region is not retngle then we need to determine the boundry urves If we think of tking vertil strips (ie, integrting dy ) then we will hve x b nd φ 1 (x) y φ 2 (x) so tht our integrl beomes b φ2 (x) φ 1 (x) f(x, y) dy When we think of tking horizontl strips (ie, integrting dy) then we will hve y d nd ψ 1 (y) x ψ 2 (y) so tht our integrl beomes d ψ2 (y) ψ 1 (y) f(x, y) dy T ometimes we might wnt to swith the order of integrtion (this n llow impossible integrls to beome possible) In doing so the first thing is to drw piture nd then express the region in terms of new bounds, ie, rewrite the bounding urves When f(x, y) = g(x)h(y) we n use properties of onstnts with respet to integrtion to onlude d b ( b g(x)h(y) dy = )( d ) g(x) h(y) dy We note tht for some funtions (for exmple ones involving bsolute vlue) it is sometimes more onvenient to brek the integrl up into piees On similr note we n use symmetry in some ses to simplify n integrl Integrtion in Polr oordintes When doing double integrls in polr oordintes we n rewrite the funtion tht we re integrting in terms of r nd θ Then we desribe our region in terms of r nd θ The most importnt thing to note is tht da = r dr dθ The extr ftor of r is very importnt! Applitions of double integrls Volume inbetween two surfes z = f(x, y) nd z = g(x, y) (usully we hve g(x, y) = 0) over the region V = (height) da where height is f(x, y) g(x, y) Mss of region with density funtion δ(x, y) M = (density) da = Moment = δ(x, y) da (distne)(density) da where distne (sine we llow for positive or negtive vlues) is with respet to n xis For exmple M y is the moment with respet to the y-xis nd is given by M y = xδ(x, y) da enter of mss is ( x, y ) ( My = M, M ) x nd written in terms of integrls orresponds to M weighted verge Physilly this orresponds to point where the mss blnes Often times we n use symmetry (of both the region nd the density funtion) to quikly solve for one of the vlues for the enter of mss
eond moment or moment of inerti is given by Inerti = (distne) 2 (density) da Where the distne is with respet to n xis we re spinning round For exmple the inerti with respet to the y xis is given by I y = x 2 δ(x, y) da The inerti with respet to the origin (ie, s we spin the plne) is I z nd stisfies I z = I x + I y urfe re of z = f(x, y) over region (hrd to integrte, esy to set up) urfe re = (f x ) 2 + (f y ) 2 + 1 da Triple integrls We n lso onsider triple integrls The generl proedure is the sme, ie, set it up s n iterted integrl nd integrte one prt t time We hve dv = dz dy (rtesin) = r dz dr dθ (ylindril) = ρ 2 sin φ dρ dφ dθ (pheril) hnging oordintes is useful when it helps simplify the desription of the region nd/or mkes the funtion tht we re integrting simpler One hint to swith to ylindril is when the bse of the region is in polr oordintes nd/or the funtion hs x 2 + y 2 whih n beome r 2 One hint to swith to spheril is if we re integrting over portion of sphere nd/or the funtion hs x 2 + y 2 + z 2 whih n beome ρ 2 When determining bounds (or swithing bounds), it is very useful to drw piture to help see wht the region looks like nd then rewrite the bounding surfes s needed It is lso helpful to see the shdow of the region to help guide setting up the bounds Applitions of triple integrls Mss of solid with density funtion δ(x, y, z) M = (density) dv = δ(x, y, z) dv Moment = (distne)(density) dv where distne (sine we llow for positive or negtive vlues) is with respet to plne For exmple M yz is the moment with respet to the yz-plne nd is given by M yz = xδ(x, y, z) dv enter of mss is ( x, y, z ) ( Myz = M, M xz M, M ) xy M nd written in terms of integrls orresponds to weighted verge Agin, it is sometimes possible to use symmetry to help us solve for some of the oordintes Jobin hnge of vribles When working in spheril or ylindril oordintes we re essentilly hnging our oordinte system This n be done more generlly nd is useful when we hve funtions inside of funtions or very diffiult boundries to work with In generl we wnt to rewrite n integrl in terms of vribles x nd y in terms or new vribles u nd v We do this by mking the substitutions u = u(x, y) nd v = v(x, y) We n tke these equtions nd solve for x nd y so tht we lso hve x = x(u, v) nd y = y(u, v) Our region tht ws in terms of x nd y beomes new region in terms of u nd v The finl thing tht we need to hndle is the da term This is done with the id of the Jobin mtrix whih is the olletion of ll possible derivtives Nmely we hve x x J(u, v) = v v = x v x v With the Jobin we hve dy = J(u, v) dv du Altogether we hve the following: f(x, y) dy = f ( x(u, v), y(u, v) ) J(u, v) dv du In three dimensions we hve something similr going on where dz dy = J(u, v, w) dw dv du nd J(u, v, w) = x x v v v x w w w
Bsi vetor lulus Divergene nd url A vetor vlued funtion is lso sometimes known s vetor field Geometrilly we n present this by drwing some vetors t few points (if we tried to drw ll of them we would turn the pper blk!) We will fous on three dimensionl vetor vlued funtions, though some of the results below lso extend to other dimensions A vetor vlued funtion is onservtive if F = f Given vetor vlued funtion we n perform some opertions involving prtil derivtives to onstrut new funtions The wy to do this is by use of the opertor whih we n think of s vetor of prtil derivtives, ie, in three dimensions this is = x i + j + k = x,, The divergene, sometimes denoted s div F, is given by using the dot produt F = x,, M, N, P = M x + N + P The url, sometimes denoted s url F, is given by using the ross produt F = x,, M, N, P i j k = x M N P P = N, M P x, N x M Line integrls A line integrl in the plne is given by f(x, y) ds where is the urve nd ds represents smll length of urve The wy tht this is tully omplished is by working with prmeteriztion of the line, ie, (x(t), y(t)), for whih ds = (x (t)) 2 + (y (t)) 2 dt, f(x, y) ds = b f ( x(t), y(t) ) (x (t)) 2 + (y (t)) 2 dt where nd b re the strt nd end for the prmeteriztion of the line More generlly we will work with integrls of the form M, N, dy = M(x, y) + N(x, y) dy }{{} =dr whih gin is done by repling every x nd y (inluding the nd dy terms) with x(t) nd y(t) respetively In the se when F = M, N = f we hve f dr = f t=β t=α = f(b) f(), ie, to evlute the line integrl we only need to evlute f t the end nd strting points uh integrls re pth independent To test if n integrl qulifies neessry nd suffiient ondition (in the plne) is M = N x If this is stisfied then f n be reovered by the following steps: 1 Integrte M with respet to x (ie, tret ys s onstnts) This gives us f up to onstnt funtion, ie, (y) (so onstnt with respet to x) 2 Differentite wht we now hve with respet to y nd set it equl to N This gives n eqution (y) = (stuff with y) whih we n use to solve for (y) nd thus reover f ompletely A similr result nd proess works in higher dimensions Green s Theorem Green s Theorem lets us relte the integrl over plnr region,, with n integrl on the boundry, We will del with regions in whih the boundries re oriented ounterlokwise (ie, we wnt to hve the interior of the region to our left if we wlked long the boundry urve) Then we hve ( N M + N dy = x M ) da This even work in regions with holes The onvenient thing is tht this llows us to tke omplited line integrl nd express it s (potentilly) less omplited integrl over region We hve the following F n ds = F da the first integrl looks t the integrl over the boundry of F dotted with the unit norml vetor (oriented
out); the seond integrl looks t the integrl over the interior of the divergene of F We similrly hve the following ( ) F T ds = F k da the first integrl looks t the integrl over the boundry of F dotted with the unit tngent vetor; the seond integrl looks t the integrl over the interior of the nontrivil prt of the url of F, where we n think of F = M, N, 0 urfe integrls In order to get generliztions of the lst two results bove we need to understnd how to del with surfe integrls, ie, f(x, y, z) d(a) G This generlizes by looking t surfe in more generl position o let G be n oriented surfe nd let G be the boundry of the surfe oriented ounterlokwise (with respet to stnding on the surfe in the diretion of n we should hve the interior of the surfe on our left hnd side s we move long the bounding urve in the diretion of its orienttion) F T ds = ( F) n d(a) G G Note tht reversing the orienttion will only hnge the vlue by sign where d(a) indites we re subdividing up the surfe into smll pthes nd this orresponds to the re of smll pth In the speil se when the surfe is z = g(x, y) over some region we n rewrite this integrl s n integrl over This is done by expressing everything, inluding the d(a) term, in terms of x nd y We hve f(x, y, z) d(a) G = f ( x, y, g(x, y) ) (g x ) 2 + (g y ) 2 + 1 da More generlly if surfe is prmeterized by the vribles u nd v, ie, x(u, v), y(u, v) nd z(u, v) then we hve f(x, y, z) d(a) G = f ( x(u, v), y(u, v), z(u, v) ) r u r v da Divergene Theorem This generlizes the two dimensionl vrition of sying tht the flux through the boundry, n be omputed by the divergene in the interior, F n d(a) = F dv As before n is unit norml vetor oriented out from the solid When the surfe is plne this n is the unit norml vetor of the plne; more generlly we n find n by using the grdient of the eqution tht defines the surfe tokes s Theorem This generlizes the two dimensionl vrition of wht hppens when we look t the integrl of F T