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
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
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
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 6 +... }{{} 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
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
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 6 +... }{{} 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 6 +... I ect = (4I trp I trp1 ) 3 + ( ) h4 3 + ( ) h6 3 +... 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
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
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 0 + = 1 ( 1) + = 1 8
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
4. get the unit norml vector s follows ˆn = F F = î f ĵ f + ˆk ( ) f ( ) f + + 1 5. 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 = 1 + + }{{} F dd }{{} da Given the surfce = f(, ), the surfce re is S = ( ) f ( ) f 1 + + 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
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
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
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
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 = 4 6 0 0 0 ( 6 =0 ) d ddd d d V = = 6 4 0 0 0 (6 )dd = 4 d 1 0 0 6 4 ( 4 ) d (4 )d = 6π 8 3 16.18 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
Volume Clcultions in Clindricl nd Sphericl Coordintes eding Trim 13.11 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
r = + + θ = tn 1 (/) ( ) φ = cos 1 + + 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
Moments of Are/ Mss / Volume eding Trim 13.5 Centres of Mss nd Moments of Inerti 13.10 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
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
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