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 Plnr Are In the limit s d the totl number of pnels A = b d = b f()d olume of Solid of evolution ) Disk Method : rotte = f() bout the is to form solid. =f() b rotte round -is The disk hs volume of = π. The totl volume between nd b cn be determined s: = 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 = (π ) The volume of the solid is determined b solving the integrl =f() = 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
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 4
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 5
Surfce Ares from Double Integrls eding Trim 13.3 Ares nd olumes 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 6
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. 7
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 = -/ 8
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 + = 9
Triple Integrls eding Trim 13.8 Triple Integrls nd Triple Iterted Integrls 13.9 olumes Assignment web pge ssignment #8 olume Clcultions in Crtesin Coordintes The triple integrl cn be identified s d d d }{{} d volume element or f d d d dd up the d 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 10
Strt with volume element t rbitrr (,, ) in spce inside d Build up slice - sum columns over, keeping, fied. d = 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 = 4 6 0 0 0 ( 6 =0 ) d ddd d d = = 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 11
olume 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 d = 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: d = (r sin φdφ) rdrdθ }{{}}{{} re height bsed on links to Crtesin coordintes d = da dr = r sin f dr d f d 1
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 d + 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, + + = 13
Moments of Are/ Mss / olume 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 14
3-D cse: We use the sme bsic ides but the bsic element is now = ddd -D Objects 3-D Objects plte of thickness t region volume d t rbitrr (,,) re element da volume is t da mss is (,) t da Quntities of interest in pplictions such s dnmics. Are: A = da ( 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 (,,) d Quntities of interest in pplictions such s dnmics. olume: = d Mss: M = ρ(,, )d where ρ(,, ) = densit of mteril in (kg/m 3 ) t point (,, ) Centroid = geometricl center of object = d 1st moment of volume bout plne = d 1st moment of volume bout plne = d 1st moment of volume bout plne Center of Mss: useful in dnmics problems ρ(,, )d c = M similr formuls for c nd c nd Moments of Are nd Mss: Polr Moments of Inerti volume moment bout: is J = ( + )d mss moment bout: is J = ( + )ρ(,, )d (similr formuls for J bout the is) nd (similr formuls for J bout the is) 15
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 16