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AER210Multiple IntegralsIntegrals as a function of a ParameterDouble integrals Rectangular and nonrectangular regionsFubini’s Theorem: can switch order of integration whether do x or y firstextension fxgydxdy= fxdxgydyCareful on limits when switching order of integrationDraw the area1.2 Differentiability of an Integral wrt its parameterTheorem: have Fx=fx,ydy, then Fxdx=??xfx,ydyCan either pass in derivative first then integrate or integrate and then take the derivative Leibniz’s Rule: for the above theorem when the bounds of integration are a function of x (cannot pass in the derivative operator and can’t integrate f easily)Fxdx=φ1(x)φ2(x)??xfx,ydy+fx,φ2xφ'2x-fx,φ1(x)φ'1(x)1.3 Formal Definition of the Double IntegralRectangular region: divide into rectangles of area ΔA and pick evaluate the function at f(x*, y*) a point in this areafx,ydA= limm, n →∞i=1mj=1nfxi*,yj*?A if limit existsNon Rectangular regionTake two approaches, one where the ΔAi are all within region but leaving gaps and some where they are overlapping out to cover all of region. Take the minimum and the maximum respectively for the two approaches of f(x,y) Then if limm →∞di→0i=1mmi?Ai= limn →∞dj→0j=1nMj?Aj=L, L=fx,ydAWhere d is the “characteristic length” of ΔA (longest line drawn in area)Can estimate value of integral using the midpoint ruleAD= DdA1.4 Polar Coordinatesx=rcosθ, y=rsinθ although not limited to be in xy-plane dxdy=rdrdθ from area of a circular wedge πr2?θ2π inner and outer wedge limit ?r→0 Rf(x,y)dxdy=R'f(rcosθ, rsinθ)rdrdθ****Don’t forget that extra factor of r****1.5 Applications of Double IntegralsMass: m= Rλ(x,y)dA λ(x,y) is density Center of Mass: xm=My=Rx λ(x,y)dA ym= My=Ry λ(x,y)dA Moment of inertia: I= Rd2λ(x,y)dA where d is distance from axis of rotationd = x for rotation about y axis Iyd = y for rotation about x axis Ixd= x2+y2 for about origin Io = Ix + Iy 1.6 Surface AreaDerive by taking a small area on the surface that is small enough to be considered planar. Project on to xy-plane. Unit normal vector of S is gradient of f/ mag of f. makes and angle a with the k unit vectorS= RdS= RdAcosα cosα= k?n= k??f?f= fzfx2+fy2+fz2S= Rfx2+fy2+fz2fzdA or if f(x, y, z) = 0 can be written as x = g(x,y) S= Rgx2+gy2+1dA1.7 Triple IntegralsFormulation:Given f(x, y, z) continuous over region E, break into n subvolumes ΔVi with characteristic length di, choose a point (xi*, yi*, zi*) in ΔVi. for the sum i=1nfxi*,yi*,zi*?Vi provided lim (ΔVi ) = V as n -> ∞ limn →∞di→0i=1nfxi*,yi*,zi*?Vi= Efx,y,z dVIn Cartesian coordinates: Efx,y,z dV= abdxh1(x)h2(x)dyg1(x,y)g2(x,y)fx,y,zdz(alternate notation (using because word sucks) not product for 3 integrals al a fubini)VE= EdVImportant to draw region when formulating or changing order so you know the bounds of integrationDouble integral applications (mass, moment, inertia) can be extended to triple integral1.8 Cylindrical and Spherical Coordinates Cylindrical:Essential Polar coordinates with z unchangedx=rcosθ, y=rsinθ, z=z dxdydz=r drdθdz Efx,y,z dV= Efrcosθ,rsinθ,z r drdθdz Useful for cylinders and conesSpherical:x=ρsinφcosθ, y=ρsinφsinθ, z=ρcosφ dxdydz=ρ2sinφ dρdθdφ comes from shrinking “cube” with arc sides to be small enough that you can consider it’s volume to be rectangularEfx,y,z dV= Efρsinφcosθ,ρsinφsinθ,ρcosφ ρ2sinφ dρdθdφ ρ2=x2+y2+z2Useful for spheres and cones1.9 Taylor Series in Two Variablesfxo+?x,xo+?y=fxo,yo+fxxo,yo+fyxo,yo+12!fxxxo,yo?x2+2fxyxo,yo?x?y+fyyxo,yo?y2+ 13!fxxxxo,yo?x3+3fxxyxo,yo?x2?y+...where ?x=x- xo, ?y=y- yoCan figure out pattern from pascal’s triangleNumber of terms considered depends on the degree of the polynomial asked to approximate to**a lot of the time terms will be zero for the questions we’re given. Make a table to keep track of all the partial derivatives and their evaluations before plugging in1.10 Change of Variables – JacobiansGiven a one to one C1 (cnt 1st orderpartial derv)transformation T by x = g(u,v), y = h(u,v), the Jacobian is defined by: ?(x,y)?(u,v)= ?x?u?x?v?y?u?y?v ***always absor (if x and y not easily solved for ?(x,y)?(u,v)= ?(u,v)?(x,y)-1=?u?x?u?y?v?x?v?y-1 or in 3 variables ?(x,y, z)?(u,v,w)= ?x?u?x?v?x?w?y?u?y?v?y?w?z?u?z?v?z?w Rf(x,y)dxdy=Rg(u,v)?(x,y)?(u,v)dudvUseful when bounds of integration are difficult to define/work with. Change of coordinates defines a new area of integration that is ideally rectangular and easier to work with (example: polar or spherical coordinates (this is where the extra factors come from))Examples: hyperbola x2 – y2 = b. let u = x2 – y2or xy = a let u = xyreally any f(x,y) = c let f(x,y) = u, gives nice rectangular regionVector CalculusLine IntegralsSimilar to single integral except will be integrating along a curve insteadFormal DefinitionLet c: rt=xti+ytj+z(t)k be a continuous curve on t [a,b]Divide the curve into n segments of length ΔsiChoose a point xi*,yi*,zi* in Δsi and form the sum i=1nfxi*,yi*,zi*?silimn →∞?si→0i=1nfxi*,yi*,zi*?si= cfx,y,zdsds= ?x?t2+?y?t2+?z?t2dtcfx,y,zds=abfx(t),y(t),z(t)?x?t2+?y?t2+?z?t2dtLine integral wrt arc lengthLine integrals of Vector Fields:cF?dr=cF?Tds=abFr(t)?r'(t)dt since T= r'(t)r'(t) and ds= r'(t)dtSometimes written cPdx+Qdy+Rdz F=(P, Q, R)Could be given two points, in which case need to parametrize a line between themFundamental Theorem of Line Integralsc?f(r)?dr=frb-fra by FTC using ddtfrt=?frtr'(t)Path independent, doesn’t matter the curve as long as end points are the sameIf a = b (ie a closed curve) then the integral is zeroCan save time and simplify integrals by check if F can be written as the gradient of some function f.If F can be written as ?f then we say its is a “conservative vector field”Each line integral is path independentEach line integral along a closed curve (every closed curve) is 0.Can check if F = (P, Q) is conservative using Py = Qx since fxy = fyx Also works for F = (P, Q, R)If F = (P, Q, R) (or F = (P, Q), need to integrate P wrt x (need to add a fn g(y,z)) then differentiate wrt y and see if gy matches with R (take extra integration constant to be 0) (choice of variable/component could be different)Green’s TheoremLet C be a positively orientated (counter clockwise), piecewise-smooth, simple (distinct orientation, doesn’t cross itself) closed curve in the plane and let D be the simple region (all horizontal and vertical lines between two boundary points lay entirely within the region). If P and Q have continuous partial derivatives on an open region contain D:cPdx+Qdy= D?Q?x-?P?ydACan be used to evaluate hard line integrals and an easier area integral or vice versaProof: Show cQdy= D?Q?xdA and cPdx= -D?P?ydA parameterize c1 and c3 as y = g1(x) and y=g2(x)The double integral becomes abPx,g2x-Px,g1xdx by FTCLine integrals c2 and c4 = 0 bc x constantc1Px,ydx= abPx,g1x, c3Px,ydx= -abPx,g2x cPdx=abPx,g1x -abPx,g2x=-D?P?ydATo prove the other term just switch x and y and repeatCan extent to non simple regions by splitting into simple regions and taking the union of the regions and the curves (see that the boundary curve has to go in the opposite direction for one region to maintain orientation, so the two integrals will cancel each other in the union and you’ll just be left with the boundary curve)To calculate area:Since A = D1dA need ?Q?x-?P?y=1Either P = 0 and Q = x, P = -y and Q = 0, or P =-y/2 and Q = x/2Thus:A = cxdy=c-ydx=12cxdy-ydxCommon question is to “verify Green’s Theorem” for a region is a curve. Have to compute both the area and line integral and show they are equal.Parametric Surfaces and Surface AreaJust like a curve, a surface can be parameterizeds:ru,v=xu,vi+yu,vj+zu,vk Example: a sphere of radius a: ru,v=acosusinvi+asinusinvj+acosvk Fixing u to v gives you a curve embedded in the surface?ru,v?u is tangent to the curve and therefore tangent to the surfaceThe vectors ru(u, v) and rv (u, v) span the tangent plane to the surfacen=ru× rv **direction will come into play laterRecall that equation of plane at point (a, b, c) is given by n? <x-a, y-b, z-c> =0Sometimes the parameterization is as simple as x = u, y= vSurface areaAS= Dru× rvdA is the projection onto uv plane (xy plane?)Comes from area of a parallelogram with sides ?uru* & ?vrv*If z = f(x, y), (x,y) in D and f has continuous partials, then can paramaterize as x = x, y = y, z = f(x,y)A(S)= D1+zx2+zy2dASurface IntegralsSfx,y,zdS=Sf(r(u,v))ru× rvdudv= Sf(xu,v, yu,v, zu,v)ru× rvdudvAnalogous to surface area where f = 1. Again, special case where z = g(x,y)Sf(x, y, gx,y)1+zx2+zy2dAFlux Integral: SF?dS=SF?ndSSince n=ru× rvru× rv and dS= ru× rvdudvSF?dS= SFru,v?ru× rvdudvNOTE the direction of n will matter depending if the surface is orientated outwards or inwards. May need to use -n after calculating Curl and DivergenceRecall Gradient of f: ?f=?f?x, ?f?y,?f?z Turns a scalar field into a vector field“Nable times f” Divergence: divF??F= ??x, ??y,??z?P, Q, R=Px+Qy+RzTurns a vector field into a scalar field “Nable dotted with F”Measure of how much near by vectors are changing in magnitudeIf divF = 0, call F “incompressible”Curl: curl F?×F= ??x, ??y,??z×P, Q, R=(Ry-Qz, Pz-Rx, Qx-Py)“Nable crossed with F”Turns vector field into another vector field If curl F = 0, call F “irrotational” curl ?f= 0, so if F is conservative, curl F = 0div curl F = 0Divergence TheoremLet E be closed region in three space and let S be the boundary of surface of E with positive (outward) orientation. Let F be a vector field which is continuous and differentiable on an open interval containing E. Then:SF?ds=E??FdVFor proof, show SR k?n dS=ERzdV and P and x^ and Q and j^Make region where top and bottom sides are surfaces defined by fn of x and y only. Other 4 sides are vertical so the surface integral is 0, others are R(x,y,u(x,y)dA projected onto the region D in the xy-plane. The sign of the bottom surface integral will be negative. Volume integral simplifies to R(x,y,u2(x,y)dA - R(x,y,u1(x,y)dA integrated over the region D by the FTCStoke’s TheoremLet S be an oriented piecewise smooth surface in three-dimensional space that is bounded by a simple, closed, piecewise smooth curve c with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region containing S. Then:CF?dr=S?×F?ds Note: the surface integral will be the same regardless of the surface as long as the boundary curve is the same Proof for case where z = g(x,y) Define c1 and D as the projection of S and C in the xy-plane Line integralUse chain rule to eliminate z’(t)Use green’s theorem to turn into area integral (simplify with chainrule and product rule)Surface integralParameterize as x=x y=y z=g(x,y)Rearranging after the dot product give the same as the line integral**have to check orientation of surface during a question to make sure normal orientated outwards Common question of Stoke’s and Divergence Theorem is to verify it for a Region and surface or surface and curve. That is evaluate both integrals and show they are equal.Fluid Mechanics IntroductionDifference between solid and fluid is response to shear stress (fluid continues to deform while solid stops after a short distance)Statistical approach: small length scales (atomic scale, single atoms or molecules); continuum approach: large length scales (atomic collisions don’t matter, number of particles inside the volume doesn’t really change)Knudsen Number: Kn = λ/L mean free path(distance between two atoms)/length scaleKn < 0.01 continuum. Kn > 1 statistical Slip flow: particles in a fluid are allowed to bounce/slide along wallKn < 0.01 no slip: V fluid at wall = V of wallKn > 1 slip: V fluid different than wallSpecific Mass: ρ (m/V ) Specific Weight: γ (W/V = ρg) Specific Gravity: ρ/ρwaterCompressible fluid density not constant, Incompressible fluid density constantBody forces: act on whole fluid element (gravity, elec mag)Surface forces: act on surface (stresses) normal stress perp to surface, shear parallel5270556451500Tensor of stress: σxxτxyτxzτyxσyyτyzτzxτzyσzz σ normal stress τ shear stressFirst index indicates the plane normal vector that contains the stress (it the normal vector of the surface its acting on)Second index refers to the direction of the axis parallel to the stressOrientated so that the stress values are always positive (ie both indexes are negative or both positive)Viscosity relates the rate of deformation to the shear stressτij= μ?ui?xj+?uj?xi i or j = 1: x and u 2: y and v 3: z and wi=j normal stress, else shearμ: dynamic/absolute viscosity366141017716500ν = μ/ρ: kinematic viscosityNewtonian fluid (linear relation between shear and deformation) Shear thinning and shear thickeningBinham plastic (deformation doesn’t start right away)Compressibility:Ev=-dPdVV=dPdρρMore incompressible -> higher EvHydrostaticsPressure at a point is independent of the orientation of the surface (ie direction)Always acts normal to the surface Only consider forces due to pressure and gravitational forces for hydrostaticsGoverning equation: derived from taylor expansion of difference in pressure on opposite faces of rectangular cube fluid element.-?P+ ρg=0 => no variation in x and y dPdz= -ρgPressure varies linearly with depthForce do to atmospheric pressure on an object is 0 All the components cancel out is 0 same for y and zForce is pressure times areaBuoyancy Archimedes Principle F = -ρgVBuoyant force acts on the center of mass of the displaced fluidStability depends on whether this centre of mass is above or below the COM for the object.If above, when titled creates a moment that continues the tipping Fluid in rigid body motion:-?P+ ρg=ρa Can determine function for P by integrating each component equation and applying boundary conditions to get the constantsFluid DynamicsLagrangian Approach: moving measurement: control massEulerian approach: stationary measurement: control volumeTotal derivative: DDt≡ ??t+V??Stream line: curve that is locally tangent to velocity vectordxu=dyv=dzw can determine velocity profilePathline: trajectory of fluid parcel for a period of time Streak line: curve that connects all the fluid parcels that passed a given point in spaceViscous flow: shear stresses important (eg flow near a wall because of no slip)Inviscid flow: shear stresses negligible (flow away from wall)Volume flowrate: Mass flowrate: Steady state: time derivative is 0Flow dimensionality: number of velocity componentsFlow inside the potential core in inviscid, outside viscousUmax > U∞Laminar vs Turbulent FlowRe ≡inertial forcesviscous forces=ρUDμ=UDνConservation of Mass or for incompressible fluidEuler and Bernoulli Equation:Assumptions: Flow is steadyStresses are negligible (inviscid)Incompressible fluidRadius of curvature for streamline is apprx constantNo work being done or heat being added parallel to streamline; perp to streamlineReynold’s Transport Theoremrelates Lagrangian approach to Eulerian approachConservation of massLet B = m. b = 1. DM is 0 for a control mass Limit dV goes to 0By product ruleIncompressible just have the nabla dot V = 0Continuity: ?u?x+?v?y+?w?z=0 Conservation of MomentumB = mV , b = V, NOTE this is the forces applied TO the CV not BY it. Open Channel FlowsFroude Number Fr2 ratio of inertial force to force due to hydrostatic pressure<1 subcritical regime, gravity dominates inertial force=1 critical >1 supercritical inertial force larger than gravitationalSpeed of wave propagationConservation of mass gives Conservation of momentum **neglect higher order termsCombining gives Flow over a bumpUse Bernoulli and governing equation for hydrostatics to derive that the hydraulic jump depends on the Froude number and the high of the hump (indent or bump)Compressible FlowsRecall First and Second Laws of thermodynamicsSpeed of sound propagationConservation of massConservation of momentum givesCombining gives which simplifies to if we assume and ideal gas and an isentropic processUsing compressibility have Why speed of sound is faster in waterQuasi one-dimensional flowsReynolds transport theorem with energyCan usually neglect the change in gravity then the constant is the stagnation enthalpyrelation of change in h to change in T gives using the definition of the Mach number (M = V/c) and definition of c for and ideal gas with the relationship between R = cp – cv and definition of γThen relationship between T P and ρDifferential Analysis of FluidsModes of fluid motion Angular velocity of fluid Curl of velocity field is called vorticityCurl = 0 => irrotational Navier-Stokes EquationsStart with F= maUtilize acceleration is the total derivative of velocityForces: gravity and shear and normal stressesSetting ν = 0 gives the Bernoulli equation (since Bernoulli assumes inviscid flow)Use with continuity equation when solvingLots of terms are usually 0 in problems based on assumptions and boundary/flow conditions ................
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