AER210Multiple Integrals1.1 Integrals as a function of a Parameter

Double integrals o Rectangular and nonrectangular regions

Fubini’s Theorem: can switch order of integration whether do x or y firsto extension ∬ f ( x ) g ( y ) dxdy=∫ f ( x )dx∫ g ( y )dyo Careful on limits when switching order of integration

Draw the area

1.2 Differentiability of an Integral wrt its parameter

Theorem: have F ( x )=∫ f ( x , y ) dy, then F ( x )dx

=∫ ∂∂ x

f ( x , y )dy

o Can 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)

o F ( x )dx

= ∫φ1 (x)

φ2 (x)∂

∂ xf ( x , y ) dy+f ( x ,φ2 ( x )) φ'

2 ( x )− f ( x , φ1(x ))φ '1(x)

1.3 Formal Definition of the Double Integral Rectangular region:

o divide into rectangles of area ΔA and pick evaluate the function at f(x*, y*) a point in this area

o ∬ f ( x , y ) dA= limm, n→∞

∑i=1

m

∑j=1

n

f (x i¿ , y j

¿) ∆ A if limit exists

Non Rectangular regiono Take 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)

o Then if limm→ ∞d i→ 0

∑i=1

m

mi ∆ A i= limn →∞d j→ 0

∑j=1

n

M j ∆ A j=L, L=∬ f ( x , y ) dA

o Where d is the “characteristic length” of ΔA (longest line drawn in area) Can estimate value of integral using the midpoint rule

A ( D )=∬D

❑

dA

1.4 Polar Coordinates x=r cosθ , y=rsin θ although not limited to be in xy-plane dxdy=rdrdθ

o from area of a circular wedge π r2 ∆ θ2 π inner and outer wedge limit ∆ r →0

∬R

❑

f (x , y)dxdy=∬R'

❑

f (r cosθ , r sin θ)rdrdθ

o ****Don’t forget that extra factor of r****

1.5 Applications of Double Integrals

Mass: m=∬R

❑

λ(x , y)dA λ (x , y ) is density

Center of Mass: x m=M y=∬R

❑

x λ(x , y)dA y m=M y=∬R

❑

y λ (x , y )dA

Moment of inertia: I=∬R

❑

d2 λ (x , y )dA where d is distance from axis of rotation

o d = x for rotation about y axis Iy

o d = y for rotation about x axis Ix

o d=√x2+ y2 for about origin Io = Ix + Iy

1.6 Surface Area Derive 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 vector

S=∬R

❑

dS=∬R

❑ dA|cos α| |cos α|=|k ∙n|=|k ∙∇ f|

‖∇ f‖=

|f z|√ f x

2+f y2+ f z

2

S=∬R

❑ √ f x2+ f y

2 +f z2

|f z|dA or if f(x, y, z) = 0 can be written as x = g(x,y)

S=∬R

❑

√gx2+gy

2 +1dA

1.7 Triple Integrals Formulation:

o 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=1

n

f ( x i¿ , y i

¿ , z i¿ )∆ V i provided lim (ΔVi ) = V as n -> ∞

limn→ ∞di → 0

∑i=1

n

f (x i¿ , y i

¿ , zi¿)∆ V i=∭

E

❑

f ( x , y , z ) dV

In Cartesian coordinates: ∭E

❑

f (x , y , z ) dV=∫a

b

dx ∫h1( x)

h2( x)

dy ∫g1 (x , y)

g2 (x , y)

f ( x , y , z ) dz

o (alternate notation (using because word sucks) not product for 3 integrals al a fubini)

o V ( E )=∭E

❑

dV

Important to draw region when formulating or changing order so you know the bounds of integration

Double integral applications (mass, moment, inertia) can be extended to triple integral

1.8 Cylindrical and Spherical Coordinates Cylindrical:

o Essential Polar coordinates with z unchangedo x=r cosθ , y=rsin θ , z=z dxdydz=r drdθdz

o ∭E

❑

f (x , y , z ) dV=∭E

❑

f (r cosθ , rsin θ , z ) r drdθdz

o Useful for cylinders and cones

Spherical:o x=ρ sin φ cosθ , y=ρsin φ sinθ , z=ρcos φo dxdydz=ρ2sin φ dρdθdφ comes from shrinking “cube” with arc sides to be

small enough that you can consider it’s volume to be rectangular

o ∭E

❑

f (x , y , z ) dV=∭E

❑

f ( ρsin φ cosθ , ρsin φ sinθ , ρ cosφ ) ρ2sin φ dρdθdφ

o ρ2=x2+ y2+z2

o Useful for spheres and cones

1.9 Taylor Series in Two Variables

f ( xo+∆ x, xo+∆ y )=f ( xo , yo )+( f x ( xo , yo )+f y ( xo , yo ))+ 12 ! ( f xx ( xo , yo ) ∆ x2+2 f xy ( xo , yo ) ∆ x ∆ y+f yy ( xo, yo ) ∆ y2 )+ 1

3 ! ( f xxx ( xo , yo ) ∆ x3+3 f xxy ( xo , yo ) ∆ x2 ∆ y+... ) where ∆ x=x−xo , ∆ y= y− yo

Can figure out pattern from pascal’s triangle Number 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 in

1.10 Change of Variables – Jacobians Given a one to one C1 (cnt 1st orderpartial derv)transformation T by x = g(u,v), y = h(u,v),

o the Jacobian is defined by: ∂(x , y )∂(u , v )

=|∂ x∂u

∂ x∂ v

∂ y∂u

∂ y∂ v

| ***always abs

or (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

o 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

| ∬

R

❑

f (x , y)dxdy=∬R

❑

g(u , v) ∂ (x , y )∂ (u , v)

dudv

Useful 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))

o Examples: hyperbola x2 – y2 = b. let u = x2 – y2

or xy = a let u = xy really any f(x,y) = c let f(x,y) = u, gives nice rectangular region

Vector Calculus2.1 Line Integrals

Similar to single integral except will be integrating along a curve instead Formal Definition

o Let c : r ( t )=x ( t ) i+ y (t ) j+z (t) k be a continuous curve on t [a,b]o Divide the curve into n segments of length Δsi

o Choose a point (x i¿ , y i

¿ , zi¿) in Δsi and form the sum ∑

i=1

n

f (x i¿ , y i

¿ , z i¿ )∆ s i

o limn→ ∞

∆ si→ 0

∑i=1

n

f ( xi¿ , y i

¿ , zi¿) ∆ si=∫

c

❑

f ( x , y , z )ds

ds=√( ∂ x∂ t )

2

+( ∂ y∂t )

2

+( ∂ z∂t )

2

dt

∫c

❑

f ( x , y , z ) ds=∫a

b

f ( x (t ), y (t) , z (t))√( ∂ x∂ t )

2

+( ∂ y∂ t )

2

+( ∂ z∂ t )

2

dt

o Line integral wrt arc length Line integrals of Vector Fields:

o ∫c

❑

F ∙ d r=∫c

❑

F ∙T ds=∫a

b

F ( r (t)) ∙ r '(t)dt

since T=r ' (t )|r ' (t )| and ds=|r ' (t)|dt

o Sometimes written ∫c

❑

Pdx+Qdy+Rdz F=(P , Q ,R)

o Could be given two points, in which case need to parametrize a line between them

2.2 Fundamental Theorem of Line Integrals

∫c

❑

∇ f (r¿)∙ d r=f ( r (b ) )−f ( r (a ) ) ¿ by FTC using ddt

f ( r ( t ) )=∇ f (r ( t ) ) r ' (t)

o Path independent, doesn’t matter the curve as long as end points are the sameo If a = b (ie a closed curve) then the integral is zeroo Can 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”

o Each line integral is path independento Each line integral along a closed curve (every closed curve) is 0.o Can check if F = (P, Q) is conservative using Py = Qx since fxy = fyx

Also works for F = (P, Q, R)o 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)

2.3 Green’s Theorem Let 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:

o ∮c

❑

Pdx+Qdy=∬D

❑ ∂ Q∂ x

−∂ P∂ y

dA

o Can be used to evaluate hard line integrals and an easier area integral or vice versa

Proof:

o Show ∫c

❑

Qdy=∬D

❑ ∂ Q∂ x

dA and ∫c

❑

Pdx=−∬D

❑ ∂ P∂ y

dA

parameterize c1 and c3 as y = g1(x) and y=g2(x)

The double integral becomes ∫a

b

[ P (x , g2 ( x ) )−P ( x , g1 ( x ) ) ]dx by FTC

Line integrals c2 and c4 = 0 bc x constant

∫c1

❑

P ( x , y ) dx=∫a

b

P (x ,g1 ( x ) ) ,∫c3

❑

P ( x , y ) dx=−∫a

b

P ( x , g2 ( x ) )

∫c

❑

Pdx=∫a

b

P ( x ,g1 ( x ))−∫a

b

P ( x , g2 ( x ) )=−∬D

❑ ∂ P∂ y

dA

To prove the other term just switch x and y and repeat Can 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:

o Since A = ∬D

❑

1 dA need ∂Q∂ x

− ∂ P∂ y

=1

Either P = 0 and Q = x, P = -y and Q = 0, or P =-y/2 and Q = x/2o Thus:

A = ∮c

❑

xdy=∮c

❑

− ydx=12∮c

❑

xdy− ydx

Common 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.

2.4 Parametric Surfaces and Surface Area Just like a curve, a surface can be parameterized

o s : r (u , v )=x (u , v ) i+ y (u , v ) j+z (u , v ) k Example: a sphere of radius a:

r (u ,v )=a cosu sin v i+a sinu sin v j+acos v k o Fixing u to v gives you a curve embedded in the surface

o ∂ r (u , v )∂ u

is tangent to the curve and therefore tangent to the surface

o The vectors ru(u, v) and rv (u, v) span the tangent plane to the surface

o n=ru ×rv **direction will come into play later Recall that equation of plane at point (a, b, c) is given by n ∙<( x−a ) , ( y−b ) , (z−c )>¿0

o Sometimes the parameterization is as simple as x = u, y= v Surface area

o A (S )=∬D

❑

‖ru × rv‖dA is the projection onto uv plane (xy plane?)

Comes from area of a parallelogram with sides ∆ u ru¿∧∆ v r v

¿

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)=∬D

❑

√1+ zx2+ z y

2 dA

2.5 Surface Integrals

∬S

❑

f ( x , y , z )dS=∬S

❑

f (r (u , v ))‖r u× rv‖dudv=∬S

❑

f ( x (u ,v ) , y (u , v ) , z (u , v ))‖ru ×r v‖dudv

o Analogous to surface area where f = 1. o Again, special case where z = g(x,y)

∬S

❑

f (x , y , g ( x , y ))√1+z x2+z y

2 dA

Flux Integral: ∫S

❑

F ∙ d S=∫S

❑

F ∙ n dS

o Since n=ru ×rv

‖ru ×rv‖ and dS=‖ru× rv‖dudv

o ∬S

❑

F ∙ d S=∬S

❑

F ( r (u , v )) ∙ ( ru ×r v) dudv

NOTE the direction of n will matter depending if the surface is orientated outwards or inwards. May need to use -n after calculating

2.6 Curl and Divergence

Recall Gradient of f: ∇ f =( ∂ f∂ x

, ∂ f∂ y

, ∂ f∂ z )

o Turns a scalar field into a vector fieldo “Nable times f”

Divergence: divF

o ∇ ∙ F=¿ ( ∂∂ x

, ∂∂ y

, ∂∂ z )∙ ( P ,Q , R )=Px+Q y+R z

o Turns a vector field into a scalar field o “Nable dotted with F”o Measure of how much near by vectors are changing in magnitudeo If divF = 0, call F “incompressible”

Curl: curl F

o ∇× F=¿ ( ∂∂ x

, ∂∂ y

, ∂∂ z )× (P , Q , R )=(Ry−Q z , P z−R x ,Q x−Py )

o “Nable crossed with F”

o Turns vector field into another vector field o If curl F = 0, call F “irrotational” o curl ∇ f = 0, so if F is conservative, curl F = 0o div curl F = 0

2.7 Divergence Theorem Let 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:

o ∯S

❑

F ∙ d s=∭E

❑

∇ ∙ F dV

For proof, show ∯S

❑

R k ∙ n dS=∭E

❑

R z dV and P and x^ and Q and j^

o 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.

o 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 FTC

2.8 Stoke’s Theorem Let 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:

o ∫C

❑

F ∙ d r=∬S

❑

∇× F ∙d s

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)o Define c1 and D as the projection of S and C in the xy-plane o Line integral

Use chain rule to eliminate z’(t) Use green’s theorem to turn into area integral (simplify with chainrule and

product rule)o Surface integral

Parameterize as x=x y=y z=g(x,y) Rearranging after the dot product give the same as the line integral

o **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 Introduction

Difference 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 scaleo Kn < 0.01 continuum. Kn > 1 statistical

Slip flow: particles in a fluid are allowed to bounce/slide along wallo Kn < 0.01 no slip: V fluid at wall = V of wallo Kn > 1 slip: V fluid different than wall

Specific Mass: ρ (m/V ) Specific Weight: γ (W/V = ρg) Specific Gravity: ρ/ρwater

Compressible fluid density not constant, Incompressible fluid density constant Body forces: act on whole fluid element (gravity, elec mag) Surface forces: act on surface (stresses) normal stress perp to surface, shear parallel

Tensor of stress: [σ xx τxy τ xz

τ yx σ yy τ yz

τ zx τ zy σ zz] σ❑ normal

stress τ shear stress

First 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 stress

o τ ij=μ ( ∂u i

∂x j+

∂ u j

∂ x i) i or j = 1: x and u 2: y and v 3: z and w

o i=j normal stress, else shearo μ: dynamic/absolute viscosityo ν = μ/ρ: kinematic viscosity

Newtonian fluid (linear relation between shear and deformation)

o Shear thinning and shear thickeningo Binham plastic (deformation doesn’t start

right away)

Compressibility:E v=

−dPdVV

=dPdρρ

o More incompressible -> higher Ev

Hydrostatics Pressure at a point is independent of the orientation of the surface (ie direction)

o Always acts normal to the surface Only consider forces due to pressure and gravitational forces for hydrostatics Governing equation: derived from taylor expansion of difference in pressure on opposite

faces of rectangular cube fluid element.

o −∇P+ ρ g=0 => no variation in x and y dPdz

=−ρg

Pressure varies linearly with depth

Force do to atmospheric pressure on an object is 0 o All the components cancel out

is 0 same for y and z Force is pressure times area Buoyancy

o Archimedes Principle

o F = -ρgVo Buoyant force acts on the center of mass of the displaced fluid

Stability 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:

o −∇P+ ρ g=ρ a Can determine function for P by integrating each component equation and

applying boundary conditions to get the constants

Fluid Dynamics Lagrangian Approach: moving measurement: control mass Eulerian approach: stationary measurement: control volume

Total derivative: DDt

≡ ∂∂ t

+V ∙∇

Stream line: curve that is locally tangent to velocity vector

o dxu

=dyv

=dzw

can determine velocity profile

Pathline: trajectory of fluid parcel for a period of time Streak line: curve that connects all the fluid parcels that passed a given point in space Viscous 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 0 Flow dimensionality: number of velocity components

o Flow inside the potential core in inviscid, outside viscouso Umax > U∞

Laminar vs Turbulent Flow

o ℜ≡ inertial forcesviscous forces

= ρUDμ

=UDν

Conservation of Mass

o

o or for incompressible fluid Euler and Bernoulli Equation:

o Assumptions: Flow is steady Stresses are negligible (inviscid) Incompressible fluid Radius of curvature for streamline is apprx constant No work being done or heat being added

o parallel to streamline; perp to streamline Reynold’s Transport Theorem

o relates Lagrangian approach to Eulerian approach

o Conservation of mass Let B = m. b = 1. DM is 0 for a control mass

Limit dV goes to 0

By product rule Incompressible just have the nabla dot V = 0

o Continuity: ∂ u∂ x

+ ∂ v∂ y

+ ∂ w∂ z

=0

o Conservation of Momentum B = mV , b = V,

NOTE this is the forces applied TO the CV not BY it.

Open Channel Flows

Froude Number Fr2 ratio of inertial force to force due to hydrostatic pressure

o <1 subcritical regime, gravity dominates inertial forceo =1 critical o >1 supercritical inertial force larger than gravitational

o Speed of wave propagation

o

o Conservation of mass gives

o Conservation of momentum **neglect higher order terms

o Combining gives Flow over a bump

o Use 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 Flows Recall First and Second Laws of thermodynamics

Speed of sound propagation

o

o Conservation of mass

o Conservation of momentum gives

o Combining gives which simplifies to if we assume and ideal gas and an isentropic process

o Using compressibility have Why speed of sound is faster in water

Quasi one-dimensional flowso Reynolds transport theorem with energy

o Can usually neglect the change in gravity

then the constant is the stagnation enthalpy

o relation of change in h to change in T gives o 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 γ

Then relationship between T P and ρ

Differential Analysis of Fluids Modes of fluid motion

o

Angular velocity of fluid o Curl of velocity field is called vorticityo Curl = 0 => irrotational

Navier-Stokes Equationso Start with F= mao Utilize acceleration is the total derivative of velocityo Forces: gravity and shear and normal stresses

oo Setting ν = 0 gives the Bernoulli equation (since Bernoulli assumes inviscid flow)o Use with continuity equation when solving

Lots of terms are usually 0 in problems based on assumptions and boundary/flow conditions