AER210
Multiple Integrals
1.1 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 first
o extension ∬ 𝑓(𝑥)𝑔(𝑦)𝑑𝑥𝑑𝑦 = ∫ 𝑓(𝑥)𝑑𝑥 ∫ 𝑔(𝑦)𝑑𝑦
o Careful on limits when switching order of integration
 Draw the area
1.2 Differentiability of an Integral wrt its parameter


have 𝐹(𝑥) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦, then
Theorem:
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
𝜑 (𝑥) 𝜕
= ∫𝜑 2(𝑥)
1
𝜕𝑥
𝑓(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
𝑛
∗ ∗
o ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = lim ∑𝑚
𝑖=1 ∑𝑗=1 𝑓(𝑥𝑖 , 𝑦𝑗 )∆𝐴 if limit exists
𝑚,𝑛 →∞

Non Rectangular region
o 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)
∑𝑚 𝑚 ∆𝐴𝑖 = 𝑛lim
∑𝑛 𝑀 ∆𝐴𝑗 = 𝐿, 𝐿 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴
o Then if 𝑚lim
→∞ 𝑖=1 𝑖
→∞ 𝑗=1 𝑗
𝑑𝑖 →0
𝑑𝑗 →0

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 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃
𝑑𝑥𝑑𝑦 = 𝑟𝑑𝑟𝑑𝜃
o

although not limited to be in xy-plane
∆𝜃
from area of a circular wedge 𝜋𝑟 2 2𝜋 inner and outer wedge limit ∆𝑟 → 0
∬𝑅 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦 = ∬𝑅′ 𝑓(𝑟 cos 𝜃 , 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

Center of Mass: 𝑥̅ 𝑚 = 𝑀𝑦 = ∬𝑅 𝑥 𝜆(𝑥, 𝑦) 𝑑𝐴

Moment of inertia: 𝐼 = ∬𝑅 𝑑 2 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 = √𝑥 2 + 𝑦 2 for about origin Io = Ix + Iy
𝜆(𝑥, 𝑦) is density
𝑦̅𝑚 = 𝑀𝑦 = ∬𝑅 𝑦 𝜆(𝑥, 𝑦) 𝑑𝐴
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

𝑆 = ∬𝑅
√𝑓𝑥2 +𝑓𝑦2 +𝑓𝑧2
|𝑓𝑧 |
𝑑𝐴
|cos 𝛼|
𝑑𝐴
|cos 𝛼| = |𝑘 ∙ 𝑛| =
|𝑘∙∇𝑓|
‖∇𝑓‖
=
|𝑓𝑧 |
√𝑓𝑥2 +𝑓𝑦2 +𝑓𝑧2
or if f(x, y, z) = 0 can be written as x = g(x,y)
𝑆 = ∬𝑅 √𝑔𝑥2 + 𝑔𝑦2 + 1 𝑑𝐴
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
∑𝑛𝑖=1 𝑓(𝑥𝑖∗ , 𝑦𝑖∗ , 𝑧𝑖∗ )∆𝑉𝑖 provided lim (ΔVi ) = V as n -> ∞
lim ∑𝑛𝑖=1 𝑓(𝑥𝑖∗ , 𝑦𝑖∗ , 𝑧𝑖∗ )∆𝑉𝑖 = ∭𝐸 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉
𝑛 →∞
𝑑𝑖 →0

𝑏
ℎ (𝑥)
𝑔 (𝑥,𝑦)
In Cartesian coordinates: ∭𝐸 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∫𝑎 𝑑𝑥 ∫ℎ 2(𝑥) 𝑑𝑦 ∫𝑔 2(𝑥,𝑦) 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧
1
1
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 unchanged
o 𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃, 𝑧 = 𝑧
𝑑𝑥𝑑𝑦𝑑𝑧 = 𝑟 𝑑𝑟𝑑𝜃𝑑𝑧
o ∭𝐸 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑉 = ∭𝐸 𝑓(𝑟 cos 𝜃 , 𝑟 sin 𝜃 , 𝑧) 𝑟 𝑑𝑟𝑑𝜃𝑑𝑧
o Useful for cylinders and cones

Spherical:
o 𝑥 = 𝜌 sin 𝜑 cos 𝜃 , 𝑦 = 𝜌 sin 𝜑 sin 𝜃, 𝑧 = 𝜌 cos 𝜑
o 𝑑𝑥𝑑𝑦𝑑𝑧 = 𝜌2 sin 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(𝜌 sin 𝜑 cos 𝜃 , 𝜌 sin 𝜑 sin 𝜃, 𝜌 cos 𝜑) 𝜌2 sin 𝜑 𝑑𝜌𝑑𝜃𝑑𝜑
o 𝜌2 = 𝑥 2 + 𝑦 2 + 𝑧 2
o Useful for spheres and cones
1.9 Taylor Series in Two Variables

1
𝑓(𝑥𝑜 + ∆𝑥, 𝑥𝑜 + ∆𝑦) = 𝑓(𝑥𝑜 , 𝑦𝑜 ) + (𝑓𝑥 (𝑥𝑜 , 𝑦𝑜 ) + 𝑓𝑦 (𝑥𝑜 , 𝑦𝑜 )) + 2! (𝑓𝑥𝑥 (𝑥𝑜 , 𝑦𝑜 )∆𝑥 2 +
2𝑓𝑥𝑦 (𝑥𝑜 , 𝑦𝑜 )∆𝑥∆𝑦 + 𝑓𝑦𝑦 (𝑥𝑜 , 𝑦𝑜 )∆𝑦 2 ) +




1
3!
(𝑓𝑥𝑥𝑥 (𝑥𝑜 , 𝑦𝑜 )∆𝑥 3 +
3𝑓𝑥𝑥𝑦 (𝑥𝑜 , 𝑦𝑜 )∆𝑥 2 ∆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),
𝜕(𝑥,𝑦)
o the Jacobian is defined by:

𝜕(𝑢,𝑣)
=
or (if x and y not easily solved for
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑢
𝜕𝑣
𝜕𝑦
𝜕𝑤
𝜕𝑦 |
𝜕𝑦
o or in 3 variables
= ||
𝜕(𝑢,𝑣,𝑤)
𝜕𝑢
𝜕(𝑥,𝑦,𝑧)
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑤|
𝜕𝑧
𝜕𝑢
𝜕𝑣
𝜕𝑤
𝜕𝑥
𝜕𝑥
|𝜕𝑢
𝜕𝑦
𝜕𝑣
|
𝜕𝑦
𝜕𝑢
𝜕𝑣
𝜕(𝑥,𝑦)
***always abs
𝜕(𝑢,𝑣) −1
= (𝜕(𝑥,𝑦))
𝜕(𝑢,𝑣)
𝜕𝑢
𝜕𝑢 −1
𝜕𝑥
𝜕𝑦
|
𝜕𝑣
𝜕𝑥
𝜕𝑦
= |𝜕𝑣
𝜕(𝑥,𝑦)

∬𝑅 𝑓(𝑥, 𝑦) 𝑑𝑥𝑑𝑦 = ∬𝑅 𝑔(𝑢, 𝑣) 𝜕(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 Calculus
2.1 Line Integrals
 Similar to single integral except will be integrating along a curve instead
 Formal Definition
o Let 𝑐: 𝑟⃑(𝑡) = 𝑥(𝑡)𝑖̂ + 𝑦(𝑡)𝑗̂ + 𝑧(𝑡)𝑘̂ be a continuous curve on t [a,b]
o Divide the curve into n segments of length Δsi
o Choose a point (𝑥𝑖∗ , 𝑦𝑖∗ , 𝑧𝑖∗ ) in Δsi and form the sum ∑𝑛𝑖=1 𝑓(𝑥𝑖∗ , 𝑦𝑖∗ , 𝑧𝑖∗ )∆𝑠𝑖
o
lim ∑𝑛𝑖=1 𝑓(𝑥𝑖∗ , 𝑦𝑖∗ , 𝑧𝑖∗ )∆𝑠𝑖 = ∫𝑐 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠
𝑛 →∞
∆𝑠𝑖 →0
2
2
2

𝜕𝑥
𝜕𝑦
𝜕𝑧
𝑑𝑠 = √( 𝜕𝑡 ) + ( 𝜕𝑡 ) + (𝜕𝑡 ) 𝑑𝑡

∫𝑐 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫𝑎 𝑓(𝑥(𝑡), 𝑦(𝑡), 𝑧(𝑡))√( 𝜕𝑡 ) + ( 𝜕𝑡 ) + ( 𝜕𝑡 ) 𝑑𝑡

o Line integral wrt arc length
Line integrals of Vector Fields:
⃑⃑ ∙ 𝑑𝒓
⃑⃑ ∙ 𝑻𝑑𝑠 = ∫𝑏 𝑭
⃑⃑(𝒓
⃑⃑ = ∫ 𝑭
⃑⃑(𝑡)) ∙ ⃑⃑⃑⃑
o ∫ 𝑭
𝒓′(𝑡)𝑑𝑡
𝜕𝑥 2
𝑏
𝑐
𝑐

since 𝑻 =
𝜕𝑦 2
𝜕𝑧 2
𝑎
⃑⃑⃑⃑(𝑡)
𝒓′
|𝒓′
⃑⃑⃑⃑(𝑡)|
⃑⃑⃑⃑(𝑡)| 𝑑𝑡
and 𝑑𝑠 = |𝒓′
o Sometimes written ∫𝑐 𝑃𝑑𝑥 + 𝑄𝑑𝑦 + 𝑅𝑑𝑧 𝐹⃑ = (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


𝑑
⃑⃑⃑⃑(𝑡)
⃑⃑(𝑡)) = ∇𝑓(𝑟⃑(𝑡))𝒓′
∫𝑐 ∇𝑓(𝑟⃑) ∙ 𝑑𝑟⃑ = 𝑓(𝑟⃑(𝑏)) − 𝑓(𝑟⃑(𝑎)) by FTC using 𝑑𝑡 𝑓(𝒓
o Path independent, doesn’t matter the curve as long as end points are the same
o If a = b (ie a closed curve) then the integral is zero
o 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 independent
o 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)
𝑏


The double integral becomes ∫𝑎 [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 𝑃(𝑥, 𝑔1 (𝑥)), ∫𝑐3 𝑃(𝑥, 𝑦)𝑑𝑥 = − ∫𝑎 𝑃(𝑥, 𝑔2 (𝑥))

∫𝑐 𝑃𝑑𝑥 = ∫𝑎 𝑃(𝑥, 𝑔1 (𝑥)) − ∫𝑎 𝑃(𝑥, 𝑔2 (𝑥)) = − ∬𝐷
𝑏
𝑏
𝑏
𝑏
𝜕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:
𝜕𝑄
𝜕𝑃
o Since A = ∬𝐷 1 𝑑𝐴 need 𝜕𝑥 − 𝜕𝑦 = 1
 Either P = 0 and Q = x, P = -y and Q = 0, or P =-y/2 and Q = x/2
o Thus:


1
A = ∮𝑐 𝑥𝑑𝑦 = ∮𝑐 −𝑦𝑑𝑥 = 2 ∮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: 𝑟⃑(𝑢, 𝑣) = 𝑥(𝑢, 𝑣)𝑖̂ + 𝑦(𝑢, 𝑣)𝑗̂ + 𝑧(𝑢, 𝑣)𝑘̂
 Example: a sphere of radius a:
 𝑟⃑(𝑢, 𝑣) = 𝑎 cos 𝑢 sin 𝑣 𝑖̂ + 𝑎 sin 𝑢 sin 𝑣 𝑗̂ + 𝑎 cos 𝑣 𝑘̂
o Fixing u to v gives you a curve embedded in the surface
o

𝜕𝑟⃑(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 𝑛⃑⃑ = 𝑟⃑𝑢 × 𝑟⃑𝑣 **direction will come into play later
 Recall that equation of plane at point (a, b, c) is given by
 𝑛⃑⃑ ∙ < (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 ‖𝑟⃑𝑢 × 𝑟⃑𝑣 ‖𝑑𝐴


is the projection onto uv plane (xy plane?)
∗
⃑⃑⃑⃑∗
Comes from area of a parallelogram with sides ∆𝑢𝑟⃑⃑⃑⃑
𝑢 & ∆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) = ∬𝐷 √1 + 𝑧𝑥2 + 𝑧𝑦2 𝑑𝐴
2.5 Surface Integrals

∬𝑆 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑆 = ∬𝑆 𝑓(𝒓(𝑢, 𝑣))‖𝑟⃑𝑢 × 𝑟⃑𝑣 ‖𝑑𝑢𝑑𝑣 =
∬𝑆 𝑓(𝑥(𝑢, 𝑣), 𝑦(𝑢, 𝑣), 𝑧(𝑢, 𝑣))‖𝑟⃑𝑢 × 𝑟⃑𝑣 ‖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 + 𝑧𝑥2 + 𝑧𝑦2 𝑑𝐴
Flux Integral: ∫𝑆 𝐹⃑ ∙ 𝑑𝑆⃑ = ∫𝑆 𝐹⃑ ∙ 𝑛̂𝑑𝑆
𝑟⃑ × 𝑟⃑
o Since 𝑛̂ = ‖𝑟⃑𝑢× 𝑟⃑𝑣‖ and 𝑑𝑆 = ‖𝑟⃑𝑢 × 𝑟⃑𝑣 ‖𝑑𝑢𝑑𝑣
𝑢
𝑣
o ∬𝑆 𝐹⃑ ∙ 𝑑𝑆⃑ = ∬𝑆 𝐹⃑ (𝑟⃑(𝑢, 𝑣)) ∙ (𝑟⃑𝑢 × 𝑟⃑𝑣 )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
𝜕𝑓
𝜕𝑓 𝜕𝑓

Recall Gradient of f: ∇𝑓 = (𝜕𝑥 ,

o Turns a scalar field into a vector field
o “Nable times f”
Divergence: div𝐹⃑
𝜕
𝜕 𝜕
o ∇ ∙ 𝐹⃑ = ( ,
, ) ∙ (𝑃, 𝑄, 𝑅) = 𝑃𝑥 + 𝑄𝑦 + 𝑅𝑧
𝜕𝑥

,
𝜕𝑦 𝜕𝑧
)
𝜕𝑦 𝜕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 magnitude
o If divF = 0, call F “incompressible”
Curl: curl 𝐹⃑
𝜕
𝜕 𝜕
o ∇ × 𝐹⃑ = ( ,
, ) × (𝑃, 𝑄, 𝑅) = (𝑅𝑦 − 𝑄𝑧 , 𝑃𝑧 − 𝑅𝑥 , 𝑄𝑥 − 𝑃𝑦 )
𝜕𝑥
o
o
o
o
o
𝜕𝑦 𝜕𝑧
“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 = 0
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 ∯ 𝐹⃑ ∙ 𝑑𝑠⃑ = ∭ ∇ ∙ 𝐹⃑ 𝑑𝑉
𝑆

𝐸
For proof, show ∯𝑆 𝑅 𝑘̂ ∙ 𝑛̂ 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 ∫ 𝐹⃑ ∙ 𝑑𝑟⃑ = ∬ ∇ × 𝐹⃑ ∙ 𝑑𝑠⃑
𝐶


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 scale
o Kn < 0.01 continuum. Kn > 1 statistical
Slip flow: particles in a fluid are allowed to bounce/slide along wall
o Kn < 0.01 no slip: V fluid at wall = V of wall
o 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

𝜎𝑥𝑥
Tensor of stress: [𝜏𝑦𝑥
𝜏𝑧𝑥
𝜏𝑥𝑦
𝜎𝑦𝑦
𝜏𝑧𝑦
𝜏𝑥𝑧
𝜏𝑦𝑧 ]
𝜎𝑧𝑧
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 stress
Orientated 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
𝜕𝑢
𝜕𝑢
o 𝜏𝑖𝑗 = 𝜇 (𝜕𝑥 𝑖 + 𝜕𝑥𝑗 ) i or j = 1: x and u 2: y and v
𝑗


𝑖
3: z and w
o i=j normal stress, else shear
o μ: dynamic/absolute viscosity
o ν = μ/ρ: kinematic viscosity
Newtonian fluid (linear relation between shear and
deformation)
o Shear thinning and shear thickening
o Binham plastic (deformation doesn’t start
right away)
Compressibility:𝐸𝑣 = −
𝑑𝑃
𝑑𝑉
𝑉
=
𝑑𝑃
𝑑𝜌
𝜌
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 + 𝜌𝑔⃑ = 0 => no variation in x and y


 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
𝑑𝑃
𝑑𝑧
= −𝜌𝑔


Force is pressure times area
Buoyancy
o Archimedes Principle
o
F = -ρgV
o 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 + 𝜌𝑔⃑ = 𝜌𝑎⃑
 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
𝐷
𝜕
⃑⃑ ∙ ∇
Total derivative: ≡ + 𝑉

Stream line: curve that is locally tangent to velocity vector
𝐷𝑡
o
𝑑𝑥
𝑢
=
𝑑𝑦
𝑣
=
𝑑𝑧
𝑤
𝜕𝑡
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 viscous
o Umax > U∞
Laminar vs Turbulent Flow
o 𝑅𝑒 ≡

𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠
𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒𝑠
=
𝜌𝑈𝐷
𝜇
=
𝑈𝐷
𝜈
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
Reynold’s Transport Theorem
o
parallel to streamline; perp to streamline
 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
𝜕𝑢
𝜕𝑣
o Continuity: 𝜕𝑥 + 𝜕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 force
o =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
ideal gas and an isentropic process

o Using compressibility have
 Why speed of sound is faster in water
Quasi one-dimensional flows
o Reynolds transport theorem with energy
if we assume and
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 vorticity
o Curl = 0 => irrotational

Navier-Stokes Equations
o Start with F= ma
o Utilize acceleration is the total derivative of velocity
o Forces: gravity and shear and normal stresses
o
o 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