4. Multiple Dimensional Integration

Question 1

Defining the Ordinary Integral

Answer 1

• we can define integration in two ways:
  i) anti derivative
  ii) Rieman sum

Question 2

Integration as the Anti Derivative

Answer 2

-integration as the inverse of differentiation
-if dg/dx = f(x),
then, ∫ f(x) dx = g(x) + C
and b,a ∫ f(x) dx = g(b) - g(a)

Question 3

Integration as a Rieman Sum

Answer 3

consider, b,a ∫ f(x) dx
-we can divide the region a≤x≤b into N sections each of length 𝛿xi:
𝛿xi = b-a / N
-then,
b,a ∫ f(x) dx = 𝛿xi->0 lim { i=1->N Σ f(xi)*𝛿xi }

Question 4

Partial Integration

Indefinite Integrals

Answer 4

• integrate with respect to the desired variable while treating the others as constants
• the constant of integration is a function of the other variables
• e.g. if f=f(x,y), then integrating with respect to x you would hold y constant and integrate as normal, the constant of integration would be A(y), a function of y

Question 5

Partial Integration

Definite Integrals

Answer 5

• integrate with respect to the desired variable while holding the other constant
• when you substitute in the limits, the ‘constant’ of integration will cancel as normal
• the limits may also be functions of the other variable

Question 6

Integration in Over Two Variables

Answer 6

• integrate with respect to x and sub in limits
• then integrate with respect to y and sub in limits
• you can also reverse the order of this

Question 7

Two Dimensional Integration/Double Intergals

Rectangular Domains

Answer 7

-integrating over the region A:
A(x,y,z) = {(x,y,z) : a≤x≤b , c≤y≤d , z=0}
-we can divide the area into N=mn rectangles by cutting it into a grid of m by n pieces
-the double integral is now defined in terms of a Rieman sum
∬f(x,y) dA = m->∞,n->∞ lim [ k=1->N=mn Σf(xi,yi)𝛿A
=𝛿x->0, 𝛿y->0 lim [i=1->m Σ [j=1->n Σf(xi,yi)𝛿x𝛿y] ]

=𝛿x->0 lim i=1->m Σ[𝛿y->0 lim j=1->nΣf(xi,yi)𝛿y]𝛿x
-this can also be reversed

Question 8

Fubini’s Theorem for Double Integrals

Answer 8

∬f(x,y)dA = b,a∫ {d.c ∫f(x,y)dy} dx = d,c∫ {b,a ∫f(x,y)dx} dy

-here the inner integrals are partial integrals

Question 9

Two Dimensional Integration/Double Intergals

Non-Rectangular Domains

Answer 9

-if integration is to be performed over a non-rectangular domain,A, we can convert this to an integral over an enclosing rectangle A’, by defining:
g(x,y) = f(x,y) for (x,y)ϵA, and 0 otherwise
-so the integral over A of f(x,y) is equal to the integral over A’ of g(x,y)
-now when you evaluate the inner integral you make sure that the limits correspond to the boundaries of region A
-if you perform the x integrations first, divide A into infinitesimal horizontal strips and integrate along each from left to right so that the limits are functions of y that describe the left and right boundaries A
-to perform the y integral first, divide A into infinitesimal vertical strips and integrate along each from top to bottom with limits that are functions of x that describe the upper and lower boundaries of A
-the outer integration is always over constant limits that define the edges of the rectangle A’ in the required direction

Question 10

Using Double Integrals to Find Area

Answer 10

-to fin the area of a domain, evaluate:

∬ dA over the domain A

Question 11

Reversing the Order of Integration

Answer 11

• if the inner integral is intractable, we may choose to reverse the order of integration
  1) sketch the area described by the original double integral
  2) find the boundaries that were constants in terms of functions of the other variable to form the new inner integral
  3) form the outer integral between constants by reading off values from the graph

Question 12

Changing Variables of Standard Integration

Answer 12

a,b ∫f(x) dx = c,d ∫F(u) dx/du du

-where F(u) = f(x(u))
and x=a -> u=c , x=b -> u=d

Question 13

Changing Variables of Double Integrals

Equation

Answer 13

∬f(x,y) dxdy = ∬F(u,v) |J(u,v)| dudv

Question 14

2D Jacobian

Equation Definition

Answer 14

J(u,v) = ∂(x,y)/∂(u,v)

Question 15

2D Jacobian

Word Definition

Answer 15

• the Jacobian is the ratio of the area elements in the two coordinate systems
  i. e. it is the scale factor of the transformation

Question 16

Jacobian for Plane Polar Coordinates in Two Dimensions

Answer 16

J = R cos²φ + R sin²φ = R

Question 17

Changing Variables of Double Integrals

Plane Polar Coordinates

Answer 17

∬f(x,y) dA = ∬f(x,y) dxdy = ∬F(R,φ) R dRdφ

Question 18

Changing Variables of Double Integrals

Derivation

Answer 18

-start with a curved shape outline A in the x-y plane
-in order to integrate over a region A in the x-y plane, integrate over a simpler corresponding region A’ in the u-v plane
-an element 𝛿A’=𝛿u𝛿v in the u-v plane maps onto some area element 𝛿A in the x-y plane that will have straight (but not perpendicular) sides if 𝛿u,𝛿v->0 i.e. a parallelogram
-find vectors for two adjacent sides of the parallelogram
-𝛿A is the cross product of these two vectors
𝛿A = |J| 𝛿u𝛿v
-in the limit 𝛿u,𝛿v->0, dA = |J| dA’

Question 19

Triple Integrals

Answer 19

∭f(x,y,z) dV = (𝛿V->0) lim (k=1->N) Σ f(|rk) 𝛿V

• where 𝛿V = 𝛿x𝛿y𝛿z and rk = (xk,yk,zk) is the position of a point in the kth volume element
• like double integrals, these are evaluated as nested integrals over x, y and z

Question 20

Triple Integrals to Find Volume

Answer 20

-the integral over the domain of ∭ dV gives the volume of the domain

Question 21

Change of Variables in Triple Integration

Answer 21

∭f(x,y,z) dxdydz = ∭F(u,v,w) |J| dudvdw

where, f(x,y,z) = F(u,v,w)

Question 22

Jacobian for Triple Integration

Equation

Answer 22

J = ∂(x,y,z)/∂(u,v,w)
-this is given by the determinant of a 3x3 matrix, first row = ∂x/∂u , ∂y/∂u, ∂z/∂u second row = ∂x/∂v, ∂y/∂v, ∂z/∂v third row = ∂x/∂w, ∂y/∂w, ∂z/∂w

Question 23

Three Dimensional Polar Coordinates

Cylindrical Polar Coordinates

Answer 23

-the z axis remains the same
-the (x,y)-position is replace by an angle and a distance
-we transform from (x,y,z) to (R,φ,z)
x = Rcosφ , y = Rsinφ, z=z
R = √(x²+y²)

Question 24

Three Dimensional Polar Coordinates

Spherical Polar Coordinates

Answer 24

-we use polar coordinates (r,θ) to describe the position in the (z,R) plane of cylindrical polar coordinates
-we obtain spherical polar coordinates (r,θ,φ)
x = r sinθ cosφ , y = r sinθ sinφ , z = r cosθ
r = √(x²+y²+z²)

Question 25

Sketching Graphs in 3D

Symmetry

Answer 25

• a function that remains the same under the transformation x->-x has mirror symmetry in the plane x=0
• a function that depends upon x and y only in the combination x²+y² is axisymmetric about the z axis

Question 26

Sketching Graphs in 3D

Interesting Features

Answer 26

• maxima
• minima
• asymptotes

Question 27

Sketching Graphs in 3D

Sections

Answer 27

• examine the intersections of surfaces with the place x=0

- and similarly y=0 & z=0

Question 28

Cylindrical Polar Coordinates

dV

Answer 28

dV = R dR dφ dz

Question 29

Spherical Polar Coordinates

dV

Answer 29

dV = r² sinθ dr dθ dφ