Defining the Ordinary Integral
- we can define integration in two ways:
i) anti derivative
ii) Rieman sum
Integration as the Anti Derivative
-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)
Integration as a Rieman Sum
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 }
Partial Integration
Indefinite Integrals
- 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
Partial Integration
Definite Integrals
- 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
Integration in Over Two Variables
- 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
Two Dimensional Integration/Double Intergals
Rectangular Domains
-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
Fubini’s Theorem for Double Integrals
∬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
Two Dimensional Integration/Double Intergals
Non-Rectangular Domains
-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
Using Double Integrals to Find Area
-to fin the area of a domain, evaluate:
∬ dA over the domain A
Reversing the Order of Integration
- 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
Changing Variables of Standard Integration
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
Changing Variables of Double Integrals
Equation
∬f(x,y) dxdy = ∬F(u,v) |J(u,v)| dudv
2D Jacobian
Equation Definition
J(u,v) = ∂(x,y)/∂(u,v)
2D Jacobian
Word Definition
- the Jacobian is the ratio of the area elements in the two coordinate systems
i. e. it is the scale factor of the transformation
Jacobian for Plane Polar Coordinates in Two Dimensions
J = R cos²φ + R sin²φ = R
Changing Variables of Double Integrals
Plane Polar Coordinates
∬f(x,y) dA = ∬f(x,y) dxdy = ∬F(R,φ) R dRdφ
Changing Variables of Double Integrals
Derivation
-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’
Triple Integrals
∭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
Triple Integrals to Find Volume
-the integral over the domain of ∭ dV gives the volume of the domain
Change of Variables in Triple Integration
∭f(x,y,z) dxdydz = ∭F(u,v,w) |J| dudvdw
where, f(x,y,z) = F(u,v,w)
Jacobian for Triple Integration
Equation
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
Three Dimensional Polar Coordinates
Cylindrical Polar Coordinates
-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)
Three Dimensional Polar Coordinates
Spherical Polar Coordinates
-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)
