Double Integrals
- Assume
- D is a bounded closed region on a plane
- f(x,y) is bounded on D⟹m≤f(x,y)≤M ∀(x,y)∈D
- Find
- R=[a,b]×[c,d]
- Rkn=[xk−1,xk]×[yk−1,yk]
- xk=a+Δxk, k=0…N1
- xn=c+Δyn, n=0…N2
- mkn=infRknf(x,y)Mkn=supRknf(x,y)
- L(f,N1,N2)=∑k=1N1∑n=1N2mknΔAU(f,N1,N2)=∑k=1N1∑n=1N2MknΔA
- Definition
- If lim(N1,N2)→∞L(f,N1,N2)=lim(N1,N2)→∞U(f,N1,N2)
- Then f is Riemann integrable on D and the limit is called the Riemann integral of f over D
- Notated: ∬Df(x,y)dA
Integrability Conditions
- f is continuous on D (for the scope of this course)
- D is bounded by a smooth curve (can be piecewise like in the case of a rectangle)
Solving Strategies
Fubini’s Theorem
- If f is continuous on [a,b]×[c,d]=R
- Then ∬Df(x,y)dA=∫ab∫cdf(x,y)dydx=∫cd∫abf(x,y)dxdy
- Means you can swap the order of integration
Factorization
∬DfdA=∫ab∫cdg(x)h(y)dxdy=∫abg(x)dx∫cdh(y)dy
Over General Regions
- General Region = D is simple relative to the direction u if any line L∥u intersects D along one line segment
- When D is vertically simple
- ∬Df(x,y)dA=∫ab∫yb(x)yt(x)f(x,y)dydx
- When D is horizontally simple
- ∬Df(x,y)dA=∫ab∫yb(x)yt(x)f(x,y)dydx
- When D is simple both directions, one way might be a lot easier
- If D is not simple either way, it can be cut into D1,D2,… then added together after individual integration
Properties
Area of D
A(D)=∬DdA
Volume Under z=f(x,y)
V=∬Df(x,y)dA when f(x,y)≥0
Integral Mean Value Theorem (IMVT)
- If f is continuous and integrable on a closed bounded D
- ∃(x∗,y∗)∈D s.t. ∬Df(x,y)dA=A(D)f(x∗,y∗)
- Where A(D) is the area of the region D
- Also
- For m≤f(x,y)≤M on D
- mA(D)≤∬Df(x,y)dA≤MA(D)
Independence of a Partition
- We can take the Riemann integral using any shaped partition
- Foundation of changing variables in integrals
Change of Variables
Functions
- F(S,F) denotes all functions f where f:S→F
- Definitions (f,g∈F(S,F))
- f and g are equal if f(s)=g(s), ∀s∈S
- F(S,F) is a vector space with (f+g)(s)=f(s)+g(s) and (cf)(s)=c[f(s)]
- Polynomials
- f(x)=anxn+an−1xn−1+a1x+a0, each ak∈F (degree n)
- If an=an−1=⋯=a0=0 then f is the zero polynomial (degree -1)
- Two polynomials of degree n are equal if all coefficients are equal
- The set of all polynomials of field F and degree n form a vector space Pn(F)
- Let f(x)=anxn+⋯+a1x+a0, g(x)=bmxm+⋯+b1x+b0 where m≤n
- Define bm+1=bm+2=⋯bn=0 (make g degree n by padding with zero coefficients)
- Define
- f(x)+g(x)=(an+bn)xn+(an−1+bn−1)xn−1+⋯+(a1+b1)x+a0+b0
- For any c∈F, cf(x)=canxn+⋯+ca1x+ca0