Proof (sketch). For a continuous function f on the compact rectangle RDefine
F(x)=∫cdf(x,y)dy
Since f is continuous, F is continuous on [a,b]. For each partition P=(x0,…,xm) of [a,b]Define Riemann sums for the outer integral:
S(P)=∑i=1mF(xi∗)Δxi=∑i=1m∫cdf(xi∗,y)dyΔxi
By Fubini’s theorem for Riemann integrals (proven via uniform continuity of f on the compact set R), As ∥P∥→0 these sums converge to both ∬RfdA and ∫abF(x)dx. The Reversal of integration order follows by symmetry. ■
2.2 General Regions
For a general region D in R2:
Type I region: D=(x,y):a≤x≤b,g1(x)≤y≤g2(x)
∬DfdA=∫ab∫g1(x)g2(x)f(x,y)dydx
Type II region: D=(x,y):c≤y≤d,h1(y)≤x≤h2(y)
∬DfdA=∫cd∫h1(y)h2(y)f(x,y)dxdy
Problem. Evaluate ∬DxydA where D is the region bounded by y=x2 and y=x+2.
Solution
The curves intersect when x2=x+2I.e., x2−x−2=0So (x−2)(x+1)=0Giving x=−1 and x=2. As a Type I region, D=(x,y):−1≤x≤2,x2≤y≤x+2.
∬DxydA=∫−12∫x2x+2xydydx=∫−12x[2y2]x2x+2dx
=∫−122x[(x+2)2−x4]dx=21∫−12[x(x+2)2−x5]dx
=21∫−12[x3+4x2+4x−x5]dx
=21[4x4+34x3+2x2−6x6]−12
=21[(4+332+8−664)−(41−34+2−61)]
=21[336−129]=21[12−43]=845
■
Problem. Evaluate ∬DxdA where D is the region bounded by y=x, y=2xAnd x+y=2.
Solution
First, find the intersections. The lines y=x and y=2x intersect at (0,0). The line x+y=2 intersects y=x at (1,1) and y=2x at (2/3,4/3).
As a Type I region, we must split: for 0≤x≤2/3, x≤y≤2x; for 2/3≤x≤1, x≤y≤2−x.
∬DxdA=∫02/3∫x2xxdydx+∫2/31∫x2−xxdydx
=∫02/3x(x−x)dx...
Wait, this is getting messy. Let me use Type II instead. For each y, x ranges from y/2 to y (for 0≤y≤4/3) and from y/2 to 2−y (for 4/3≤y≤1). Actually, the simplest approach is to split D at y=4/3.
For 0≤y≤1: y/2≤x≤y (between y=x and y=2xBut only up to x+y=2). Actually y=2x gives x=y/2And y=x gives x=y. But x+y=2 gives x=2−y. For y≤1: both y≤2−y (since y≤1) and y/2≤ySo the right boundary is y. But we also need x+y≤2I.e., x≤2−y. For y≤1: y≤2−ySo the constraint x≤y is tighter.
Theorem 2.2 (Change of Variables). Let T:D⊆Rn→Rn be a C1 diffeomorphism with Jacobian determinant JT. Then
∫T(D)f(u)du=∫Df(T(x))∣JT(x)∣dx
Derivation of the Jacobian factor (for n=2). Let T(x,y)=(u(x,y),v(x,y)) be a C1 Diffeomorphism. Partition D into small rectangles Rij of area ΔxΔy. The image T(Rij) is approximately a parallelogram spanned by the vectors
a=T(x+Δx,y)−T(x,y)≈(∂x∂uΔx,∂x∂vΔx)
b=T(x,y+Δy)−T(x,y)≈(∂y∂uΔy,∂y∂vΔy)
The area of this parallelogram is ∣a×b∣Which equals
∂x∂u∂y∂v−∂y∂u∂x∂vΔxΔy=∣JT∣ΔxΔy
Summing over all subrectangles and taking the limit gives the change of variables formula. ■
Problem. Evaluate ∬De−(x2+y2)dA where D is the entire R2 plane.
Solution
Use polar coordinates. The region D′ is 0≤r<∞, 0≤θ≤2π.
∬De−(x2+y2)dA=∫02π∫0∞e−r2rdrdθ
The inner integral: ∫0∞re−r2dr=[−21e−r2]0∞=21.
=∫02π21dθ=π
■
Remark. This is the classic Gaussian integral computation, yielding ∫−∞∞e−x2dx=π.
Problem. Evaluate ∭EzdV where E is the solid bounded above by the sphere x2+y2+z2=2 and below by the paraboloid z=x2+y2.
Solution
The surfaces intersect when x2+y2+(x2+y2)2=2. Let r2=x2+y2. Then r2+r4=2I.e., (r2+2)(r2−1)=0So r=1 (positive root). Use Cylindrical coordinates. The region E′ is
Remark. This integral cannot be evaluated in the original order because ey2 has no elementary Antiderivative with respect to y. Swapping the order was essential.
2.7 Common Pitfalls
:::caution Common Pitfalls
Order of integration limits. When setting up ∫ab∫g1(x)g2(x)fdydxVerify that g1(x)≤g2(x) for all x∈[a,b]. If the region is described as “between two curves,” determine which curve is above the other.
Forgetting the Jacobian. In a change of variables, the Jacobian determinant ∣J∣ must be included. For polar coordinates, this factor is r; omitting it is one of the most common errors.
Spherical coordinate conventions. Different texts use different conventions for ϕ and θ. Here, ϕ∈[0,π] is the polar angle (from the positive z-axis) and θ∈[0,2π] is the azimuthal angle.
Region description. When swapping integration order, carefully redraw the region and re-derive the bounds. The new bounds may require splitting the integral into multiple pieces.