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. ■
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=0 So (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
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Problem. Evaluate ∬DxdA where D is the region bounded by y=x, y=2x And 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=2x But only up to x+y=2). Actually y=2x gives x=y/2 And 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≤y So the right boundary is y. But we also need x+y≤2I.e., x≤2−y. For y≤1: y≤2−y So 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
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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)=0 So 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.
A multiple integral extends the idea of area under a curve to higher dimensions. A double integral computes the signed volume under a surface z=f(x,y) over a region in the plane, while a triple integral computes the integral of a function over a three-dimensional region. Fubini’s theorem says you can evaluate these integrals one variable at a time, converting a double integral into iterated single integrals, which is the computational workhorse.
The key to setting up multiple integrals is describing the region correctly. Type I and Type II regions in the plane correspond to fixing x and integrating over y, or vice versa. When the region has a complicated shape, you may need to split the integral or change the order of integration. Changes of variables, such as polar or spherical coordinates, simplify the integrand or the region at the cost of introducing a Jacobian determinant that accounts for how the coordinate transformation stretches or compresses space. The Jacobian is the multivariable analogue of the chain rule: it measures how much a small volume element is distorted by the transformation.