Math S21a: Multivariable calculus Oliver Knill, Summer 2011Lecture 15: Double integralsHere is a one paragraph summary of single variable calculus: if f (x) is a differentiable function,then the Riemann integralRbaf(x) dx is defined as the limit of the Riemann sum Snf(x) =1nPk/n∈[a,b]f(k/n) for n → ∞. The derivative is the limit of difference quotients Dnf(x) =n[f(x + 1/n) − f (x)] as n → ∞. The integralRbaf(x) dx is the signed area under the graph off, where ”signed” indicates that it can become negative too. The function F (x) =Rx0f(y) dy iscalled an anti-derivative of f and determined up a constant. The fundamental theorem ofcalculus statesF′(x) = f(x),Zx0f(x) = F (x) − F (0) .This theorem is obtained from the quantum fundamental theorem DF (k/n) = f(k/n) −f(0), Sf(k/n) = F (k/n) (which holds for all functions!) in the limit n → ∞ and allows tocompute integrals by inverting differentiation. Differentiation rules become integration rules: theproduct rule becomes integration by parts, the chain rule becomes partial integration.k nxy0SfHk nL = FHk nL- FH0Lk nxy0fHknL= DFHk nLIf f (x, y) is differentiable on a r egion R, the integralRRf(x, y) dxdy is defined asthe limit of the Riemann sum1n2X(in,jn)∈Rf(in,jn)when n → ∞. We write alsoRRf(x, y) dA and think of dA as an area element.1 If we integrate f(x, y) = xy over the unit square we can sum up the Riemann sum for fixedy = j/n and get y/2. Now perform the integral over y to get 1/4. This example shows howwe can reduce double integrals to single variable integrals.2 If f(x, y) = 1, then the integral is the area of the region R. The integral is the limit L(n)/n2,where L(n) is the number of lattice points (i/n, j/n) inside R.3 The integralRRRf(x, y) dA divided by the area of R is the average value of f on R.4 One can interpretR RRf(x, y) dydx as the signed volume of the solid below the graph o ff and above R in the x − y plane. As in 1D integration, the volume of the solid below thexy-plane is counted negatively.Fubini’s theorem allows to switch the order of integration over a rectangle if thefunction f is continuous:RbaRdcf(x, y) dxdy =RdcRbaf(x, y) dydx.Pro of. We have for every n the ”quantum Fubini identity”Xin∈[a,b]Xjn∈[c,d]f(in,jn) =Xjn∈[c,d]Xin∈[a,b]f(in,jn)which holds for all functions. Now divide both sides by n2and take the limit n → ∞.Fubini’s theorem only holds for rectangles. We extend the class of regions now to so called TypeI and Type II regions:A type I region is of the f ormR = {(x, y) | a ≤ x ≤ b, c(x) ≤ y ≤ d(x) } .An integral over such a region is called a type I integralZZRf dA =ZbaZd(x)c(x)f(x, y) dydx .bacHxLdHxLA type II region is of the formR = {(x, y) | c ≤ y ≤ d, a(y) ≤ x ≤ b(y) } .An integral over such a region is called a type II inte-gralZZRf dA =ZdcZb(y)a(y)f(x, y) dxdy .dcaHyLbHyL5 Integrate f(x, y) = x2over the region bounded aboveby sin(x3) and bounded below by the graph of −sin(x3)for 0 ≤ x ≤ π. The value of this integral has a physicalmeaning. It is called moment of inertia.Zπ1/30Zsin(x3)−sin(x3)x2dydx = 2Zπ1/30sin(x3)x2dxWe have now an integral, which we can solve by substi-tution= −23cos(x3)|π1/30=43.6 Integrate f (x, y) = y2over the region bound by the x-axes, the lines y = x + 1 and y = 1 −x. The problem isbest solved as a type I integral. As you can see from thepicture, we would have to compute 2 different integralsas a type I integral. To do so, we have to write thebounds as a function of y: they are x = y − 1 andx = 1 − yZ10Z1−yy−1y3dx dy = 2Z10y3(1 −y) dy = 2(14−13) =110.7 Let R be the triangle 1 ≥ x ≥ 0, 0 ≤ y ≤ x. What isZ ZRe−x2dxdy ?The type II integralR10[R1ye−x2dx]dy can not be solvedbecause e−x2has no anti-derivative in terms of elemen-tary functions.The type I integralR10[Rx0e−x2dy] dx however can besolved:=Z10xe−x2dx = −e−x22|10=(1 − e−1)2= 0.316... .8 The area of a disc of radius R isZR−RZ√R2−x2−√R2−x21 dydx =ZR−R2√R2− x2dx .This int egr al can be solved with the substitution x =R sin(u), dx = R cos(u)Zπ/2−π/22qR2− R2sin2(u)R cos(u) du =Zπ/2−π/22R2cos2(u) du .Using a double angle formula we getR2Rπ/2−π/22(1+cos(2u)2du = R2π. We will now seehow to do that better in polar coordinates.Remark: The Riemann integral just defined works well for continuous functions. In otherbranches of mathematics like probability theory, a better integral is needed. The Lebesgueintegral fits the bill. Its definition is close to the Riemann integral which we have given as thelimit n−2R(xk,yl)∈Rf(xk, yl) where xk= k/n, yl= l/n. The Lebesgue integral replaces the r egularlyspaced (xk, yl) g rid with random points xk, yland uses the same formula. The following Mathe-matica code computes the integralR10R10x2y using this Monte Carlo definition of the Lebesgueintegral.M=10000; R:=Random [ ] ; f [ x , y ]:= xˆ2 y ; Sum[ f [ R,R] , {M}] /M It is as elegant than the numerical Riemann sum computationM=100; f [ x , y ]:= xˆ2 y ; Sum[ f [ k/M, l /M] , {k ,M},{l ,M}] /Mˆ2 but the Lebesgue integral is usually closer to the actual answer 1/6 than the Riemann integral.Note that for all continuo us functions, the Lebesgue integral gives the same results than theRiemann integral. It does not change calculus. But it is useful for example to compute nastyintegrals like the area of the Mandelbrot set.Homework1 Find the double integralR41R20(3x −√y) dxdy.2 Find the area of the regionR = {(x, y) | 0 ≤ x ≤ 2π, sin(x) − 1 ≤ y ≤ cos(x) + 2}and use it to compute the average valueR RRf(x, y) dxdy/area(R) of f(x, y) = y over thatregion.3 Find the volume of the solid lying under the paraboloid z = x2+ y2and above the rectangleR = [−2, 2] × [−3, 3] = {(x, y) | − 2 ≤ x ≤ 2, −3 ≤ y ≤ 3 }.4 Calculate the iterated integralR10R2−xx(x2−y) dydx. Sketch the correspo nding type I region.Write this integral as integral over a type II region and compute the integral again.5 Evaluate the double integralZ20Z4x2xey2dydx
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