Lecture 17: 3/19/2004, 2D INTEGRALS Math21a, O. KnillHOMEWORK. Section 12.1: 16,26,30,44. Section 12.4: 441D INTEGRATION IN 100 WORDS. If f(x) is a continuous function thenRbaf(x) dx can be defined as alimit of the Riemann sum fn(x) =1nPxk∈[a,b]f(xk) for n → ∞ with xk= k/n. This integral divided by|b − a| is the average of f on [a, b]. The integral can be interpreted as an signed area under the graphof f . If f(x) = 1, the integral is the length of the interval. The function F (x) =Rxaf(y) dy is called ananti-derivative of f. The fundamental theorem of calculus states F0(x) = f(x). Unlike the derivative,anti-derivatives can not always be expressed in terms of known functions. An example is: F (x) =Rx0e−t2dt.Often, the anti-derivative can be found: Example: f(x) = cos2(x) = (cos(2x) + 1)/2, F (x) = x/2 − sin(2x)/4.F(x)F(x+dx)-F(x)AVERAGES=MEAN. www.worldclimate.com gives the follow-ing data for the average monthly rainfall (in mm) for Cam-bridge, MA, USA (42.38 North 71.11 West,18m Height).Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec93.9 88.6 83.3 67.0 42.9 26.4 27.9 83.8 35.5 61.4 166.8 82.8The average 860.3/12 = 71.7 is a Rieman sum integral.1 2 3 4 5 6 7 8 9 10 11 122550751001251502D INTEGRATION. If f(x, y) is a continuous function of two variables on a region R, the integralRRf(x, y) dxdy can be defined as the limit1n2Pi,j,xi,j∈Rf(xi, yj) with xi,j= (i/n, j/n) when n goes toinfinity. If f(x, y) = 1, then the integral is the area of the region R. The integral divided by the areaof R is the average value of f on R. For many regions, the integral can be calculated as a doubleintegralRba[Rd(x)c(x)f(x, y) dy]dx. In general, the region must be split into pieces, then integrated seperately.One can interpretR RRf(x, y) dydx as the volume of solid below the graph of f and above R in the x −y plane.(As in 1D integration, the volume of the solid below the x-y plane is counted negatively).EXAMPLE. CalculateR RRf(x, y) dxdy, where f (x, y) = 4x2y3and where R is the rectangle [0, 1] × [0, 2].Z10[Z204x2y3dy] dx =Z10[x2y4|20] dx =Z10x2(16 − 0) dx = 16x3/3|10=163.FUBINI’S THEOREM.RbaRdcf(x, y) dxdy =RdcRbaf(y, x) dydx.TYPES OF REGIONS.R RRf dA =RbaRg2(x)g1(x)f(x, y) dydxtype I region.R RRf dA =RbaRh2(y)h1(y)f(x, y) dxdytype II region.EXAMPLE. Let R be the triangle 1 ≥ x ≥ 0, 1 ≥ y ≥ 0, y ≤ x. CalculateR RRe−x2dxdy.ATTEMPT.R10[R1ye−x2dx]dy. We can not solve the inner integral because e−x2has no anti-derivative in terms of elementary functions.IDEA. Switch order:R10[Rx0e−x2dy] dx =R10xe−x2dx = −e−x22|10=(1−e−1)2=0.316....If you can’t solve adouble integral, try tochange the order of in-tegration!A special case of switching the order of integration is Fubini’s theorem.QUANTUM MECHANICS. In quantum mechanics, the motion of a particle (like an electron) in the plane isdetermined by a function u(x, y), the wave function. Unlike in classical mechanics, the position of a particleis given in a probabilistic way only. If R is a region and u is normalized so thatR|u|2dxdy = 1, thenRR|u(x, y)|2dxdy is the probability, that the particle is in R.EXAMPLE. Unlike a classical particle, a quantum particle in a box [0, π] × [0, π] can have a discrete set ofenergies only. This is the reason for the name ”quantum”. If −(uxx+ uyy) = λu, then a particle of mass m hasthe energy E = λ¯h2/2m. A function u(x, y) = sin(kx) sin(ny) represents a particle of energy (k2+ n2)¯h2/(2m).Let us assume k = 2 and n = 3 from now on. Our aim is to find the probability that the particle with energy13¯h2/(2m) is in the middle 9’th R = [π/3, 2π/3] × [π/3, 2π/3] of the box.SOLUTION: We first have to normalize u2(x, y) = sin2(2x) sin2(3y), sothat the average over the whole square is 1:A =Zπ0Zπ0sin2(2x) sin2(3y) dxdy .To calculate this integral, we first determine the inner integralRπ0sin2(2x) sin2(3y) dx = sin2(3y)Rπ0sin2(2x) dx =π2sin2(3y) (the fac-tor sin2(3y) is treated as a constant). Now, A =Rπ0(π/2) sin2(3y) dy =π24, so that the probability amplitude function is f (x, y) =4π2sin2(2x) sin2(3y).0123012300.10.20.30123The probability that the particle is in R is slightly smaller than 1/9:1AZRf(x, y) dxdy =4π2Z2π/3π/3Z2π/3π/3sin2(2x) sin2(3y) dxdy=4π2(4x − sin(4x))/8|2π/3π/3(6x − sin(6x))/12|2π/3π/3= 1/9 − 1/(4√3π)The probability is slightly smaller than 1/9.WHERE DO DOUBLE INTEGRALS OCCUR?- compute areas.- compute averages. Examples: average rain fall or average population in some area.- probabilities. Expectation of random variables.- quantum mechanics: probability of particle being in a region. - find moment of inertiaR RR(x2+y2)ρ(x, y)dxdy- find center of mass (R RRxρ(x, y) dxdy/M,R RRyρ(x, y) dxdy/M), with M =R RRdxdy.- compute some 1D integrals.TRIPLE INTEGRALS are defined similarly and covered later in detail. Fubinis theorem generalizes.R R RR1 dxdydz is a
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