11/12/2002, 2D INTEGRATION Math 21a, O. KnillHomework: Section 12.2: 2, 14, 20 12.3 30, 401D INTEGRATION IN 100 WORDS. If f (x) is a continuous function of one variable, thenR10f(x) dx canbe defined as a limit of the Riemann sum fn(x) =1nPnk=1f(xk) for n → ∞ with xk= k/n. The in-tegral is the average of f on the interval [a, b]. It 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 calledan anti-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: Example: F (x) =Rx0e−x2dx. Of-ten, the anti-derivative can be found: Example: f(x) = sin2(x) = (cos(2x) + 1)/2, F (x) = x/2 − sin(2x)/4.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. This integral divided by the area is the average value of f on the region R. If f(x, y) = 1, thenthe integral is the area of the region R. For many regions, the integral can be calculated as a double in-tegralRba[Rd(x)c(x)f(x, y) dy]dx. In general, the region must be split into pieces, then integrated seperately.EXAMPLE. CalculateRRf(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.TYPES OF REGIONS.R RRf dA =RbaRg2(x)g1(x)f(x, y) dydx type I region.R RRf dA =RbaRh2(y)h1(y)f(x, y) dxdy type II region.R RRf(x, y)dxdy =RβαRbaf(r cos(θ), r sin(θ)) rdrdθ integral in polar coordinates.EXAMPLE. Let R be the triangle 1 ≥ x ≥ 0, 1 ≥ y ≥ 0, y ≤ x. CalculateRRe−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:RbaRdcf(x, y) dxdy =RdcRbaf(y, x) dydx.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 of energiesonly. This is the reason for the name ”quantum”. If −(uxx+uyy) = λu, then a particle of mass m has the energyE = λ¯h2/2m. A function u(x, y) = sin(kx) sin(ny) represents a particle of energy (k2+ n2)¯h2/(2m). Our aim isto find the probability that the particle with energy 13¯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.MOMENT OF INERTIA. Compute the kinetic energy of a square iron plate R = [−1, 1] × [−1, 1] of densityρ = 1 (about 10cm thick) rotating around its center with a 60000rpm (rounds per minute). The angular velocityspeed is ω = 2π 60000/60 = 100 2π. Because E =R RR(rω)2/2 dxdy, where r =px2+ y2, we have E = ω2I/2,where I =R RR(x2+ y2) dxdy is the moment of inertia. For the square, I = 4/3. Its energy of the plate isω24/6 = 4π210024/6Joule ∼ 0.43KW h. You can run with this energy a 60 Watt bulb for 7 hours.WHERE DO DOUBLE INTEGRALS OCCUR?- areas.- averages. Examples: average rain fall or average population in some area.- probabilities. Expectation of random variables. - quantum mechanics: probability of particle.- moment of inertiaR RR(x2+ y2)ρ(x, y)dxdy- center of mass (R RRxρ(x, y) dxdy/M,R RRyρ(x, y) dxdy/M), with M =R RRdxdy.- 1D integrals (see challenge
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