7/25/2006 DOUBLE INTEGRALS O.Knill,Maths21a This is part 1 (of 2) of theweekly homework. It is due August 1 at the beginning of class.SUMMARY. dA = dxdy is called area element.•R RRf dA =RbaRdcf(x, y) dydx is called a double integral over a rectangle R.•R RRf dA =RbaRg2(x)g1(x)f(x, y) dydx double integral over a type I region.•R RRf dA =RbaRh2(y)h1(y)f(x, y) dxdy double integral over a type II region.• A(R) =R RR1 dA is called the area of R.•1A(R)R RRf dA is called the average value or the mean of f o n R.• For f ≥ 0, the integralRRf dA is the volume of the solid over R bounded below by thexy-plane and b ounded above by the graph of f.Homework Problems1) (4 points) Calculate the iterated integralR41R20(2x−√y) dxdy. Can you interpret it as a volumeof a solid? If not, can you express the result in terms of two volumes?Solution:Start with the inner integralR20(2x −√y) dx = 4 − 2√y. Integrating this fr om 1 to 4gives8/3 .2) (4 points) 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 value of f(x, y) = y over that region.Remark. You will use here the integralR2π0sin2(x) dx treated in class.Solution:In this problem, it helps to see thatR2π0cos(x) dx = 0 andR2π0cos2(x) dx = π and thesame for sin.The area is A =R2π0Rcos(x)+2sin(x)−11 dxdy =R2π0(cos(x)+2)−(sin(x)−1) dx = 6π. The averagevalue isR2π0Rcos(x)+2sin(x)−1y dxdy/A =R2π0(cos(x) + 2)2−(sin(x) −1)2dx/A = (4 −1)π/(2π) =3π/(6π) =1/2 .3) (4 points) Find the volume of the solid lying under the paraboloid z = x2+ y2and above therectangle R = [−2, 2] × [−3, 3] = {(x, y) | − 2 ≤ x ≤ 2, −3 ≤ y ≤ 3 }.Solution:We have to compute the double integral of f(x, y) = x2+ y2over R. The inner integralisR3−3(x2+ y2) dx = 18 + 6y2so thatR2−2R3−3(x2+ y2) dydx =R2−2(18 + 6y2) dy = 104.4) (4 points) Calculate the iterated integralR10R2−xx(x2−y) dydx. Sketch the corresponding typeI region. Write this integral as integral over a type II region and compute the integral again.Solution:R10R2−xx(x2−y) dydx = −5/6. The region is a triangle bound by the lines y = x, the liney = 2 − x and the y axis. The inner integral is −2 + 2x + 2x2− 2x3.As a type II region, the region has to be splitR10Ry0(x2−y) dxdy +R21R2−x0(x2− y) dxdy = −1/4 − 7/12 = −5/6.5) (4 points) Compute the probability that a quantum particle with energy (k2+ n2)~2/(2m) =5~2/(2m) is in the region R = [0, π/2] × [0, π/2] = {(x, y ) | 0 ≤ x ≤ π/2, 0 ≤ y ≤ π/2 } of thesquare box [0, π] × [0, π].Check. The result will involve integralsA =Zπ0Zπ0sin2(kx) sin2(ny) dxdyas well asB =Zπ/20Zπ/20sin2(kx) sin2(ny) dxdyIn the notes, you find a n analogues computation for a different region R and a different energylevel. Identities like 1 − 2 sin2(x) = cos(2x) are useful here.Solution:The integral A = π2/4 is the integral of the wave function over the whole box, the integralB = π2/16 is the integral over the region R. The result is B/A = 1/4. This result makessense due to the symmetry of the wave.Remarks(You don’t need to read these remarks to do the problems.)A quantum mechanical particle confined to a region X in two dimensions is represented by afunction f (t, x, y) satisfyingR RXf2(t, x, y) dxdy = 1. For each time t, the probability that theparticle is in some subregion R is the double integralZ ZRf2(t, x, y) dxdy .According to the classical interpretation of quantum mechanics, particles don’t have determinedpo sitions any more, the probabilities are all an experimenter can measure. If a particle isexposed to an external field F (x, y) = ∇V (x, y ), then the evolution of the particle is given bythe equationi~mft(t, x, y) = fxx(x, y) + fyy(x, y) + V (x, y)f(x, y)which is called the Schr¨odinger equation. A special case is if there is no force F = ∇V . Inthat case, one can assume that V (x, y) = 0. Important are solutions f(x, y) which satisfy thepartial differential equationfxx(x, y) + fyy(x, y) + V (x, y)f(x, y) = Ef(x, y) ,where E is a number called the energy. In this case, the evolution of the particle isi~mft(t, x, y) =Ef(t, x, y) which has a solution ei~Et/mf(0, x, y). It is a mathematical fact that for a boundedregion X, not all energies E are allowed. They come in discrete steps, energies appear ”quan-tized”. Quantized energies also appear for potentials like V (x, y) = x2+ y2, which is calledthe quantum mechanical oscillator or V (x, y) =1√x2+y2which is called the Coulomb po-tential. All what has b een said works also in three dimensions. It is just that f dependsnow o n three space variables and the double integrals will be replaced by triple integrals. IfV (x, y, z) = 1/|(x, y, z)|, then the mathematics of the solutions to fxx(x, y, z) + fyy(x, y, z) +fzz(x, y, z) + V (x, y, z)f(x, y) = Ef(x, y) is the story of the hydrogen atom. The possibleenergy levels explains to a great deal the build-up of the periodic elements and so the stuff weare made of.Challenge Problems(Solutions to these problems are not turned in with the homework.)1) Let M be a polygon in the plane where each edge is at a lattice point. Verify that the areaA of the polygon satisfies A = I + B/2 − 1, where I is the number of lattice points inside thepo lygon a nd B is the number of lattice points at the boundary.Solution:2) The integralR10arccos(√x) dx can be written as a double integralR10Rarccos(√x)0dydx. Calculatethis
View Full Document
Unlocking...