11/26/2002, 3D INTEGRATION Math 21a, O. KnillHOMEWORK:12.7: 4, 6, 8, 30, 343D INTEGRATION. If f(x, y, z) is a function of three variables and R is a region in space, thenR R RRf(x, y, z) dxdydz is defined as the limit of Riemann sum1n3P~xijk)∈Rf(~xijk) for n → ∞, where~xijk= (in,jn,kn).TRIPLE INTEGRALS. As in two dimensions, triple integrals cab be evaluated through iterated 1D integrals.EXAMPLE. If R is the box [0, 1] × [0, 1] ×[0, 1] and let f(x, y, z) = 24x2y3z.R10R10R1024x2y3z dxdy dz .CALCULATION. We start from the coreR1024x2y3z dx= 12x3y3, then integrate the middle layer:R1012x3y3dy= 3x2and finally handle the outer layer:R103x2dx=1.WHAT DID WE DO? When we calculate the most inner integral, we fix z and y. The integral is the average off(x, y, z) along a line intersected with the body. After completing the second integral, we have computed theaverage on the plane z = const intersected with R. The most outer integral averages all these two dimensionalsections.VOLUME OF THE SPHERE. (We will do this more elegantly later). The volume isV =Z Z ZRdxdydz =Z1−1[Z√1−x2−√1−x2[Z√1−x2−y2−√1−x2−y2dz]dy]dxAfter computing the inner integral, we have V = 2R1−1[R√1−x2−√1−x2(1 − x2− y2)1/2dy]dx.To resolve the next layer, call 1 − x2= a2. The task is to findRa0√a2− x2dx. Make the substitutionx/a = sin(u), dx = a cos(u) to write this as aRarcsin(a/a)0q1 − sin2(u)a cos(u)du = a2Rπ/20cos2(u)du = a2π/2.At this stage we have computed the area of a disc of radius a (which is the intersection of the sphere with theplane x = const. And we can finished up by calculating the last, the most outer integralV = 2π/2Z1−1(1 − x2) dx = 4π/3 .TOTAL MASS. Compute the mass of a body which is bounded by theparabolic cylinder z = 4 − x2, and the planes x = 0, y = 0, y = 6, z = 0.The density of the body is 1.Z20Z60Z4−x20dz dy dx =Z20Z60(4 − x2) dydx= 6Z20(4 − x2) dx = 6(4x − x3/3|20= 32CENTER OF MASS. Compute the center of mass of the same body. The center of mass is(24/32, 96/32, 256/180) = (3/4, 3, 8/5):Z20Z60Z4−x20x dz dy dx =Z20Z60x(4 − x2) dydx = 6Z20x(4 − x2) dx = 24x2/2 − 6x4/4|20= 24Z20Z60Z4−x20y dz dy dx =Z20Z60y(4 − x2) dydx =Z2018(4 − x2) dx = 18(4x − x3/3)|20= 96Z20Z60Z4−x20z dz dy dx =Z20Z60(4 − x2)2/2 dydx = 6Z20(4 − x2)2/2 dx = 3(16x − 8x3/3 + x5/5)|20= 256/5SOME HISTORY OF COMPUTING VOLUMES. How did people come up calculating the volumeR R RR1 dxdydz of a body?Archimedes ((-287)-(-212): Archimedes’s method of integration al-lowed him to find areas, volumes and surface areas in many cases. Hismethod of exhaustion paths the numerical method of integration byRiemann sum. The Archimedes principle states that any body sub-merged in a water is acted upon by an upward force which is equal to theweight of the displaced water. This provides a practical way to computevolumes of complicated bodies.Cavalieri (1598-1647): Cavalieri could determined area and volumeusing tricks like the Cavalieri principle. Example: to get the volumethe half sphere of radius R, cut away a cone of height and radius R froma cylinder of height R and radius R. At height z this body has a crosssection with area R2π − r2π. If we cut the half sphere at height z, weobtain a disc of area (R2− r2)π. Because these areas are the same, thevolume of the half-sphere is the same as the cylinder minus the cone:πR3− πR3/3 = 2πR3/3 and the volume of the sphere is 4πR3/3.Newton (1643-1727) and Leibniz(1646-1716): Newton and Leibniz,developed calculus independently. Thenew tool made it possible to computeintegrals through ”anti-derivation”.Suddenly, it was possible to computeintegrals using analytic tools.MONTE CARLO COMPUTATIONS. Here is an other way to compute integrals: Suppose we want to calculatethe volume of some body R inside the unit cube [0, 1] × [0, 1] × [0, 1]. The Monte Carlo method is to shootrandomly onto the unit cube and count the fraction of times, we hit the region. Here is an experiment withMathematica and where the body is one eigths of the unit ball:R := Random[]; k = 0; Do[x = R; y = R; z = R; If[x2+ y2+ z2< 1, k + +], {10000}]; k/10000Assume, we hit 5277 of 10000 times. The volume is so measured as 0.5277. The volume of 1/8’th of the sphereis π/6 = 0.524WHERE CAN TRIPLE INTEGRALS OCCUR?• Calculation of volumes, masses.• Finding averages. Examples: average algae concentration in a swimming pool.• Determining probabilities. Example: quantum probability• Moment of inertiaR RRr(x, y, z)2ρ(x, y)dxdydz, where r(x, y, z) is the distance to the axes of rotation.• Center of mass calculation (R R RRx dxdydz/M,R R RRy dxdydz/M,R R RRz
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