Lecture 18: 7/22/2003, APPLICATIONS Math21a, O. KnillHOMEWORK. Section 12.2: 26,36,44, Section 12.5: 14THINGS TO KEEP IN MIND.• Double integrals can often be evaluatedthrough iterated integrals”Integrals have layers” .•RRR1 dxdy =RR1 dA is the area of theregion R.•RRRf(x, y) dxdy is the volume of the solidhaving the graph of f as the ”roof” and the Rin the xy− plane as the ”floor”.TYPES OF REGIONS.R RRf dA =RbaRdcf(x, y) dydx rectangle.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.A general region, we try to cut it into pieces, where each piece is a Type I or Type II region.RectangleType I Type IITo cutAREA A =RRR1 dAMASS M =RRRρ(x, y) dAAVERAGERRRf(x, y) dA/A.CENTROID (RRRx dA/A,RRRy dA/A).CTR MASS (RRRxρ(x, y)dA,RRRyρ(x, y)dA)/M .MOMENT OF INERTIA I =RRR(x2+ y2) dARADIUS OF GYRATIONpI/MVOLUME V =RRRR1 dVMASS M =RRRρ(x, y, z) dVAVERAGERRRRf(x, y, z) dV /V .CENTROID (RRRRx dV/V,RRRRy dV/V,RRRRz dV /V ).C.O.M. (RRRRxρdV,RRRRxρdV,RRRyρdV )/M.MOMENT OF INERTIA I =RRRR(x2+ y2) dVRADIUS OF GYRATIONpI/MAREA OF CIRCLE. To compute the area of the circle of radius r, we integrateA =Zr−rZ√r2−x2−√r2−x2dydx .The inner integral is 2√r2− x2so thatA =Zr−r2pr2− x2dx .This can be solved with a substitution: x = r sin(u), dx = r cos(u). With the new bounds a = −π/2, b = π/2and√r2− x2=qr2− r2sin2(u) = r cos(u) we end up withA =Zπ/2−π/22r2cos2(u) du =Zπ/2−π/2r2(1 + cos(2u)) du = r2π .MOMENT OF INERTIA. Compute the kinetic energy of a square iron plate R = [−1, 1]×[−1, 1] of density ρ = 1(about 10cm thick) rotating around its center with a 60000rpm (rounds per minute). The angular velocity speedis ω = 2π 60000/60 = 100 2π. Because E =R RR(rω)2/2 dxdy, where r =px2+ y2, we haveE = ω2I/2 ,where I =R RR(x2+ y2) dxdy is the moment of inertia. For the square, I = 8/3. Its energy of the plate isω28/6 = 4π210028/6Joule ∼ 0.86KW h. You can run with this energy a 60 Watt bulb for 14 hours.A FLOWER. What isR RRx2+ y2dxdy, where R is a flower obtained by ro-tating the region enclosed by the curves y = x2and y = 2x − x2by addingmultiples of the angles 2π/12?SOLUTION. The moment of inertia of all the petals add up: I =12R10R2x−x2x2(x2+ y2) dydx =R10[x2y + y3/3]y=2x−x2y=x2dx = 1243/210 = 86/35.PROBLEM: BERTRAND’S PARADOX (Bertrand 1889)We throw randomly lines onto the disc. What is the probability that theintersection with the disc is larger than the length√3 of the equilateral triangleinscribed in the unit circle?Answer Nr 1: take an arbitrary point P in the disc. The set of lines whichpass through that point is parametrized by an angle φ. In order that the chordis longer than√3, the angle has to fall within an angle of 60◦of a total of 180◦.The probability is1/3 .Answer Nr 2: consider all lines perpendicular to a fixed diameter. The chordis longer than√3, when the point of intersection is located on the middle halfof the diameter. The probability is1/2 .Answer Nr. 3: if the midpoint of the intersection with the disc is located inthe disc of radius 1/2 with area π/4, then the chord is longer than√3. Theprobability is1/4 .The paradox comes from the choice of the probability density function f(x, y). In each case, there is adistribution function f(x, y) which is radially symmetric.The constant distributionf(x, y) = 1/π is obtained when we throw the center of the line into the disc. Thedisc Arof radius r has probability r2/π. The density in the r direction is 2r.The distributionf(x, y) = 1/r is obtained when throwing parallel lines. This will put more weight to center.The disc Arof radius r has probability of Aris bigger than the area of Ar. The density in the r direction isconstant equal to 1.Lets compute the distribution when we rotate the line around a point at the boundary. We hit a disc Arofradius r with probability F (r) = arcsin(r)2/π. The density in the radial direction isf(r) = 2/(πp(1 − r2)) .COMPARISON OF THE DEN-SITY FUNCTIONS. A plot of theradial distribution functions f(r) aswell as F (r), the probability of Arshows why we get different resultsfor F (1/2).0.2 0.4 0.6 0.8 112340.2 0.4 0.6 0.8 10.20.40.60.81What happens if we really do an experiment and throw randomly lines onto a disc? The outcome of theexperiment will depend on how the experiment will be performed. If we would do the experiment by hand,we would probably try to throw the center of the stick into the middle of the disc. Since we would aim to thecenter, the distribution would probably be different from any of the three solutions discussed above.STATISTICS. If f(x, y) is a probability distribution on R: f(x, y) ≥ 0,RRRf(x, y) dA = 1, then E[X] =RX(x, y)f(x, y) dA is called the expectation of X, Var(X) = E[(X − E[X])2] is called the variance andσ(X) =p(Var[X]) the standard
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