4/5/2004, SPHERICAL INTEGRALS Math21a, O. KnillHOMEWORK: Section 12.6: 14,18,32,34,48,62REMINDER: POLAR AND CYLINDRICAL COORDINATES.Z ZRf(r, θ)r dθdr .Cylindrical coordinates are obtained by taking polar coordinates inthe x-y plane and leave the z coordinate. With T (r, θ, z) =(r cos(θ), r sin(θ), z). the integration factor r is the same as in polarcoordinates.Z Z ZT (R)f(x, y, z) dxdydz =Z Z ZRf(r, θ, z)r drdθdzSPHERICAL COORDINATES. Spherical coordinates use ρ, the distance to the origin as well as two angles: θthe polar angle and φ, the angle between the vector and the z axis. The coordinate change isT : (x, y, z) = (ρ cos(θ) sin(φ), ρ sin(θ) sin(φ), ρ cos(φ)) .The integration factor can be seen by measuring the volume of a spher-ical wedge which is dρ, ρ sin(φ) dθ, ρdφ = ρ2sin(φ)dθdφdρ.Z Z ZT (R)f(x, y, z) dxdydz =Z Z ZRf(ρ, θ, z)ρ2sin(φ) dρdθdφT : (ρ, θ, φ)7→(ρ cos(θ) sin(φ),ρ sin(θ) sin(φ),ρ cos(φ))VOLUME OF SPHERE. A sphere of radius R has the volumeZR0Z2π0Zπ0ρ2sin(φ) dφdθdρ .The most inner integralRπ0ρ2sin(φ)dφ = −ρ2cos(φ)|π0= 2ρ2.The next layer is, because φ does not appear:R2π02ρ2dφ = 4πρ2.The final integral isRR04πρ2dρ = 4πR3/3.MOMENT OF INERTIA. The moment of inertia of a bodyG with respect to an axis L is defined as the triple integralR R RGr(x, y, z)2dzdydx, where r(x, y, z) = R sin(φ) is thedistance from the axes L. For a sphere of radius R weobtain with respect to the z-axis:I =ZR0Z2π0Zπ0ρ2sin2(φ)ρ2sin(φ) dφdθdρ= (13sin3(φ)π0)(ZR0ρ4dr)(Z2pi0dθ)=43·R55· 2π =8πR515=V R25If the sphere rotates with angular velocity ω, then Iω2/2is the kinetic energy of that sphere. Example: the mo-ment of inertia of the earth is 81037kgm2. The angularvelocity is ω = 1/day = 1/(86400s). The rotational en-ergy is 81037kgm2/(7464960000s2) ∼ 1028J = 1025kJ ∼2.51024kcal.How long would you have to run on a treadmill to accumulate this energy if you could make 2’500 kcal/hour?We would have to run 1021hours = 3.61024seconds. Note that the universe is about 1017seconds old. If all the6 million people in Massachusetts would have run since the big bang on a treadmill, they could have producedthe necessary energy to bring the earth to the current rotation. To make classes pass faster, we need to spinthe earth more and just to add some more treadmills ... To the right you see a proposal for the science center.DIAMOND. Find the volume and the center of mass of a diamond, theintersection of the unit sphere with the cone given in cylindrical coordinatesas z =√3r.Solution: we use spherical coordinates to find the center of mass (x, y, z):V =R10R2π0Rπ/60ρ2sin(φ) dφdθdρ =(1−√32)32πx =R10R2π0Rπ/60ρ3sin2(φ) cos(θ) dφdθdρ/V = 0y =R10R2π0Rπ/60ρ3sin2(φ) sin(θ) dφdθdρ/V = 0z =R10R2π0Rπ/60ρ3cos(φ) sin(φ) dφdθdρ/V =2π32VPROBLEM FindR R RRz2dV for the solid obtained by inter-secting {1 ≤ x2+y2+z2≤ 4} with the double cone {z2≥ x2+y2}.Solution: since the result for the double cone is twice the resultfor the single cone, we work with the diamond shaped region R in{z > 0} and multiply the result at the end with 2.In spherical coordinates, the solid R is given by 1 ≤ ρ ≤ 2 and0 ≤ φ ≤ π/4. With z = ρ cos(φ), we haveZ21Z2π0Zπ/40ρ4cos2(φ) sin(φ) dφdθdρ = (255−155)2π(−cos3(φ))3|π/40= 2π(31/5)(1−2−3/2) .The result for the double cone is4π(31/5)(1 − 1/√23)
View Full Document
Unlocking...