4/8/2003, CYLINDRICAL/SPHERICAL COORDINATES Math 21a, O. KnillHOMEWORK. Section 12.8: 4, 8, 10, 20, 32REMINDER: INTEGRATION POLAR COORDINATES.Z ZRf(r, θ)r dθdr .EXAMPLE 1. Area of a disk of radius RZR0Z2π0r dθdr = 2πr22|R0= R2π .WHERE DOES THE FACTOR ”r” COME FROM?1. EXPLANATION. A small rectangle with dimensions dθdr in the (r, θ) plane is mapped to a sector segmentin the (x, y) plane. It has approximately the area rdθdr. It is small for small r.2. EXPLANATION (more details on Friday). Look at the map (r, θ) 7→ (r cos(θ, r sin(θ))(f(r, θ), g(r, θ)) whichchanges from Cartesian coordinates to polar coordinates. The Jacobean is T0=cos(θ) sin(θ)r sin(θ) r cos(θ)=frfθgrgθwith determinant frgθ− fθgr= r. This is a special case of a more general formula.T : (r, θ)7→(r cos(θ), r sin(θ))CYLINDRICAL COORDINATES. Use polar coordinates in the x-yplane and leave the z coordinate. Take T (r, θ, z) = (r cos(θ), r sin(θ), z).The integration factor r is the same as in polar coordinates.Z Z ZT (R)f(x, y, z) dxdydz =Z Z ZRf(r, θ, z)r drdθdzCOORDINATES OF CAMBRIDGE. On the website http : //cello.cs.uiuc.edu/cgi−bin/slamm/ip2ll/ you canenter a host like www.math.harvard.edu and get latitude and longitude of the host: (lat, lon) = (42.365, −71.1).Using (r, θ, φ) coordinates, we obtain the position (r, 90 − 42.365, −71.1) of the host in spherical coordinates.The site does not give the height, but we are about on see-level, so that r = 6365km.EXAMPLE. Calculate the volume bounded by the parabolic z = 1−(x2+y2) and the x−y plane. In cylindricalcoordinates, the paraboloid is z(r, φ) = 1 − r2:Z10Z2π0Z1−r20rdzdφdr =Z10Z2π0(r − r3) dφdr = 2π(r2/2 − r4/4)10= π .SPHERICAL COORDINATES. Spherical coordinates use the radius ρ as well as two angles: θ the polar angleand φ, the angle between the vector and the z axes. 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 from the dimensions of a sphericalwedge with dimensions 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 body G with respect to an axes L is the triple integralR R RGr(x, y, z)2dzdydx, where r(x, y, z) = R sin(φ) is the distance from L. Problem: calculate the moment ofinertia of a sphere of radius R with respect to the z-axes:I =ZR0Z2π0Zπ0ρ2sin2(φ)ρ2sin(φ) dφdθdρ = (13sin3(φ)π0)(ZR0ρ4dr)(Z2pi0dθ) =43·R55·2π =8πR515=V R25.If a sphere spins around the z-axes with angular velocity ω, then Iω2/2 is the kinetic energy of that sphere.Example: the moment of inertia of the earth is 81037kgm2. The angular velocity is ω = 1/day = 1/(86400s) sothat the energy of the earth rotation is 81037kgm2/(7464960000s2) ∼ 1028J = 1025kJ ∼ 2.51024kcal.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
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