MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable CalculusFall 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� 18.02 Lecture 26. – Tue, Nov 13, 2007 Spherical coordinates (ρ, φ, θ). ρ = rho = distance to origin. φ = ϕ = phi = angle down from z-axis. θ = same as in cylindrical coordinates. Diagram drawn in space, and picture of 2D slice by vertical plane with z, r coordinates. Formulas to remember: z = ρ cos φ, r = ρ sin φ (so x = ρ sin φ cos θ, y = ρ sin φ sin θ). 2 2ρ = � x2 + y2 + z = √r2 + z . The equation ρ = a defines the sphere of radius a centered at 0. On the surface of the sphere, φ is similar to latitude, except it’s 0 at the north pole, π/2 on the equator, π at the south pole. θ is similar to longitude. φ = π/4 is a cone (asked using flash cards) (z = r = x2 + y2). φ = π/2 is the xy-plane. Volume element: dV = ρ2 sin φ dρ dφ dθ. To understand this formula, first study surface area on sphere of radius a: picture shown of a “rectangle” corresponding to Δφ, Δθ, with sides = portion of circle of radius a, of length aΔφ, and portion of circle of radius r = a sin φ, of length rΔθ = a sin φΔθ. So ΔS ≈ a2 sin φ ΔφΔθ, which gives the s urface elem ent dS = a2 sin φ dφdθ. The volume element follows: for a small “box”, ΔV = ΔS Δρ, so dV = dρ dS = ρ2 sin φ dρdφdθ. Example: rec all the complicated example at end of Friday’s lecture (region sliced by a plane inside unit sphere). After rotating coordinate system, the question becomes: volume of the portion of unit sphere above the plane z = 1/√2? (picture drawn). This can be set up in cylindrical (left as exercise) or spherical coordinates. For fixed φ, θ we are slicing our region by rays straight out of the origin; ρ ranges from its value on the plane z = 1/√2 to its value on the sphere ρ = 1. Spherical coordinate equation of the plane: z = ρ cos φ = 1/√2, so ρ = sec φ/√2. The volume is: � 2π � π/4 � 1 ρ2 sin φ dρ dφ dθ. 0 0 √12 sec φ (Bound for φ explained by looking at a slice by vertical plane θ = constant: the edge of the region is at z = r = √12 ). 2π 5πEvaluation: not done. Final answer: .3 − 6√2 Application to gravitation. Gravitational force exerted on mass m at origin by a mass ΔM at (x, y, z) (picture shown) is given by F�= G ΔM m , dir(F�) = �x, y, z�, i.e. F�= G ΔM m �x, y, z�. (G = gravitational | | ρ2 ρ ρ3 constant). If instead of a point mass we have a solid with density δ, then we must integrate contributions to gravitational attraction from small pieces ΔM = δ ΔV . So ��� ��� F�= Gm �x, y, z� δ dV, i.e. z-component is Fz = Gm z δ dV, . . . ρ3 ρ3 R R If we c an set up to use symmetry, then Fz can be computed nicely using spherical coordinates. General setup: place the mass m at the origin (so integrand is as above), and place the solid so that the z-axis is an axis of symmetry. Then F�= �0, 0, Fz� by symmetry, and we have only one 1� �� �� �� 2 component to compute. Then ��� ��� ��� z ρ cos φ Fz = Gm ρ3 δ dV = Gm ρ3 δ ρ2 sin φ dρ dφ dθ = Gm δ cos φ sin φ dρ dφ dθ. R R R Example: Newton’s theorem: the gravitational attraction of a spherical planet with uniform density δ is the same as that of the equivalent point mass at its center. [[Setup: the sphere has radius a and is centered on the positive z-axis, tangent to xy-plane at the origin; the test mass is m at the origin. Then ��� z � 2π � π/2 � 2a cos φ 4 GMm Fz = Gm δ dV = Gm δ cos φ sin φ dρ dφ dθ = = Gmδ πa = ρ3 ··· 3 a2 R 0 0 0 where M = mass of the planet = 43 πa3δ. (The bounds for ρ and φ need to be explained carefully, by drawing a diagram of a vertical slice with z and r coordinate axes, and the inscribed right triangle with vertices the two poles of the sphere + a point on its surface, the hypothenuse is the diameter 2a and we get ρ = 2a cos φ for the spherical coordinate equation of the sphere).]] 18.02 Lecture 27. – Thu, Nov 15, 2007 Handouts: PS10 solutions, PS11 Vector fields in space. At every point in space, F�= P ˆı + Qˆj + Rkˆ, where P, Q, R are functions of x, y, z. Examples: force fields (gravitational force F�= −c�x, y, z�/ρ3; electric field E, magnetic field B); velocity fields (fluid flow, v = v(x, y, z)); gradient fields (e.g. temperature and pressure gradients). Flux. Recall: in 2D, flux of a vector field F�across a curve C = C F�nˆds.· In 3D, flux of a vector field is a double integral: flux through a surface, not a curve! F�vector field, S surface, nˆunit normal vector: Flux = F�nˆdS.· Notation: dS�= nˆdS. (We’ll see that dS�is often easier to compute than nˆand dS). Remark: there are 2 choices for nˆ(choose which way is counted positively!) Geometric interpretation of flux: As in 2D, if F�= velocity of a fluid flow, then flux = flow per unit time across S. Cut S into small pieces, then over each small piece: what passes through ΔS in unit time is the contents of a parallelepiped with base ΔS and third side given by F�. Volume of box = base × height = (F�nˆ) ΔS.· • Examples: 1) F�= xˆı + yˆj + zkˆthrough sphere of radius a centered at 0. nˆ= a 1 �x, y, z� (other choice: −a 1 �x, y, z�; traditionally choose nˆpointing out). F�nˆ= �x, y, z� · nˆ= 1 (x2 + y2 + z2) = a, so F�nˆdS = a dS = a (4πa2).· a S S·�� �� �� ������ ������ � � 3 z2) Same sphere, H�= zkˆ: H�nˆ= a 2 .· z2 � 2π � π a2 cos2 φ � π 4 H�dS�= dS = a 2 sin φ dφdθ = 2πa3 cos2 φ sin φ dφ = πa3 .· a a 3S S 0 0 0 Setup. Sometimes we have an easy geometric argument, but in general we must compute the surface integral. The s etup requires the use of two parameters to describe the surface, and F�nˆdS· must be expressed in terms of them. How to do this depends on the type of surface. For now, formulas to remember: …
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