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lec_week11

# lec_week11 - MIT OpenCourseWare http/ocw.mit.edu 18.02...

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MIT OpenCourseWare http://ocw.mit.edu 18.02 Multivariable Calculus Fall 200 7 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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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 2 ρ = x 2 + y 2 + z = r 2 + 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 = x 2 + y 2 ). φ = π/ 2 is the xy-plane. Volume element: dV = ρ 2 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 Δ φ , Δ θ , with sides = portion of circle of radius a , of length a Δ φ , and portion of circle of radius r = a sin φ , of length r Δ θ = a sin φ Δ θ . So Δ S a 2 sin φ Δ φ Δ θ , which gives the surface element dS = a 2 sin φ dφdθ . The volume element follows: for a small “box”, Δ V = Δ S Δ ρ , so dV = dρ dS = ρ 2 sin φ dρdφdθ . Example: recall 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: 2 π π/ 4 1 ρ 2 sin φ dρ dφ dθ. 0 0 1 2 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 = 1 2 ). 2 π 5 π Evaluation: not done. Final answer: . 3 6 2 Application to gravitation. Gravitational force exerted on mass m at origin by a mass Δ M at ( x, y, z ) (picture shown) is given by F = G Δ M m , dir ( F ) = x, y, z , i.e. F = G Δ M m x, y, z . ( G = gravitational | | ρ 2 ρ ρ 3 constant).
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