By Mejlbro L.
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Additional resources for Calculus 2c-6, Examples of Space Integrals
D I have here found four variants: 1) Reduction in spherical coordinates. 2) Reduction in semi-polar coordinates. 3) Reduction by the slicing method. 4) Reduction in rectangular coordinate. These methods are here numbered according to their increasing diﬃculty. The fourth variant is possible, but it is not worth here to produce all the steps involved, because the method cannot be recommended i this particular case. I First variant. Spherical coordinates. The set A is described in spherical coordinates by r ∈ [a, 2a], ϕ ∈ [0, 2π], θ ∈ 0, (r, ϕ, θ) π 2 , hence by the reduction of the space integral, A z dΩ = 2 2 c + x + y2 + z2 π 2 = 2π 0 sin2 θ 2 = 2π = π 2 cos θ sin θ dθ · 4a2 a2 π 2 0 1− · 4a2 a2 c2 c2 + t π 2 2π 0 a 0 2a a c2 2a r cos θ · r2 sin θ dr dθ c2 + r2 r2 · r dr + r2 dϕ [t = r 2 ] t + c 2 − c2 1 · dt c2 + t 2 dt = π t − c2 ln c2 + t 2 4a2 t=a2 = π 2 3a2 − c2 ln 4a2 + c2 a2 + c2 .
C2 −x2 0 |y| ≤ √ c2 −z 2 c J z· d z dΩ = 2 1 z(c2 − z 2 ) dz · · 1 − t2 dt dz 1 − t2 dt = c4 · 0 π , 4 where there are lots of similar variants. 2) First variant. A symmetric argument. The set A is symmetric with respect to the planes y = 0 and x = 0, and the integrand x, resp. y, is an odd function. Hence, x dΩ = 0 and A y dΩ = 0. A Second variant. Spherical coordinates. By insertion, π 2 2π x dΩ = 0 A 0 2π = 0 c 0 cos ϕ dϕ · r sin θ cos ϕ · r 2 sin θ dr dθ π 2 0 sin2 θ dθ · c 0 dϕ r3 dr = [sin ϕ]2π 0 · π c4 · = 0, 4 4 and similarly.
Reduced by the factor 1 − h area(B(z)) = 1 − z h 2 area B = 1 − z h 2 A. com 63 Calculus 2c-6 Volume 2) Using the result from 1) we get by the slicing method, h vol(K) h dΩ = = dx dy 0 K h = 0 1− z h dz = area((B(z)) dz 0 B(z) 2 A dz = Ah − 3 h z 1 1− 3 h = 0 1 hA. 3 3) Let the cone be homogeneously coated (density μ > 0). Then the mass is M = μ vol(K) = 1 μhA. 3 The z-coordinate ζ of the centre of gravity is given by M ·ζ =μ z dΩ, K thus ζ = μ M z dΩ = K h z z 1− h h = 3 0 1 3 1 4 t − t 3 4 = 3h 1 3 μ μhA 2 h 0 1 dz = 3h 0 1 (1 − t)t2 dt = 3h 1 1 − 3 4 = 3h 0 = h 3 hA z · areal(B(z)) dz = 1 0 0 z 1− z h 2 A dz (t2 − t3 ) dt h .