Calculus 2c-6, Examples of Space Integrals by Mejlbro L.

By Mejlbro L.

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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 .

Find the volume and the space integral. I The volume is vol(Ω) π (x2 + y 2 ) dx dy = = 0 B = π· a4 + 4 0 −a 1 3 x + y2 x 3 a 0 a+y · d dϕ + dy = x=−a−y 4 = 2 1 πa 2 1 + (a + y)4 + ay 3 + y 4 4 6 3 2 = a4 π 1 2 1 + + − 4 6 3 2 = a4 a+y −a −a−y πa4 + 4 0 0 −a (x2 + y 2 ) dx dy 2 (a + y)3 + 2y 2 (a + y) dy 3 4 = −a π 1 + 4 3 0 1 πa 2 1 + a4 + a4 − a4 4 6 3 2 . 5 –1 Figure 35: The projection B of Ω for a = 1. Due to the symmetry with respect to the Y -axis, we get for the space integral that xy(x2 + y 2 ) dx dy + vol(Ω) = 0 + vol(Ω) = a4 (xy + 1) dΩ = Ω B π 1 + 4 3 .

1. 2. 2) The set A is the unit ball, so ϕ ∈ [0, 2π], and B ∗ = B ∗ (ϕ) is the unit half circle in the right half plane which does not depend on ϕ, B ∗ = B ∗ (ϕ) = {(r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π}. Then by the reduction theorem in spherical coordinates, (x2 + y 2 + z 2 )2 dΩ = 2π B∗ A r4 · r2 sin θ dr dθ = 2π 1 0 r6 · π sin θ dθ = 0 4π . 7 3) The domain of integration is that part of the unit ball which lies in the first octant, thus π 0 ≤ ϕ ≤ and 2 B ∗ (ϕ) = (r, θ) 0≤θ≤ π 2 for 0 ≤ ϕ ≤ π . 3. 2 By the reduction theorem in spherical coordinates, π 2 xyz dΩ = A 0 π 2 = 0 = B∗ r3 sin2 θ cos θ · sin ϕ cos ϕ · r 2 sin θ dr dθ dϕ 1 sin ϕ · cos ϕ dϕ · 1 sin2 ϕ 2 π 2 0 r2 6 · 0 1 0 r dr · 1 sin4 θ 4 · π 2 5 sin3 θ cos θ dθ 0 π 2 = 0 1 1 1 1 · · = .

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