Esercizi Analisi 2

Exercise 10.1

Let Γ be the union of the parabolic arc y = 4 - x² going from A (-2, 0) to C (2, 0) and the circle x² + y² = 4 from C to A.

Integrate the function f(x, y) = x along the closed curve Γ.

Let's have a look at Γ.

∫_Γ f ds = ∫_Γ1 f ds + ∫_Γ2 f ds

Γ₁ parametrisation

γ(t) = (t, 4 - t²), t ∈ [-2, 2]

γ̇(t) = (1, -2t)

‖γ̇(t)‖ = √(1 + 4t²)

∫_Γ1 f ds = ∫_-2² t · √(1 + 4t²) dt = 1/8 ∫_-2² p t(1 + 4t²)^1/2 dt

= -1/8 [3/2 (1 + 4t²)^3/2 ]_-2² = 0

Γ₂ parametrisation

γ(t) = (2 cos t, 2 sin t), t ∈ [π, 2π]

γ̇(t) = (-2 sin t, 2 cos t)

‖γ̇(t)‖ = 2

∫_Γ2 f ds = ∫_π^2π 2 cos t · √2 dt = 2√2 [sin t]_π^2π = 0

Therefore: ∫_Γ f ds = ∫_Γ1 f ds + ∫_Γ2 f ds = 0

Exercise 10.2

Consider the cycloid parametrized by γ(t) = (r(t-sin t), r(1-cos t)), t ∈ [0, 2π],

with r > 0 and constant linear density μ = 1.

Find the centre of mass.

We have to find the mass:

ℓ(δ) = ∫₀^2π 1 ds = ∫₀^2π √(2(1-cos t)) dt = 2r ∫₀^2π √(1-cos t)/2 dt

f(t) = (r(1-cos t), r sin t)

||f'(t)|| = (r²(1-cos t)² + r²sin²t)^1/2

= (r²+r²cos²t + r²sin²t - 2r²cos t)^1/2

= r (2 - 2 cos t)^1/2

= 2r ∫₀^2π sin(t/2) dt = 4r ∫₀^2π 1/2 sin(t/2) dt =

= 4r [–cos(t/2)]₀^2π = 4r [1+1] = 8r

x_G = 1/ℓ(δ) ∫₀^2π x μ ds = 1/8r ∫₀^2π r(t–sin t) r√(2–2 cos t) dt

= 1/8r ∫₀^2π 2r²(t–sin t) √(1–cos t)/2 dt

= r/4 ∫₀^2π (t–sin t) sin(t/2) dt = 4π²/4 = πr

∫₀^2π t sin t/2 dt = [2t(− cos t/2)]₀^2π + ∫₀^2π 2 cos t/2 dt = 4π + 4 [sin t/2]₀^2π + 0

= 4π + 4 [sin t/2]₀^2π

= 4π

by parts

f g' = [F g]-F' g

f(t)=t, g'(t)=sin t/2

f'(t)=1, g(t)= ∫ sin t/2 = 2∫ sin t/2 = 2 (– cos t/2)

∫₀^2π sin t sin(t/2) dt = ∫₀^2π sin t (1–cos t/2)/√2 dt = √2/2 [3/2 (1–cos t)^3/2 ]₀^2π = 0

Σ_e parametrized by b_e(θ, h) = (cosθ, sinθ, h)

θ ∈ [0, 2π], 0 ≤ h ≤ 3 - cosθ - ¹⁄₂sinθ

area(Σ_e) = ∫_Σe 1  dS = ∫₀^2π ∫₀^{3-cosθ-1⁄₂sinθ}  1  dθ dh = ∫₀^2π 3 - cosθ - ¹⁄₂sinθ dθ

<∂_θb_e∧∂_hb_e∇^ = || || ||                         ‍‍‍‍‍         ‍    ‍‍‍‍ ‍‍‍‍             ‍‍‍‍‍‍‍‍‍-sinθ  cosθ   ‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍‍b_e = (cosθ,sinθ,0)   ‍‍‍‍‍0             ‍‍‍‍‍‍‍‍‍‍‍1

||∂_θb_e∧∂_hb_e|| = 1

        =[3θ - sinθ + ¹⁄₂cosθ]₀^2π

        = 6π + ¹⁄₂ - ¹⁄₂ = 6π

      Σ − π + ^3π⁄₂ + 6π = ¹³⁄₂ π