Esercizi risolti Integrali doppi

Integrali doppi (Giordano)

Ex1

∫₀¹ ∫₀^x (x + y²) dx dy

D è il triangolo di vertici [0,0] [0,1] e [1,1]

D = { (x,y) ∈ ℜ² | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }

∫₀¹ ∫_y¹ √(y + x²) dx dy = ∫₀¹ ∫_y¹ x³/3 | dx - ∫₀¹ x^5/2 / 5/2 == &frac{2}{3} &frac{5}{3} [%]₀¹

Ex2

∫_D x dx dy

D è il quadrilatero di vertici [0,1] [2,0) [2,2] e [1,2]

D₁ = { (x,y) ∈ ℜ² | 1 ≤ y ≤ 2, -1 ≤ x ≤ 3-y }

D₂ = { (x,y) ∈ ℜ² | 0 ≤ x ≤ 2, x/3 }

∫_D y dx dy = ∫_D1 ∫_x dy + ∫_D2 y dx dy =2 [ &frac{2}{3} ] [ &frac{3}{3} ] [ 2 ] [ &frac{3}{2} ]

Ex 1

D è il triangolo di vertici (0,0) (1,0) e (1,1)

D = { (x,y) ∈ R² | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }

∬_D (√ + ²) = ∫₀¹ ∫₀ (√ + ²) = ∫₀¹ [y√ + ²]₀ = ∫₀¹ (^5/2 + ³) == ∫₀^{1 5/2} + ∫₀^{1 3} = [^7/2/^7/2] + [⁴/4]₀¹ == 2/7 [^7/2]₀¹ + 1/4 [⁴]₀¹ = 2/7 + 1/4 = 15/28

Ex 2

D è il quadrilatero di vertici (0,1) (2,0) (2,2) e (1,2)

D₁ = { (x,y) ∈ R² | 0 ≤ x ≤ 2, 2-x ≤ y ≤ 2 }

D₂ = { (x,y) ∈ R² | 1 ≤ x ≤ 2, 1 ≤ y ≤ x }

∫∫_D = ∫∫_D1 + ∫∫_D2 = ∫₀² ∫₂₋² d + ∫₁² ∫₁ d d == ∫₀² []₂₋² d + ∫₁² []₁ d = ∫₀² (2−(2−)) d + ∫₁² (−1) d = ∫₀² d + ∫₁² −1 d = [²/2]₀² + [−1]₁² = 4/2 + [1] = 2 + 1 = 3

AD:DC:AB:BC:D₁:x_g dx dy

D₂:x_g dx dy

∫₀^2y x dx =∫₀^2y (x² / 2)_|0^2y =[x² / 2_|0^2y] =(4y² / 2) - 0= 24y³ / 3 -∫₀² (4x² / 2) dx +8y3(4y²) = 4(4y² / 2) - (16y / 2)∫₀²( x² + 16x + 8x) - (4 / 7) (16y) =16 / 33 / 8 16 3= [y³ / 3]_|0³ =83 / 2 38 / 3 - 16 * 3 / 2 = (43 / 2)21 / 2 + 13 / 2 - 3l / 2

D = D₁ + D₂ = 21 / 2 + 13 / 2 - 3l / 2 = 1F∬(x-3y)dxdy

A(0,0) B(2,1) D(1,2)

D₁ = {(x,y)∈ℝ² | 0≤x≤1 ½x≤y≤2x}

∫_½x^2x(x-3y)dy dx

∫x-3y dy = x∫dy - 3∫y dy = x[y]^2x_½x - 3[y²/2]^2x_½x∫₀¹ -33/2x³ dx = -33/8

D₂= {(x,y)∈ℝ² | 1≤x≤2 ½x≤y≤-x+3}

∫_½x^-x+3(x-3y)dy dx

∫x-3y dy = x[y]^-x+3_½x - 3[y²/2]^-x+3_½x= -3/2x²+3x-12x-72x+108-3x-3x² + 3x - 9x⁴ + 7x + 108 = -12x⁴ + 24x - 9x² + 72x - 108-27x² + 96x - 108 = -27(x² + 96x + 8)-21/8 x^[3/2] |₁² x dx - 2/2 ∫₁² dx =-24/8 [x^3/3]₁² + 12 [(x^2/2]₁² - 2± (x)^7/2 |₁² =-1/8 (8-1/3) +12 (2-1/2) - 27/2 (2-1) =-x . 7/3 + 12^(3/2) - 2± 2/2 = -49/8 + 16 - 2± = 49 + 144 - 108-13/8 - 11/8 = -3∬_D (u x² + 4 y² − xy) dx dy

D = { (x,y) ∈ R² | y > 0, y ≤ √3x, x² + 4y² ≤ 1 }

y = √3x = 0

Punto spia: (√3,0) =>z(√3,0) = 0 - √3√3 < 0 al di sotto sono negativi

x² + 4y² = 1 = 0

Punto spia: (0,0)z(0,0) = -11 < 0 => all'interno sono negativi

Dominioπ = {(θ,p) ∈ R² | 0 ≤ p ≤ π, 0 ≤ θ ≤ π/3 }

∬_T (4p²cos²θ + 4p²sin²θ - (p sin θ cos θ)p dp dθ)

∬_V (4p² (cos²θ + sin²θ) - p sin θ cos θ)p dp dθ

∬_U (4p³ - p³ sin θ cos θ) dp dθ

∫₀^π/3 4p³ - p³ sin θ cos θ dθ dp

∫₀^π/3 4p³ - p³ sin θ cos θ dθ = 4p³ ∫₀^π/3 4 - sen θ cos θ dθ = p³ [4θ + cos θ]= ρ³ ( ₄/₃ π + ₁/₈ - ₁/₂) = ρ³ ( ₄/₃ π - ₃/₈) = ∫₀¹ ρ³ ( ₄/₃ π - ₃/₈) dρ = (₄/₃ π - ₃/₈) [_ρ⁴/₄]₀¹ =(₄/₃ π - ₃/₈) ⋅ ₁/₄ = ₁/₃ π - ₃/₃₂

D⌈∫∫_D x y² dx dy⌠

D = {(x, y) ∈ ℝ² | y ≤ √2x , x ≥ 0 , x 2y ≤ 1 ≤ 0}

Domenio⌈∫∫_L (p²cos³θ p³sen²θ) p dρ dθ = ⌈∫∫₀ p¹ r² cos²θ sen²θ p dρ dθ•₀ ∈ ∞ p² cos² ∈ † sen² dρ p⁰ ⌈√(cos³θp dρ = p⁵† ∫₀ sin³θ co p dθ∮p⁵ cos³θ sin³θ dρ = p⁵ ∫₀ sin³θ cos³θ p dθp⁷ ∇ e^dθp sqr(p dθ dρ

⌈ ∫_-∞ ∫_θ π ∞ ∫Es.20∫∫_D √(x² + y²)^1/2 dx dy

D = {(x,y) ∈ ℜ² | x ≥ 0, y ≥ 0, x² + y² - x ≥ 0, x² + y² ≤ 1 }

Studio del Dominio x ≥ 0 y ≥ 0 x² + y² ≤ 1 C (0,0) C ≠ 1 x² + y² - x ≥ 0 Q(-½, -b²) = Q (½, 0) r₂ = √(1-(1/2)²) = ½ Q (1,1) : 1 + 1 - 1 = 1 > 0 Disegno e Dominio

OA ≤ ρ ≤ OB OA = OD cos θ = cos θ OB = 1 cos θ ≤ ρ ≤ 1 Passaggio a Coord. Polari 0 ≤ θ ≤ π/2(ρ cos θ)² + (ρ sin θ)² - ρ cos θ ≥ 0 ρ² cos θ + ρ² imu θ - ρ cos θ ≥ 0 ρ² - ρ cos θ ≥ 0 ρ (ρ - cos θ) ≥ 0 ρ > 0 (p cos θ)² + (p sin θ)² ≤ 1 p²(cos²θ + sin²θ) ≤ 1 p² - 1 ≤ 0 → (p+1)(p-1) ≤ 0 p≤1 ✓ cos θ ≤ p ≤ 1

T: { (p,θ) ∈ ℝ² | 0 ≤ θ ≤ π/2 cos θ ≤ p ≤ 1 }

I = ∫₀^π/2 ∫_cosθ^{1 1}√(p cos θ)²(p sin θ) ² dp d θ = ∫₀^π/2 ∫_cosθ¹ √p² p dp d θ I₂ = ∫₀^π/2 ∫_cosθ¹ p² dp d θ* = ∫_cosθ¹ p² dp = [^p3/₃]_cosθ¹ = 1/3 - cos³θ / 3 I = ∫₀^π/2 (1/3 - cos³θ / 3) dθ = ∫₀^π/2 1/3 dθ - 1/3 ∫₀^π/2 cos³θ dθ J = ∫₀^π/2 cos³θ dθ = ∫₀^π/2 cosθ . cos²θ dθ J = ∫₀^π/2 cosθ (1 - sin²θ) dθ J = ∫₀^π/2 cosθ dθ + ∫₀^π/2 -cosθ sin²θ dθ = ∞ J = [-Δωμθ]₀^π/2 = [^sinθ3/₃]₀^π/2 = 1 - 1/3 - 2/3 3/3