Integrali doppi e tripli - analisi

INTEGRALI DOPPI

1. ∬_{[-π,π]×[0,π]} senx seny dxdy = ∫₀^π seny dy ∫_-π^π senx dx = 0 per simmetria (senx seny è una funzione dispari)

2. ∬_{[0,π]×[0,π]} senx seny dxdy = ∫₀^π seny dy ∫₀^π senx dx = ∫₀^π senx dx [-cosx]₀^π =

   [senx dx] 2 = 2 [-cosx]₀^π. 2 ⋅ 2 = 4

3. ∬_{[-π,π]×[0,π]} |senx| seny dxdy = ∫₀^π seny dy ∫_-π^π |senx| dx =

   ∫₀^π |senx| dx = -cosy|₀^π = ∫_-π^π |senx| dx ⋅ 2 = 2 ∫₀^π senx dx ⋅ 2 =

   2 ⋅ 2 [-cosx]₀^π = 2 ⋅ 2 ⋅ 2 = 2³ = 8

4. ∬_Ω (x+y) dxdy Ω = {(x,y) ∈ ℝ² : 0 < y < √2/2 , y < x < √(1-y²)}

   ∫₀^√2/2 dy ∫_y^√(1-y2) (x+y) dx = ∫₀^√2/2 dy [ x²/2 + xy]^√(1-y2)_y =

   ∫₀^√2/2 dy [(1-y²)/2 - y²/2 ) + (y √(1-y²) - y²) ] =

   = ∫₀^√2/2 (1/2 - y²/2 + y√(1-y²) - y²) dy =

= ∫₀^√3/2 dy - 2 ∫₀^√3/2 y² dy + ∫₀^√3/2 y √(1 - y²) dy =

= [1/2 y - 2 y³/3 + (1 - y²)^3/2]₀^√3/2 = 1/3

5.

∫∫_Ω (x² + y²) dx dy, Ω = { (x, y) ∈ ℝ² : 0 ≤ x ≤ 1, 1 ≤ y ≤ 2 }

∫₀¹ ∫₁² (x² + y²) dy dx = ∫₀¹ [x² y + y³/3]²₁ dx = ∫₀¹ [2x² - x²] + (8/3 - 1/3) dx =

= ∫₀¹ (x² + 7/3) dx = [x³/3 + 7/3 x]¹₀ = 1/3 + 7/3 = 8/3

11. \(\int\int_{\Omega} \log(xy) dxdy\)

\(\Omega = \{(x,y) \in \mathbb{R}^2: -1 \leq x \leq -\frac{1}{2},\ 4x \leq y \leq \frac{1}{x}\}\)

\(\int_{-1}^{-\frac{1}{2}} dx \int_{4x}^{\frac{1}{x}} \log(xy) dy\)

\(\int_{-\frac{1}{2}}^{-1} \left( \left[ y \log(xy) \right]_{4x}^{\frac{1}{x}} - \int_{4x}^{\frac{1}{x}} \frac{1}{x} dy \right) dx =\)

= \(\int_{-\frac{1}{2}}^{-1} (-4x \log 4x^2 - \frac{1}{x} + 4x) dx = -4 \left( \left[ \frac{1}{2} x^2 \log x^2 \right]_{-1}^{-\frac{1}{2}} - \int_{-1}^{-\frac{1}{2}} x dx + \left[ -\log |x| + 2x \right]_{-1}^{-\frac{1}{2}} \right)\)

= -4 \left( -\log 2 + \frac{3}{2} \right) + \log \frac{1}{27} = 5 \log 2 - 3

12. \(\int \int_{\Omega} \frac{1}{(x+y)^2} dxdy\)

\(\Omega = \{(x,y) \in \mathbb{R}^2: 1 \leq x \leq 2, 3 \leq y \leq 4\}\)

\(\int_{1}^{2} dx \int_{3}^{4} \frac{1}{(x+y)^2} dy = \int_{1}^{2} dx \left[ -\frac{1}{x+y} \right]_{3}^{4} =\)

\(\int_{1}^{2} \left( \frac{1}{x+3} - \frac{1}{x+4} \right) dx =\)

\(\left[ \log |x+3| - \log |x+4| \right]_{1}^{2} = (\log 5 - \log 4) - (\log 6 - \log 5)\)

= 2 \log 5 - \log 4 - \log 6 = \log(5^2) - \log(4 \times 6) = \log 25 - \log 24

INTEGRALI DOPPI

1.

∬ (x|y| + ¹/₂ x²|y| + x e^x) dxdy

= ∬ x|y| dxdy + ¹/₂ ∬ x²|y| dxdy + ∬ x e^xy dxdy = I₁ + I₂ + I₃

I₁ = 0 per simmetria

I₂ = ¹/₂ ∫_-1¹ x² dx ∫_-2² |y| dy = ¹/₂ ∫₀² ydy ∫₀¹ x² dx = ∫₀¹ x² dx ∙ ¹/₂ ∫₀² y dy =

= [^x3/₃]₀¹[^y2/₂]₀² = ¹/₃ ∙ ⁴/₂ = ⁴/₃

I₃ = ∬_-1¹ dx ∬₀¹ x e^xy dy = 2∫_-1¹ [e^x]² dx = 2∫_-1¹ (e^x - 1) dx =

= 2 [^e2x/₂ - x]_-1¹ = 2 [ ( ^e2/₂ - 1) - ( ^e-2/₂ + 1) ] = e² - 4 - e^-2

I₄ + I₂ + I₃ = 0 + ⁴/₃ + e² - 4 - e^-2 = e² - e^-2 - ⁸/₃

3.

∬ |x|y² dxdy 4x² + y² ≤ 1

x = ½ ρcosθ y = ρsenθ dxdy = ½ ρdρdθ

D = {ρ,θ} ; ρ ∈ [0,1], θ ∈ [0,2π]

∫_D 1/2 ρ|ρcosθ|ρsenθ ρdθ = (1/2)∫₀¹ ρ⁴ dρ ∫₀^2π |cosθ| sen²θ dθ = = (1/2) ∫₀¹ ρ⁵ dρ) (4 ∫₀^π/2 cosθ sen²θ dθ) = 1/10 * 4 [sen³θ/3]₀^π/2 = 1/10 * 4/3 = 2/15

4.

∬ |y-x| dxdy D = {(x,y): x² + y² ≤ 4}

D* = {ρ,θ} ; ρ ∈ [0,1], θ ∈ [π/4, 5π/4]

2 ∫₀¹ dρ ∫_π/4^5π/4 (ρsenθ - ρcosθ) dθ

= 2 ∫₀¹ ρ² dρ ∫_π/4^5π/4 (senθ - cosθ) dθ = = 2 [ρ³/3]₀¹ [-cosθ - senθ]_π/4^5π/4 =

= 2 * 1/3 (1/√2 + 1/√2 + 1/√2 + 1/√2) = 2/3 * 4/√2 = 8/3√2

Rombo

1: Rispetto ad un asse passante per il centro e perpendicolare al suo piano:

Ω = { (x,y): -a ≤ x ≤ 0, -b - b/a x ≤ y ≤ b + b/a x } ∪

{ (x,y): 0 ≤ x ≤ a, -b + b/a x ≤ y ≤ b - b/a x }

|Ω| = 1/2 ab

Per simmetria il momento di inerzia è uguale a quello di uno dei 4

triangoli uguali in cui è suddiviso dalle diagonali, scegliendo

il triangolo del primo quadrante:

T = { (x,y): 0 ≤ x ≤ a, 0 ≤ y ≤ b - b/a x }

I = m/2ab ∫₀^a dx ∫₀^{b-b/a x} (x² + y²) dy =

= 2m/ab ∫₀^a dx [x² y + y³/3 ]₀^{b-b/a x} = 2m/ab [x²(b-b/a x) + (b-b/a x)³/3 ] dx =

= 2m/ab ∫₀^a (x²(b-b/a x) + b³(1-x/a)³/3) dx = 2m/ab [x³b/3 - b^x/4 + ab³(1-x/a)³/4 ] ₀^a =

= 2m/ab (a³b/3 - ab⁴/4 + ab³/12) = 2m/ab (a³b/3 - ab³/4 + ab³/12) =

= 2m/ab ( 1/12 a³b + 1/12 ab³ ) = 2m/ab ( 1/12 a³ + 1/12 b³ ) =

= 2m/12⁶ (a² + b²) = m/6 (a² + b²)

5. SIA T IL TRIANGOLO, NEL PIANO x,y, DI VERTICI (-4,0),(0,2),(3,0), CALCOLARE:

∬_T x² y dy dx

T = \left\{ \begin{array}{ll} (x,y) : -4 \le x \le 0 ; 0 \le y \le 2x+2 \\ \cup \\ (x,y) : 0 \le x \le 3 ; 0 \le y \le 2- \frac{2x}{3} \end{array} \right\}

= \int_{-4}^{0} x^{2} dx \int_{0}^{2x+2} y dy + \int_{0}^{3} x^{2} dx \int_{0}^{2-\frac{2x}{3}} y dy =

= \int_{-4}^{0} x^{2} dx \left[ \frac{y^{2}}{2} \right]_{0}^{2x+2} + \int_{0}^{3} x^{2} dx \left[ \frac{y^{2}}{2} \right]_{0}^{2-\frac{2x}{3}} =

= \int_{-4}^{0} \frac{x^{2}(2x+2)^{2}}{2} dx + \int_{0}^{3} \frac{x^{2}\left(2-\frac{2x}{3}\right)^{2}}{2} dx =

= \int_{-4}^{0} x^{2}(x+1) dx + \int_{0}^{3} x^{2} \left( \frac{36+4x^{2}-24x}{18} \right) dx = \int (x^{3} + x^{2}) dx + \int \left( x^{3}+\frac{2}{9} x^{2} + \frac{4}{3} x \right) dx

= \left[ \frac{x^{4}}{4} + \frac{x^{3}}{3} \right]_{-4}^{0} + \left[ \frac{2x^{3}}{3} + \frac{2x^{5}}{45} - \frac{x^{4}}{3} \right]_{0}^{3} = \left[ \left( \frac{1}{4} + \frac{1}{3} \right) + \left( 18 + \frac{54}{5} - 27 \right) \right] =

= -\frac{1}{12} + \frac{9}{5} =