Integrali

  • Materia: Integrali
  • Visto: 1763
  • Data: 30/06/2007
  • Di: Redazione StudentVille.it

$A = \\{(x,y,z) \\in \\mathbb{R}^3: x^2 + y^2 + 3 \\le z \\le 4 \\sqrt{x^2 + y^2}\\}$

esercizio svolto o teoria

A cura di: Gianni Sammito

Calcolare il volume dell'insieme

 

$A = \{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + 3 \le z \le 4 \sqrt{x^2 + y^2}\}$

 


Il volume dell'insieme $A$ vale

 

 

$\int \int \int_A dxdydz = \int \int_{B} (\int_{x^2 + y^2 + 3}^{4\sqrt{x^2 + y^2}} dz) dxdy$

 

con $B = \{(x,y) \in \mathbb{R}^2: x^2 + y^2 + 3 \le 4 \sqrt{x^2 + y^2}\}$.

 

$\int \int_{B} (\int_{x^2 + y^2 + 3}^{4\sqrt{x^2 + y^2}} dz) dxdy = \int \int_{B} (4 \sqrt{x^2 + y^2} - x^2 - y^2 - 3) dxdy$ (1)

 

Conviene passare in coordinate polari

 

$\{(x = \rho \cos(\theta)),(y = \rho \sin(\theta)):}$

 

$\rho \in [0, +\infty) \quad \quad \theta \in [0, 2 \pi]$ 

 

e la matrice Jacobiana vale

 

$J(\rho, \theta) = [(\frac{\partial}{\partial \rho} x, \frac{\partial}{\partial \theta} x),(\frac{\partial}{\partial \rho} y, \frac{\partial}{\partial \theta} y)] = [(\cos(\theta), - \rho \sin(\theta)),(\sin(\theta), \rho \cos(\theta))]$

 

Pertanto

 

$dxdy = |\det(J(\rho, \theta))|d \rho d \theta = |\rho \cos^2(\theta) + \rho \sin^2(\theta)|d \rho d \theta = |\rho|d \rho d \theta = \rho d \rho d \theta$

 

$x^2 + y^2 + 3 \le 4 \sqrt{x^2 + y^2} \quad \implies \quad \rho^2 + 3 \le 4 \rho \quad \implies \quad 1 \le \rho \le 3$

 

Quindi (1) equivale a

 

$\int_{0}^{2 \pi} \int_1^3 (4 \rho - \rho^2 - 3) \rho d \rho d \theta =  \int_{0}^{2 \pi} \int_1^3 (4 \rho^2 - \rho^3 - 3 \rho) d\rho d \theta = \int_{0}^{2 \pi} [\frac{4}{3} (\rho^3)_1^3 - \frac{1}{4} (\rho^4)_1^3 - \frac{3}{2} (\rho^2)_1^3]d \theta =$

$= 2 \pi (\frac{104}{3} - 20 - 12) = 2 \pi (\frac{104}{3} - 32) = 2 \pi (\frac{104 - 96}{3}) = \frac{16}{3} \pi$

 

FINE