Integrali

  • Materia: Integrali
  • Visto: 2183
  • Data: 28/06/2007
  • Di: Redazione StudentVille.it

$\\int \\int \\int_A (z+1) dxdydz$

esercizio svolto o teoria

A cura di: Gianni Sammito

Calcolare

 

$\int \int \int_A (z+1) dxdydz$

 

con $A = \{(x,y,z) \in \mathbb{R}^3: 1 \le x^2 + y^2 + z^2 \le 4\}$

 


Conviene passare in coordinate sferiche, mediante la trasformazione

 

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

 

inizialmente con $\rho \in [0, +\infty)$, $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\phi \in [0, 2 \pi]$.

La matrice Jacobiana è

 

$J(\rho, \theta, \phi) = [(\frac{\partial}{\partial \rho}x, \frac{\partial}{\partial \theta}x, \frac{\partial}{\partial \phi}x),(\frac{\partial}{\partial \rho}y, \frac{\partial}{\partial \theta}y, \frac{\partial}{\partial \phi}y),(\frac{\partial}{\partial \rho}z, \frac{\partial}{\partial \theta}z, \frac{\partial}{\partial \phi}z)] = [(\cos(\theta) \cos(\phi), -\rho \sin(\theta) \cos(\phi), -\rho \cos(\theta) \sin(\phi)),(\cos(\theta) \sin(\phi), -\rho \sin(\theta) \sin(\phi), \rho \cos(\theta) \cos(\phi)),(\sin(\theta), \rho \cos(\theta), 0)]$

 

Calcolando il determinante usando la regola di Sarrus, si ottiene

 

$\det(J(\rho, \theta, \phi)) = \rho^2 \sin^2(\theta) \cos(\theta) \cos^2(\phi) - \rho^2 \cos^3(\theta) \sin^2(\phi) - \rho^2 \sin^2(\theta) \cos(\theta) \sin^2(\phi) - \rho^2 \cos^3(\theta) \cos^2(\phi) =$

$ =-\rho^2 \cos^3(\theta) - \rho^2 \sin^2(\theta) \cos(\theta) = - \rho^2 \cos(\theta)$

 

Dato che

 

$dxdydz = |\det(J(\rho, \theta, \phi))|d \rho d \theta d \phi$

 

allora

 

$dxdydz = |- \rho^2 \cos(\theta)| d\rho d \theta d \phi = \rho^2 \cos(\theta) d \rho d \theta  d \phi$

 

considerando che

 

$\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \implies \cos(\theta) \ge 0$

 

Considerando il vincolo $1 \le x^2 + y^2 + z^2 \le 4$, sostituendo i valori di $x,y,z$ in funzione di $\rho, \theta, \phi$, si trova $1 \le \rho^2 \le 4$, ovvero $\rho \in [1,2]$. 

 

Dunque l'integrale diventa

 

$\int_{0}^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{1}^{2} (\rho \sin(\theta) + 1) (\rho^2 \cos(\theta)) d \rho d \theta d \phi =$

$ =  \int_{0}^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{1}^{2} (\rho^3 \sin(\theta) \cos(\theta) + \rho^2 \cos(\theta)) d \rho d \theta d \phi =$

$ = \int_{0}^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} [\frac{\sin(2 \theta)}{8} (\rho^4)_1^2 + \frac{\cos(\theta)}{3} (\rho^3)_1^2] d \theta d \phi =$

$= \int_{0}^{2 \pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\frac{15}{16} 2\sin(2 \theta) + \frac{7}{3} \cos(\theta)) d \theta d \phi =$

$ = \int_{0}^{2 \pi} (\frac{-15}{16} (\cos(2 \theta))_{-\frac{\pi}{2}}^{\frac{\pi}{2}} + \frac{7}{3} (\sin( \theta))_{-\frac{\pi}{2}}^{\frac{\pi}{2}}) d \phi =$

$ = \frac{14}{3} \int_{0}^{2 \pi} d \phi = \frac{28}{3} \pi$

 

FINE