Calcolare
[math]\int \int \int_A (z+1) dxdydz[/math]
con
[math]A = {(x,y,z) \in \mathbb{R}^3: 1 \le x^2 + y^2 + z^2 \le 4}[/math]
Conviene passare in coordinate sferiche, mediante la trasformazione
[math]\egin{cases} x = \\rho \\cos(\theta) \\cos(\phi) \\ y = \\rho \\cos(\theta) \\sin(\phi) \\ z = \\rho \\sin(\phi) \ \end{cases}[/math]
inizialmente con
[math]\\rho \in [0, +\infty)[/math]
, [math]\theta \in [-\frac{\\pi}{2}, \frac{\\pi}{2}][/math]
, [math]\phi \in [0, 2 \\pi][/math]
. La matrice Jacobiana è
[math]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)][/math]
Calcolando il determinante usando la regola di Sarrus, si ottiene
[math]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) =[/math]
[math] =-\\rho^2 \\cos^3(\theta) - \\rho^2 \\sin^2(\theta) \\cos(\theta) = - \\rho^2 \\cos(\theta)[/math]
Dato che
[math]dxdydz = |det(J(\\rho, \theta, \phi))|d \\rho d \theta d \phi[/math]
allora
[math]dxdydz = |- \\rho^2 \\cos(\theta)| d\\rho d \theta d \phi = \\rho^2 \\cos(\theta) d \\rho d \theta d \phi[/math]
considerando che
[math]\theta \in [-\frac{\\pi}{2}, \frac{\\pi}{2}] \implies \\cos(\theta) \ge 0[/math]
Considerando il vincolo
[math]1 \le x^2 + y^2 + z^2 \le 4[/math]
, sostituendo i valori di [math]x,y,z[/math]
in funzione di [math]\\rho, \theta, \phi[/math]
, si trova [math]1 \le \\rho^2 \le 4[/math]
, ovvero [math]\\rho \in [1,2][/math]
.
Dunque l'integrale diventa
[math]\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 =[/math]
[math] = \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 =[/math]
[math] = \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 =[/math]
[math]= \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 =[/math]
[math] = \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 =[/math]
[math] = \frac{14}{3} \int_{0}^{2 \\pi} d \phi = \frac{28}{3} \\pi[/math]
FINE
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