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

 

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

 

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 =$

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

$ = 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