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

 

 

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

 

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

 

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

 

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

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

 

FINE