[m]0 ≤ z ≤ \frac{6y}{11}[/m]
[m]= ∫ ∫ _{ D }(z)|^{\frac{6y}{11}}_{0}dxdy= ∫ ∫ _{ D } \frac{6y}{11}dxdy[/m]
D:
x^2+y^2=50 ⇒ y= ± sqrt(50-x^2) нам нужна правая полуокружность y=sqrt(50-x^2)
x=sqrt(5y) ⇒ x^2=5y ⇒ y=x^2/5
х=0
cм. рис.
0 ≤ x ≤ 5
y=x^2/5
y=sqrt(50-x^2)
[m]=∫_{0}^{5}( ∫_{x^2/5}^{ \sqrt{50-x^2}} \frac{6y}{11}dy)dx=[/m]