Задача 59421 ...
Условие
f(z,y) =x^3y^2 D — круг x^2+y^2 ≤ R^2
Решение
⇒
f(x_(o);y_(o))=[m]\frac{ ∫ ∫_{D}f(x;y)dxdy }{ S_{D}}[/m]
[m]∫ ∫_{D}f(x;y)dxdy=∫ ∫_{x^2+y^2 ≤R^2 }x^3y^2dxdy[/m] переход к полярным координатам
[m] = ∫_{0} ^{2π} (∫_{0}^{R} ρ^3cos^3 θ \cdot ρ^2sin^2 θ \cdot ρ d ρ)d θ =[/m]
[m]=∫_{0} ^{2π} cos^3 θ sin^2 θ ∫_{0}^{R} ρ^3 \cdot ρ^2 \cdot ρ d ρ)d θ =[/m]
[m]=∫_{0} ^{2π} cos^3 θ sin^2 θ ( ∫_{0}^{R} ρ^6 d ρ)d θ =[/m]
[m]=∫_{0} ^{2π} cos^3 θ sin^2 θ ( ρ^7/7)|_{0}^{R})d θ =[/m]
[m]=(1/7)R^7∫_{0} ^{2π} cos^2 θ \cdot sin^2 θ \cdot cos θ d θ =[/m]
cos^2 θ =1-sin^2 θ
d(sin θ )=(sin θ )`d θ =cos θ d θ
[m]=(1/7)R^7∫_{0} ^{2π} (1-sin^2 θ) \cdot sin^2 θ \cdot d(sin θ) =[/m]
[m]=(1/7)R^7∫_{0} ^{2π} (sin^3 θ/3) -(sin^4 θ/4) =0[/m] ⇒
f(x_(o);y_(o))=0/πR^2=0
