Задача 30954 ...

alekseevadaria

2018-11-12 14:38:03

Вычислить двойной интеграл ∫ ∫ y*sin2xydx, где D - прямоугольник, ограниченный прямыми x = 1/2, x=3, y=π/2, y=3π/2

sova

2018-11-12 15:45:05

★

∫ ∫ y*sin(2xy)dx dy =∫^(3π/2)_(π/2)(∫^(3) _(1/2)y*sin2xy [b]dх[/b]) dу=

[так как d(2xy)=2ydx, то ∫y*sin2xy dx= (1/2) ∫ sin(2xy)d(2xy)=
=(1/2)*(-cos(2xy))]

=∫^(3π/2)_(π/2)( (1/2)*(-cos(2xy)))|^(x=3) _(x=1/2)dy=

=(-1/2)*∫^(3π/2)_(π/2)(cos2*3y-cos2*(1/2)y)dy=

=(-1/2)* ∫^(3π/2)_(π/2)(cos6y-cosy)dy=

=(-1/2)*((1/6)sin6y-siny)|^(3π/2)_(π/2)=

=(-1/2)*(1/6)sin6*(3π/2)-(-1/2)*(1/6)sin6*(π/2) + (1/2)sin(3π/2)-(1/2)sin(π/2)=

= (-1/12)*sin9π+(1/12)sin3π+(1/2)sin(3π/2)-(1/2)sin(π/2)=

=0+0+(1/2)*(-1)-(1/2)*1= -1