Задача 60693 Выбирая подходящие замены переменных,...
Условие
математика ВУЗ
363
Решение
★
y^2/x=u
a ≤ u ≤ b
xy=v
p ≤ v ≤ q
Считаем якобиан
u`_(x)=(y^2/x)`_(x)=(y^2)*(-1/x^2)=-(y^2/x^2)
u`_(y)=(y^2/x)`_(y)=(1/x)*2y=(2y/x)
v`_(x)=(xy)`_(x)=y
v`_(y)=(xy)`_(y)=x
[m]\begin {vmatrix} -(y^2/x^2)&(2y/x)\\y&x\end {vmatrix}=-(y^2/x)\cdot x-y\cdot(2y/x)=-3y^2/x[/m]
∫∫_(D)sqrt(xy)dxdy= ∫^(b)_(a) ∫_(p)^(q) sqrt(v)* ( -3[blue]u[/blue]) [blue]du[/blue]dv=
=∫^(b)_(a) (-3u*v^(3/2)/(3/2) |_(p)^(q)[blue]du[/blue]=
=-(1/2)(q^(3/2)-p^(3/2))∫^(b)_(a)udu=-(1/2)(q^(3/2)-p^(3/2))*(u^2/2)|^(b)_(a)=-(1/4)(q sqrt(q)-p sqrt(p))*(b^2-a^2)