Задача 59888 Кто решит тому сразу дам лучший ответ!!!...
Условие
Решение
Вычислить [b]три двойных интеграла[/b] интеграла по области D:
0 ≤ x ≤ 1
2x^2 ≤ y ≤ 2 sqrt(x)
M= ∫∫_(D) [blue] 6xy[/blue] dxdy= ∫ _(0)^(1)[b]([/b] ∫_(2x^2)^(2 sqrt(x)) 6xy dy[b])[/b]dx=∫ _(0)^(1)6x*(y^2/2)|_(2x^2)^(2 sqrt(x))dx=
= ∫ _(0)^(1)3x*((2 sqrt(x))^2-(2x^2)^2)dx=∫_(0)^(1)3x*(4x-4x^4)dx=∫_(0)^(1)(12x^2-12x^5)dx=(12*(x^3/3)-12*(x^6/6))|^(1)_(0)=
=(4x^3-2x^6)|^(1)_(0)=2
M_(x)= ∫∫_(D) [blue] 6xy[/blue] * [b]x[/b] dxdy=∫_(0)^(1)[b]([/b] ∫_(2x^2)^(2 sqrt(x)) 6x^2ydy[b])[/b]dx=∫_(0)^(1)6x^2[b]([/b] ∫_(2x^2)^(2 sqrt(x)) ydy[b])[/b]dx=
=∫_(0)^(1)6x^2*[b] ([/b] ([m]\frac{y^2}{2}[/m])|_(2x^2)^(2 sqrt(x)) [b])[/b]dx=
=∫_(0)^(1) 3x^2*[b]([/b][m]\frac{(2\sqrt{x})^2}{2}-\frac{(2x^2)^2}{2}[/m] [b])[/b]dx=
=∫_(0)^(1) 3x^2*[b]([/b]2x-2x^4[b])[/b]dx=
=∫_(0)^(1) [b]([/b]6x^3-6x^6[b])[/b]dx=
=(6*(x^4/4)-6*(x^7/7))|^(1)_(0)=(6/4)-(6/7)=6*(7-4)/28=[blue]9/14[/blue]
[b]x_(C)[/b]=M_(x)/M=([blue]9/14[/blue])/2=[b]9/28[/b]
M_(y)= ∫∫_(D) [blue] 6xy[/blue] * [b]y[/b] dxdy=∫_(0)^(1)[b]([/b] ∫_(2x^2)^(2 sqrt(x))6xy ^2dy[b])[/b]dx=
=∫_(0)^(1)6x*[b] ([/b] ([m]\frac{y^3}{3}[/m])|_(2x^2)^(2 sqrt(x)) [b])[/b]dx=
=∫_(0)^(1) 2x*[b]([/b](2\sqrt(x})^3-(2x^2)^3 [b])[/b]dx=
=∫_(0)^(1) 2x*[b]([/b]8xsqrt(x)-8x^6[b])[/b]dx=
=∫_(0)^(1) [b]([/b]16*x^(5/2)-16x^7[b])[/b]dx=
=16*(x^(7/2)/(7/2))|_(0)^(1) -16*(x^8/8)|_(0)^(1)=16*(2/7)-2=18/7
y_(C)=(18/7)/2=[b]18/14=9/7[/b]