Задача 62346 ...

5fa90879322e6f41f4eb5311

26.11.2020 09:16:34

∭ (2x^3 + 3y + z)dxdydz | V: 2 ≤ x ≤ 3, -1 ≤ y ≤ 2, 0 ≤ z ≤ 4

5f3ea7e3faf909182968ddd9

28.12.2021 15:38:28

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∫∫∫_(V)(2x^2+3y+z)dxdydz= ∫ _(2) ^(3) [b]([/b]∫_(-1) ^(2) [blue](∫_(0) ^(4)(2x^2+3y+z)dz)[/blue]dy[b])[/b]dx=

=∫ _(2) ^(3) [b]([/b]∫_(-1) ^(2) [blue](2x^2z+3yz+(z^/2))|_(0) ^(4)[/blue]dy[b])[/b]dx=


=∫ _(2) ^(3) [b]([/b]∫_(-1) ^(2) [blue](2x^2*4+3y*4+(4^/2))|_(0) ^(4)[/blue]dy[b])[/b]dx=


=∫ _(2) ^(3) [b]([/b]∫_(-1) ^(2) (8x^2+12y+8)dy[b])[/b]dx=


=∫ _(2) ^(3) (8x^2y+(12y^2/2)+8y)|_(-1) ^(2)dx=


=∫ _(2) ^(3) (8x^2*2+(12*2^2/2)+8*2-(8x^2*(-1)+(12(-1)^2/2)+8*(-1))dx=


=∫ _(2) ^(3) (16x^2+24+16+8x^2-6+8)dx=∫ _(2) ^(3) (24x^2+42)dx=((24x^3/3)+42x)|_(2) ^(3)=

=8(27-8)+42*(3-2)=8*19+42=.... считайте