x+3y–12=0, 2z=y2, x>=0, y>=0, z>=0.
z=[m]\frac{1}{2}y^2[/m]
x+3y-12=0 ⇒ x=12-3y
[m]V= ∫ ∫ _{D}\frac{1}{2}y^2dxdy[/m]
D:
0 ≤y ≤ 4
0 ≤ x≤ 12-3y
[m]V= ∫^{4}_{0} (∫^{12-3y} _{0}\frac{1}{2}y^2dx)dy=∫^{4}_{0} (\frac{1}{2}y^2) ∫ ^{12-3y} _{0}dx)dy=∫^{4}_{0} (\frac{1}{2}y^2) (x)| ^{12-3y} _{0}dy=∫^{4}_{0}(\frac{1}{2}y^2)\cdot (12-3y-0)dy=[/m]
[m]=\frac{1}{2}∫^{4}_{0}(12y^2-3y^3)dy=\frac{1}{2}\cdot (12\frac{y^3}{3}-3\frac{y^4}{4})|^{4}_{0}=(2y^3-\frac{3}{8}y^4)|^{4}_{0}=[/m] считайте