Find the volume of the region bounded above by the plane z = y/2 and below by the rectangle.
R ∶ 0 ≤ x ≤ 4,0 ≤ y ≤ 2
V= ∫04dx∫02dy∫0y/2dz=∫04dx∫02y2dy=\int_0^4 dx \int_0^2 dy \int_0^{y/2}dz= \int_0^4 dx \int_0^2 \frac y2 dy =∫04dx∫02dy∫0y/2dz=∫04dx∫022ydy=
∫04dxy24∣02=∫04dx=4\int_0^4 dx \frac{y^2}{4}|_0^2= \int_0^4 dx =4∫04dx4y2∣02=∫04dx=4
Answer: V=4.
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