Question

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

Answer 1

ANSWER:The volume of the region is $V=4$

EXPLANATION

If $V$ is the volume of the solid lying vertically above $R$ and below the surface $z=f(x,y)$ and $f(x,y)\geq0$ on $R$ , then

$V=\iint_{R}f(x,y) dA$

So , $V=\int_{0}^{2} \int_{0}^{4}\frac{y}{2}dxdy$ $V=\int_{0}^{2} \int_{0}^{4}\frac{y}{2}dxdy=\int_{0}^{2}\left ( \frac{y}{2 } \right )\cdot \left ( \int_{0}^{4} dx\right )dy=4\int_{0}^{2}\left ( \frac{y}{2} \right )dy=2\left [ \frac{y^{2}}{2} \right ]_{0}^{2}=\left ( 2^{2} \right )-0=4$