# Find the volume of the solid that lies under the plane 4x+10y-2z+19=0 and above the rectangle R=...

## Question:

Find the volume of the solid that lies under the plane {eq}4x+10y-2z+19=0 {/eq} and above the rectangle {eq}R=\left \{ (x,y) \ | \ -1 \leq x \leq 2, \ -1 \leq y \leq 1 \right \} {/eq}.

## Volume of the Region:

Using the double integrals formula for the volume of the region which is {eq}V=\int \int zdydx {/eq} where {eq}z {/eq} represents the length and {eq}\int \int dydx {/eq} represents the area. Note that the given limits for integration are {eq}-1 \leq x \leq 2, \ -1 \leq y \leq 1 {/eq}

The volume of the region is,

{eq}V=\int_{-1}^{2}\int_{-1}^{1}zdydx {/eq}

{eq}V=\int_{-1}^{2}\int_{-1}^{1}\left ( 2x+5y+\frac{19}{2} \right )dydx {/eq}

Integrate with respect to {eq}y {/eq}

{eq}V=\int_{-1}^{2}\left [ \frac{19}{2}y+2xy+\frac{5y^2}{2} \right ]^{1}_{-1}dx {/eq}

{eq}V=\int_{-1}^{2}\left ( 4x+19 \right )dx {/eq}

Integrate with respect to {eq}x {/eq}

{eq}V=\left [ 2x^2+19x \right ]^{2}_{-1} {/eq}

{eq}V=63 {/eq}