# Use a double integral to find the volume of the solid bounded by the plane 2x+y+z in the...

## Question:

Use a double integral to find the volume of the solid bounded by the plane {eq}2x+y+z =4 {/eq} in the first octant.

## Volume of the Region:

The double integrals formula for the volume of the region in rectangular coordinates is {eq}A=\int \int zdydx {/eq} where {eq}z {/eq} is the length and {eq}\int \int dydx {/eq} represents the area.

Based on the given plane equation the limits are the following,

{eq}0\leq x\leq 2,\:0\leq y\leq 4-2x {/eq}

The volume of the region is,

{eq}V=\int_{0}^{2}\int_{0}^{4-2x}zdydx {/eq}

{eq}V=\int_{0}^{2}\int_{0}^{4-2x}\left ( 4-2x-y \right )dydx {/eq}

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

{eq}V=\int_{0}^{2}\left [ 4y-\frac{y^2}{2}-2xy \right ]^{4-2x}_{0}dx {/eq}

{eq}V=\int_{0}^{2}\left ( 2x^2-8x+8 \right )dx {/eq}

Integrate,

{eq}V=\left [ \frac{2x^3}{3}-4x^2+8x \right ]^{2}_{0} {/eq}

{eq}V=\frac{16}{3} {/eq}

{eq}V\approx \:5.333 {/eq}