Question

Find the volumes of the solids described in Exercises 41–44. The portion of the first octant bounded by the coordinate planes and the plane$3x+4y+6z=12$

Short answer

The volume of the described solid is: 4 units

Step-by-step solution

Step 1. Given Information  

There is a portion of the first octant bounded by the coordinate planes and the plane,

$3x+4y+6z=12$

Step 2. Finding the volume of given portion of octant 

The double integral to find the volume of portion of octant is given by,

$$\begin{gathered} {\int }_0^4{\int }_0^{\frac{12-3x}{4}}\frac{\left(12-3x-4y\right)}{6}dydx \\ ={\int }_0^4{\overline{)\left(\frac{12y}{6}-\frac{3xy}{6}-\frac{4y^2}{12}\right)}}_0^{\frac{12-3x}{4}}dx \\ ={\int }_0^4{\overline{)\left(2y-\frac{xy}{2}-\frac{y^2}{3}\right)}}_0^{\frac{12-3x}{4}}dx \\ ={\int }_0^4\left(\frac{2\left(12-3x\right)}{4}-\frac{x(12-3x)}{8}-\frac{{(12-3x)}^2}{48}\right)dx \\ ={\int }_0^4\left(6-\frac{3x}{2}-\frac{3x}{2}+\frac{3x^2}{8}-\frac{3(16+x^2-8x)}{16}\right)dx \\ ={\int }_0^4\left(6-\frac{3x}{2}-\frac{3x}{2}+\frac{3x^2}{8}-3-\frac{3x^2}{16}+\frac{3x}{2}\right)dx \\ ={\int }_0^4\left(3-\frac{3x}{2}+\frac{3x^2}{16}\right)dx \\ ={\overline{)\left(3x-\frac{3x^2}{4}+\frac{x^3}{16}\right)}}_0^4 \\ =\left(3(4-0)-\frac{3\left(4^2-0^2\right)}{4}+\frac{\left(4^3-0^3\right)}{16}\right) \\ =12-12+4 \\ =4 \end{gathered}$$