The solid is a tetrahedron with the base on the [latex]xy[/latex]-plane and a height [latex]z=6-2x-3y[/latex] The base is the region [latex]D[/latex] bounded by the lines, [latex]x=0[/latex], [latex]y=0[/latex], and [latex]2x+3y=6[/latex] where [latex]z=0[/latex] (Figure 1). Note that we can consider the region [latex]D[/latex] as Type I or as Type II, and we can integrate in both ways.

Figure 1. A tetrahedron consisting of the three coordinate planes and the plane [latex]\small{z=6-2x-3y}[/latex], with the base bound by [latex]\small{x=0}[/latex], [latex]\small{y=0}[/latex], and [latex]\small{2x+3y=6}[/latex].

First, consider [latex]D[/latex] as a Type I region, and hence [latex]{D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {3,0} \ {\leq} \ {y} \ {\leq} \ {2-{\frac{2}{3}}x} \right \}}[/latex].

Therefore, the volume is
[latex]\hspace{3cm}\begin{align} V&=\displaystyle\int_{x=0}^{x=3}\displaystyle\int_{y=0}^{y=2-(2x/3)}(6-2x-3y)dydx=\displaystyle\int_{x=0}^{x=3}\left[\left(6y-2xy-\frac32y^2\right)\Bigg|_{y=0}^{y=2-(2x/3)}\right]dx \\ &=\displaystyle\int_{x=0}^{x=3}\left[\frac23(x-3)^2\right]dx=6 \end{align}[/latex]

Now consider [latex]D[/latex] as a Type II region, so [latex]{D} = {\left \{{(x,y)}{\mid}{0} \ {\leq} \ {y} \ {\leq} \ {2,0} \ {\leq} \ {x} \ {\leq} \ {3-{\frac{3}{2}}y} \right \}}[/latex]. In this calculation, the volume is
[latex]\hspace{3cm}\begin{align} V&=\displaystyle\int_{y=0}^{y=2}\displaystyle\int_{x=0}^{x=3-(3y/2)}(6-2x-3y)dxdy=\displaystyle\int_{y=0}^{y=2}\left[(6x-x^2-3xy)\bigg|_{x=0}^{x=3-(3y/2)}\right]dy \\ &=\displaystyle\int_{y=0}^{y=2}\left[\frac94(y-2)^2\right]dy=6 \end{align}[/latex]

Therefore, the volume is [latex]6[/latex] cubic units.

We just have to integrate the constant function [latex]f(x, y)=1[/latex] over the region. Thus, the area [latex]A[/latex] of the bounded region is [latex]\displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=x^2}^{y=2x}dydx[/latex] or [latex]\displaystyle\int_{y=0}^{x=4}\displaystyle\int_{x=y/2}^{x=\sqrt{y}}dxdy[/latex]:

[latex]A=\underset{D}{\displaystyle\iint}1dxdy=\displaystyle\int_{x=0}^{x=2}\displaystyle\int_{y=x^2}^{y=2x}1dydx=\displaystyle\int_{x=0}^{x=2}\left[y\bigg|_{y=x^2}^{y=2x}\right]dx=\displaystyle\int_{x=0}^{x=2}(2x-x^2)dx=x^2-\frac{x^3}3\bigg|_0^2=\frac43.[/latex]

First find the area [latex]A(D)[/latex] where the region [latex]D[/latex] is given by the figure. We have

[latex]A(d)=\underset{D}{\displaystyle\iint}1dA=\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y}^{x=\sqrt{y}}1dxdy \displaystyle\int_{y=0}^{y=1}\left[x\bigg|_{x=y}^{x=\sqrt{y}}\right]dy=\frac23y^{3/2}-\frac{y^2}2\Bigg|_0^1=\frac16.[/latex]

Then the average value of the given function over this region is

[latex]\begin{align} f_{ave} &= \frac1{A(D)}\underset{D}{\displaystyle\iint}f(x,y)dA = \frac1{A(D)}\displaystyle\int_{y=0}^{y=1}\displaystyle\int_{x=y}^{x=\sqrt{y}}yxy^2dxdy=\frac1{1/6}\displaystyle\int_{y=0}^{y=1}\left[\frac72x^2y^s\Bigg|_{x=y}^{x=\sqrt{y}}\right]dy \\ &=6\displaystyle\int_{y=0}^{y=1}\left[\frac72y^2(y-y^2)\right]dy=6\displaystyle\int_{y=0}^{y=1}\left[\frac72(y^3-y^4)\right]dy=\frac{42}2(\frac{y^4}4-\frac{y^5}5)\bigg|_0^1=\frac{42}{40}=\frac{21}{20}. \end{align}[/latex]