$I=\iiint_V (1-x) \,dV$ where V is the region in which $x,y,z \ge 0$ and below the plane $3x+2y+z=6$
Evaluate $I$.
Attempt: since $z\ge 0$, $ z$ will go from $x-y$ plane to $z= 6-3x-2y$
Then, by using the trace of this surface on the $x-y$ plane, we can see that the region is bounded by the lines $y=0, x=0,$ and $ y=\frac{6-3x}{2}$
So $y$ goes from $y=0$ to $y=\frac{6-3x}{2}$
Then $x$ will go from $ 0$ to $2$.
My question is: are my limits accurate? I will be integrating in this order: $dzdydx$
Thank you.
yes, you limits seems corrects, you have
$$I=\iiint_V1-xdV$$
where V is region limited by $x\ge0,y\ge0,z\ge0$ and plane $z=6-3x-2y$
the limits on $z$ will be from $z=0$ to $z=6-3x-2y$
on the plane $xy$ you have $z=0$ wich lead to $6-3x-2y=0$ so you can get the lmits on y from $y=0$ to $y=3-\frac{3x}{2}$
seting $y=0$ above we get $0=3-\frac{3x}{2}$ wich leads to $x=2$ so giving limits from $x=0$ to $x=2$
also $x=0\Rightarrow y=3$, so the region have vertices on $(0,0,0),(0,3,0),(2,0,0),(0,0,6)$
solving the integral you get
$$\begin{align} I&=\int_{0}^{2}\int_{0}^{3-\frac{3x}{2}}\int_{0}^{6-3x-2y}1-xdzdydx\\ &=\int_{0}^{2}\int_{0}^{3-\frac{3x}{2}}(1-x)\int_{0}^{6-3x-2y}dzdydx\\ &=\int_{0}^{2}(1-x)\int_{0}^{3-\frac{3x}{2}}\int_{0}^{6-3x-2y}dzdydx\\ &=\int_{0}^{2}(1-x)\int_{0}^{3-\frac{3x}{2}}6-3x-2ydydx\\ &=\int_{0}^{2}(1-x)\left(6\int_{0}^{3-\frac{3x}{2}}dy-3x\int_{0}^{3-\frac{3x}{2}}dy-2\int_{0}^{3-\frac{3x}{2}}ydy\right)dx\\ &=\int_{0}^{2}(1-x)\left[6\left(3-\frac{3x}{2}\right)-3x\left(3-\frac{3x}{2}\right)-\left(3-\frac{3x}{2}\right)^2\right]dx\\ &=\int_{0}^{2}(1-x)\left(18-9x-9x+\frac{9x^2}{2}-9+9x-\frac{9x^2}{4}\right)dx\\ &=\int_{0}^{2}(1-x)\left(9-9x+\frac{9x^2}{4}\right)dx\\ &=9\int_{0}^{2}(1-x)\left(1-x+\frac{x^2}{4}\right)dx\\ &=9\int_{0}^{2}1-x+\frac{x^2}{4}-x+x^2-\frac{x^3}{4}dx\\ &=9\int_{0}^{2}1-2x+\frac{5x^2}{4}-\frac{x^3}{4}dx\\ &=9\left(\int_{0}^{2}dx-2\int_{0}^{2}xdx+\frac{5}{4}\int_{0}^{2}x^2dx-\frac{1}{4}\int_{0}^{2}x^3dx\right)\\ &=9\left(2-2\cdot\frac{2^2}{2}+\frac{5}{4}\cdot\frac{2^3}{3}-\frac{1}{4}\cdot\frac{2^4}{4}\right)\\ &=9\left(2-4+\frac{10}{3}-1\right)\\ &=9\cdot\frac{1}{3}=3 \end{align}$$