Volume of a defined region example

Question

I was given this region M = {$(x,y,z)\in \mathbb{R}^3: x+y+z\leq 2, 2x+y+8z\leq 8,(x,y,z)\geq (0,0,0)$} and asked to calculate the volume of the region.

I can picture geometrically what the region is but I would like to solve the problem more algebraically.For that, first, I guess I'm supposed to calculate the intersection of the $x+y+z=2$ and $2x+y+8z=8$ wich gives the line $(6-7z,6z-4,z)$.

I know that this line is the one which "separates" the regions where one is defined by one condition. For this specific case the region "below" (using $z$ axis) is defined by $x+y+z\leq2$ and the other above by the other condition ($2x+y+8z\leq8).$

The only thing I'm having quite a trouble is how to define the integral using the facts above (or not if there is a smarter way to approach this :D)

Answer 1

As you rightly mentioned, we can use intersection of
Plane $1$: $x+y+z = 2$ and Plane $2$: $2x+y+8z=8$ as reference line to divide the region into sub-regions with defined bounds for setting up integration.

We can use projection of the intersection line on any of coordinate planes to divide into appropriate sub-regions. All of them will divide it into three sub-regions.

Solving $8x+8y+8z = 16$ and $2x+y+8z=8$, we get $6x+7y = 8$ which is the projection of the intersection line in XY-plane. The projection intersects x-axis $(y=0)$ at $x = \dfrac{4}{3}$. So that leads to the bounds (also see the diagram which uses projection in XY-plane to understand how we are dividing the regions for integral).

• Sub-region $1$: for $0 \leq y \leq (8-6x)/7, 0 \leq x \leq \dfrac{4}{3}$, we are bound above by plane $2$.

• Sub-region $2$: for $(8-6x)/7 \leq y \leq 2-x, 0 \leq x \leq \dfrac{4}{3}$, we are bound above by plane $1$.

• Sub-region $3$: For $0 \leq y \leq 2-x, \dfrac{4}{3} \leq x \leq 2$, we are bound above by plane $1$.

So the integral to find volume is,

$\displaystyle \int_0^{4/3}\int_0^{(8-6x)/7}\int_0^{(8-2x-y)/8} dz \ dy \ dx \ + $

$\displaystyle \int_0^{4/3}\int_{(8-6x)/7}^{2-x}\int_0^{(2-x-y)} dz \ dy \ dx \ + $

$\displaystyle \int_{4/3}^2\int_{0}^{2-x}\int_0^{(2-x-y)} dz \ dy \ dx $

Last but not least, as the volume of the region is subset of the region bound by $x+y+z = 2$ in first octant, we can also find volume of the region by finding volume of the region bound between the two given planes, above $2x+y+8z = 8$ and then subtracting from the total volume bound by $x+y+z = 2$ in the first octant.

To find volume bound between $2x+y+8z \geq 8$ and $x+y+z \leq 2$ in first octant, we can again use the projection $6x+7y = 8$ in XY plane as reference. So the volume will be given by,

$\displaystyle \int_0^2 \int_0^{2-x} \int_0^{2-x-y} dz \ dy \ dx \ - $

$\displaystyle \int_0^{4/3} \int_0^{(8-6x)/7} \int_{(8-2x-y)/8}^{2-x-y} dz \ dy \ dx$

Answer 2

Using your equation, two points on the line of intersection are: $$P=({4\over3},0,{2\over3}) \ \ \text{and}\ \ Q=(0,{8\over7},{6\over7}).$$ The volume equals $V_1-V_2,$ where $V_1={4\over3}$ is the volume under $x+y+z=2$ (triangular pyramid!) and (here I'm using $P$ and $Q$) $$V_2= \int_0^{4\over3} \int_0^{{8\over7}-{6\over7}x} f(x,y)\,dy\,dx,$$ where $\ f(x,y)=(2-x-y)-{1\over8}(8-2x-y).$
Note that $P$ is on the $X$-$Z$ plane and $Q$ is on the $Y$-$Z$ plane. I recommend a sketch that shows $P$ and $Q$.