solid volume using double integral $|x+y|+|x-2y|+|x+y+z|\leqslant1$

Question

I have some problems with this exercise: Caculate the solid volume using double integral $$|x+y|+|x-2y|+|x+y+z|\leqslant1$$ My teacher gave me the key value: $\frac{4}{9}$, but when I try to solve it, I don't know did something wrong in some step and go to wrong answer.

My attempt:

1. $$1\geq|x+y|+|x-2y|+|x+y+z|\geq|x+y+x-2y+x+y+z|=|z+3x|$$ so, $1-3x\geq z \geq -1-3x$

2. We have $1\geq|x+y|+|x-2y|$, hence we have D: $\frac{1}{3}\geq y\geq \frac{-1}{3}$ and $\frac{1+y}{2}\geq x\geq \frac{-1+y}{2}$

So, $V=\iint_{D}{(1-3x)-(-1-3x)}dxdy=\iint_{D}2dxdy=\frac{4}{3}$

I think in step 2, when I use $1\geq|x+y|+|x-2y|$, I was wrong (cause it just $1\geq|x+y|+|x-2y|+|x+y+z| \rightarrow 1\geq|x+y|+|x-2y|$
not $1\geq|x+y|+|x-2y| \rightarrow 1\geq|x+y|+|x-2y|+|x+y+z|$

So, could you guys solve this exercise for me or give me hint to do with the abs(), it make me confuse so much. Thanks for your help

Answer 1

This solid is obtained by applying a linear transformation to the union of two pyramids. First, we find the Jacobian of the linear transformation $$ u=x+y+z, v=x-2y, w=x+y. $$ We have by computing the determinant using the third column $$ \frac{\partial (u,v,w)}{\partial (x,y,z)} = 1\cdot (1- (-2)\cdot 1) = 3. $$ Thus, we have $$ \frac{\partial (x,y,z)}{\partial (u,v,w)}=\frac13. $$ After applying the change of variables, the solid becomes the union of two pyramids defined by $$ |u|+|v|+|w|\leq 1. $$ Both pyramid have the square base with vertices $(1,0,0),(0,1,0),(-1,0,0),(0,-1,0)$, and both have heights $1$. Hence, the volume is $$ 2\cdot \frac{\sqrt 2^2 \cdot 1}3 \cdot \frac{\partial (x,y,z)}{\partial (u,v,w)}=\frac49. $$

Remark) If you solve inequalities with the triangle inequality, you are likely to lose information. If you lose information, that results in larger solid, and thus larger volume.

Answer 2

I change over to the variables $u=x+y$ and $v=-x+2y$ (which will give one-third of the original volume). By symmetry, I can restrict to the first quadrant of the $U$-$V$ plane. $$\text{For}\ z\ge-u, \ \ u+v+u+z\le1 $$ $$\text{For}\ z\le-u,\ \ u+v-u-z\le1 $$

One quarter of the volume equals

$$\int_0^1\int_{-u}^1 (1-2u-z) \,dz\,du +\int_0^1\int_{-1}^{-u} (1+z) \,dz\,du $$ The final answer is twelve times the value in the displayed formula.