Calculating $\iiint_V g(x+y+z)dxdydz$

Question

I got this question on my HW and am struggling with how to approach it:

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function such that $\int^1_{-1}g(t)dt=3$. Let $f(x,y,z)=g(x+y+z)$, calculate $$\iiint_V f(x,y,z) \, dx \, dy \, dz$$ where $V=\{(x,y,z): |x|+|y|+|z| \leq 1\}$.

I know this domain is comprised of 8 symmetric pyramids, but I'm struggling with how to use this symmetry, and if I should use the Antiderivative $G$ of $g$.

Would appreciate help.

Answer 1

There are various ways to try this. My own "non-rigorous" way would be like this :

We have Constant Answer which is not Dependant on $g(x)$ , as long as that Integral is $3$.

(1) Let $g(x)=Constant=3/2$ , then $\int_{-1}^{+1}g(x)dx=3x/2|_{-1}^{+1}=3/2+3/2=3$.

Then we have $f(x,y,z)=g(x+y+z)=3/2$

Then we have $\int\int\int_{V}f(x,y,z)dxdydz=\int\int\int_{V}g(x+y+z)dxdydz=\int\int\int_{V}(3/2)dxdydz=(3/2)|_{V}=3V/2$

The Volume $V$ is a "Diamond" in 3D , where the Diagonals are the 3 Axis , between $-1$ & $+1$ , in other words , Diagonal length is $2$. Hence Cube Side "Diamond" Side is $\sqrt{2}$.

Image :

Purple lines are Equal Sides of the "Diamond" , Black lines are Equal Diagonals.

The "Diamond" is made of 2 Pyramids , each with height $2$ & Square Base with Side $\sqrt{2}$.

Blue Area is the Square Base of the Pyramids.

https://www.varsitytutors.com/hotmath/hotmath_help/topics/volume-of-a-pyramid : "volume of a pyramid = one-third the area of the base times the height" : $\sqrt{2}\times\sqrt{2}\times{2}/3=4/3$

Volume Of Our "Diamond" made of 2 Pyramids is $V=2\times(4/3)=8/3$

Hence the Volume $V$ is $8/3$ which is the Integral Domain we want.

That then gives Integral Value $(3/2)V=(3/2)(8/3)=4$

(2) We can take $g(x)=9x^2/2$ , then $\int_{-1}^{+1}g(x)dx=9x^3/6|_{-1}^{+1}=3/2+3/2=3$.

Integrating $f(x,y,z)=9(x+y+z)^2/2$ thrice & using the limits will be a little Cumber-Some , yet , we know that we have to get the Same Answer $4$.

Answer 2

The answer cannot be a mere number. It will obviously depend on $g.$

Let $t=x+y+z.$ Replacing $z$ with $t-x-y$ and using many times the rule $|a|\le b\iff-b\le a\le b,$ we find that $$|x|+|y|+|z|\le1\iff\begin{cases}-1\le t\le 1\\\frac{-1+t}2\le x\le\frac{1+t}2\\\frac{-1+|x|+t-x}2\le y\le\frac{1-|x|+t-x}2.\end{cases}$$ Therefore, $$\begin{align}\iiint_V g(x+y+z)dxdydz&=\int_{-1}^1\left(\int_{(-1+t)/2}^{(1+t)/2}\left(1-|x|\right)dx\right)g(t)dt\\&=\int_{-1}^1\left(1+\left[\frac{x^2}2\right]_{(-1+t)/2}^0-\left[\frac{x^2}2\right]_0^{(1+t)/2}dx\right)g(t)dt\\&=\int_{-1}^1\left(1-\frac{(1-t)^2}8-\frac{(1+t)^2}8\right)g(t)dt\\&=\int_{-1}^1\frac{3-t^2}4g(t)dt \end{align}$$ So, if we only know that $\int^1_{-1}g(t)dt=3,$ the best we can say is $$\iiint_V g(x+y+z)dxdydz=\frac94-\frac14\int_{-1}^1t^2g(t)dt .$$ Note that when $g$ is constant, $g(t)=\frac32,$ knowing that the volume of $V$ is $\frac43,$ we obtain directly $\iiint_V g(x+y+z)dxdydz=\frac32\frac43=2,$ which is consistent with the general case above since $\frac32\int_{-1}^1\frac{3-t^2}4dt=2.$