Integrate from $-b$ to $b$ in $\left\{(x,y,z) \mid |x|+|y|+|z| \leq b\right\}$

Question

So I know that $g:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, and I also know that $\int_{-b}^{b}g(t)dt=a$ where $a\in\mathbb{R}$ is some constant.
If we denote $V=\left\{(x,y,z) \mid |x|+|y|+|z| \leq b\right\}$ where $b>0$ is a real constant, I'm trying to calculate the following integral (as a function of $a$ and $b$):
$$ \iiint_{V}g(x+y+z)dxdydz $$

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To detail some of my previou efforts:

I tried a ton of different things but nothing seems to get me there, even though it seems to be close, somehow.
For instance: I listed out all the various planes we get by separating x,y,z into their cases (by octants) and composed them as simpler conditions, leaving me with:
$$ -b\leq x+y+z\leq b\\ -b\leq-x+y+z\leq b\\ -b\leq x-y+z\leq b\\ -b\leq x+y-z\leq b $$ At this point I had attempted a few transformations, such as, for instance:
$$ u=x+y+z\\ v=-x+y+z\\ w=x-y+z $$ Or equivalently (assuming no algebra shenanigans on my end), $$ x=\frac{w-v}{2}\\ y=\frac{u-w}{2}\\ z=\frac{u+v}{2} $$ Yielding a Jacobian determinant of $\frac{1}{4}$ and being continuously differentiable and everything (so a valid transformation – it's just a simple linear one but I made sure just in case/to practice, either way).
In this case, the new volume we're integrating over is:
$$ -b\leq u\leq b\\ -b\leq v\leq b\\ -b\leq w\leq b\\ -b\leq u-v-w\leq b $$

My reasoning for attempting this was that I was trying to get the "$x+y+z$" ($g$'s argument) to become one variable that I can bound from $-b$ to $b$, but I could see no way to do this that didn't require me to integrate $g$ in some other, variable-dependant bounds before that, and that I couldn't think of how to do when all I know about $g$ is that it is continuous and that its integral specifically from $-b$ to $b$ is $a$.
I also tried drawing this post-transformation and saw that it was the volume trapped inside a cube with sides sized $2b$ and between two slopes ($-b\leq u-v-w\leq b$). This seemed promising, since I figured with another transformation I might potentially be able to somehow integrate over this seemingly-tame volume in a mannner that sums from $-b$ to $b$ so that I may use what I know about $g$'s integral, but... to no avail.

Answer 1

For a fixed $a,b$, the value of $\iiint_V g(x+y+z)\,{\rm d}x\,{\rm d}y\,{\rm d}z$ can be any real number, so in particular, this integral cannot be written as a function purely of $a,b$ (it still depends on $g$ in ways that aren't determined by only $a,b$). To see this, consider the following:

First, fix $a,b$, let $g_0$ denote a continuous function such that $\int_{-b}^b g_0(w)\,{\rm d}w = a$. Define a function $I$ by $I(g) = \iiint_V g(x+y+z)\,{\rm d}x\,{\rm d}y\,{\rm d}z$ for any continuous $g$, and define $S_w = \{(x,y,z) \in V \mid x+y+z=w\}.$

The key observations are that $$I(g) = \frac{1}{\sqrt{3}}\int_{-b}^b g(w)\cdot \operatorname{area}(S_w)\,{\rm d}w,$$ and $w \mapsto \operatorname{area}(S_w)$ is continuous with $$\operatorname{area}(S_0) = \frac{3\sqrt{3}}{4}b^2 \\ \operatorname{area}(S_b) = \frac{\sqrt{3}}{2}b^2.$$

The rest of the answer is to show how these observations show that the restriction of $I$ to the set of continuous functions $g$ such that $\int_{-b}^b g(w)\,{\rm d}w = a$ is surjective onto $\mathbb{R}$.

For arbitrary $\varepsilon \in (0,b)$, we consider the functions $$\varphi_0(w) = \begin{cases}(\varepsilon - |w|)/\varepsilon^2 & \text{ if } |w| < \varepsilon \\ 0 & \text{ otherwise}\end{cases}$$ $$\varphi_b(w) = \begin{cases}2(\varepsilon+w-b)/\varepsilon^2 & \text{ if } w >b-\varepsilon \\ 0 & \text{ otherwise}\end{cases}$$ which we note satisfy $$\int_{-b}^b \varphi_0(w)\,{\rm d}w = 1 = \int_{-b}^b \varphi_b(w)\,{\rm d}w.$$ For any real $t$, we can define $g_t(w) = g_0(w) + t(\varphi_0(w)-\varphi_b(w))$, and note it's a continuous function that satisfies $\int_{-b}^b g_t(w)\,{\rm d}w = a$.

Now, for a fixed $\varepsilon$ sufficiently small, $$\begin{align*}I(\varphi_0)-I(\varphi_b) &= \frac1{\sqrt{3}}\int_{-b}^b \varphi_0(w)\operatorname{area}(S_w)\,{\rm d}w - \frac1{\sqrt{3}}\int_{-b}^b \varphi_b(w)\operatorname{area}(S_w)\,{\rm d}w \\ &\approx \tfrac1{\sqrt{3}}\operatorname{area}(S_0) - \tfrac1{\sqrt{3}}\operatorname{area}(S_b) \\ &= \frac{b^2}4 > 0\end{align*}$$

Finally, $$I(g_t) = I(g_0) + t(I(\varphi_0)-I(\varphi_b)),$$ so $t \mapsto I(g_t)$ is a non-constant linear function of $t$, which means it's surjective onto $\mathbb{R}$.

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Afterthoughts: I forgot to mention, but with a little more work we can compute $$\operatorname{area}(S_{w}) = \frac{(3b^2-w^2)\sqrt{3}}{4}$$

This gives a way to easily compute the integral as $$\iiint_V g(x+y+z)\,{\rm d}x\,{\rm d}y\,{\rm d}z = \frac14\int_{-b}^b g(w)(3b^2-w^2)\,{\rm d}w = \frac{3ab^2}4 - \frac14\int_{-b}^b w^2g(w)\,{\rm d}w$$

Answer 2

[This is not a solution. This is to explain my comment about the 2-D case, since OP was unable to prove the independence of $g$.]

In the 2 D case, with $ b = 1$ (We can always scale as needed), testing with the constant function $g(x) = \frac{a}{2}$, and integrating over the diamond of $ | x| + |y| \leq 1$ with side length $\sqrt{2}$ gives us $ \sqrt{2} ^2 \times \frac{a}{2} = a $.

How can we show the answer is always equal to $a$ for an arbitrary $ \int_{-1}^1 g(x) \, dx = 1$?
Hint: Do the change of variables so that we're integrating along $ y = x + k$, with $ -1 \leq k \leq 1$.
For this line, the starting point is $(- \frac{1+k}{2} , - \frac{1-k}{2} )$ and the ending point is $(\frac{1-k}{2},\frac{ 1+k}{2})$.
On this line, we have $x+y$ varying from $-1$ to $1$, and hence we can apply the $ \int_{-1}^1 g(x+y) d(x+y) = a$.
Finally, integrating across all of these lines (with a suitable volume change), show that the integral is indeed $a$.

Now approach the 3-D case in a similar manner. (To be fair, I've not thought through the details, but I believe it's not that different. OP said that they were unable to come up with a suitable transformation matrix.)