This is more of a conceptual question, I'm having trouble visualizing this. Under what conditions besides $f(x,y,z)=0$ is $$\iiint f (x,y,z)\,{\rm d}x\,{\rm d}y\,{\rm d}z=0$$ What would the limits have to be? A simple example would and a picture would kind of clear this up.
Triple integral volume.
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$\begingroup$
calculus
integration
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1Do you understand how it can happen for $f(x)$? – 2017-02-13
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0Yes but it depends if f is odd or even – 2017-02-13
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0There is no general constructive answer for saying when this will be the case. One necessary condition (though trivial) is that $f$ takes on both signs in the domain. There are some special cases for which we can say that this will happen and that is when the integration region has some form of symmetry and $f$ has some form of anti-symmetry over this region (e.g. if the domain is $[-a,a]\times [-b,b]\times [-c,c]$ and $f$ is odd wrt each argument). – 2017-02-13
2 Answers
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What if you evenly split a cube $X = [-1,1]^3 \subset \mathbb{R}^3$ about the origin into octants $I \times J \times K$ of the same volume, where $I,J,K$ could be either of the intervals $[-1,0]$ or $[0,1]$. For example, an octant could be
$[0,1] \times [-1,0] \times [0,1]$.
Here's a picture of what this splitting may look like: https://upload.wikimedia.org/wikipedia/commons/6/60/Octant_numbers.svg
Now, what if you just defined $f$ on $X$ by making it constantly $1$ on four of these octants, and constantly $-1$ on the remaining four octants. Then the integral of $f$ over $X$ could be split into eight integrals over the eight octants, and these eight integral values would cancel each other out.
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Try $$ \int \int_{\text{whatever}} \int_{-a}^{a}x^3dxdydz $$ with $f(x,y,z)=x^3$.
For a more complicated example, try a plane or some other well known symmetric shape and integrate around the origin appropriately