Triple integral between spheres $\iiint \sin (xy) - \sin (xz) + \sin (yz) \,dx \,dy \,dz$

Question

I'm hoping someone can tell me if this seems like a legitimate argument.

Problem: compute $I:=\iiint_E \sin (xy) - \sin (xz) + \sin (yz) \,dx \,dy \,dz$ where $E = \{(x,y,z) \in \mathbb{R}^3 : 1 \le x^2 + y^2 + z^2 \le 4 \}$.

Claim: $I = 0$.

First we split it up $I = \iiint_E \sin (xy) \,dx \,dy \,dz - \iiint_E \sin (xz) \,dx \,dy \,dz + \iiint_E \sin (yz) \,dx \,dy \,dz$. Let's examine just the first of these three: we show $\iiint_E \sin (xy) \,dx \,dy \,dz =0$. The key is to decompose the region $E$ and exploit symmetry. $E$ can be decomposed into several "pieces:" let

• $E_{x=0} = \{ (x,y,z) \in E : x = 0 \}$
• $E_{y=0} =\{(x,y,z) \in E : y = 0\}$
• $E_{++} = \{ (x,y,z) \in E : x,y >0 \}$
• $E_{-+} = \{(x,y,z) \in E : x <0 , y >0 \}$
• $E_{+-} = \{(x,y,z) \in E : x >0 , y <0 \}$
• $E_{--} = \{(x,y,z) \in E : x <0 , y <0 \}$

and the union of the above is precisely $E$. We establish a bijection between $E_{++}$ and $E_{-+}$ so that the values of the integrand at corresponding points differ by $-1$: given $(a,b,z) \in E_{++}$ so that $a,b >0$ we associate the point $(-a,b,z) \in E_{-+}$ and this defines a bijection. Furthermore $\sin (ab) = - \sin (-ab)$ because $\sin$ is odd. Thus $\sin (ab) + \sin (-ab) = 0$. We establish a bijection between $E_{+-}$ and $E_{--}$ with the same feature by sending $(a,-b,z) \in E_{+-}$ to $(-a,-b,z)\in E_{--}$. We note that on $E_{x=0}$ and $E_{y=0}$ the integrand $\sin(xy)$ is $0$. All of this together shows that $\iiint_E \sin (xy) \,dx \,dy \,dz =0$.

Symmetric arguments show that the other two integrals are $0$, although the subdivisions into pieces will be oriented differently.

Therefore $I=0$.

Does this seem correct? I really hope so, because otherwise I have no idea how to solve this problem. Actually if someone is aware of another way I'd be interested to hear it. Thank you!

Answer 1

You have absolutely the right idea, but are working a bit too hard. In particular, there is no need to consider your sets $E_{x=0}$, $E_{y=0}$, and $E_{z=0}$, as a triple integral over a plane will always vanish (formally, these are all sets of measure zero). Similarly, you don't need to break $E$ into quite so many regions. It is enough to note (and aim towards) the following:

Theorem: Suppose $f$ is a continuous function on $\mathbb{R}^3$ and $E\subseteq \mathbb{R}^3$ is compact. If $f$ is odd with respect to a variable and $E$ is symmetric with respect to the same variable, then $\iiint\limits_E f ~\mathrm{d}V = 0$.

We say that $f(x,y,z)$ is odd with respect to $x$ if $f(-x,y,z) = - f(x,y,z)$. Note that we only negate one component at a time. We say that $E$ is symmetric with respect to $x$ if $(x,y,z) \in E \iff (-x, y, z) \in E$.

My approach

Consider the integral $I_1=\iiint\limits_E \sin(x y) ~\mathrm{d}x \, \mathrm{d} y \, \mathrm{d} z$. The integrand is odd with respect to $x$, as $\sin((-x) y) = \sin(- xy) = - \sin(xy)$ and $E$ is symmetric with respect to $x$ as $$(-x,y,z) \in E \iff 1 \leq (-x)^2 + y^2 + z^2 = x^2 + y^2 + z^2 \leq 4 \iff (x,y,z) \in E.$$ Therefore $I_1 = 0$.

By similar argument, we have that the two other integrals vanish, and thus $I=0$.