On symmetries in multiple integrals

Question

If the calculation of the triple integral is required:

$$ I = \iiint\limits_A f(x,\,y,\,z)\,\text{d}x\,\text{d}y\,\text{d}z\,, \quad \text{with} \; f(x,\,y,\,z) = x^1+y^3+z^5\,, \quad A = \{x^2+y^4+z^6 \le 1\} $$

I would notice that:

• domain $A$ enjoys the symmetry $S(x,\,y,\,z) = (-x,\,-y,\,-z)$, i.e. $S(A) = A$;

• the function $f$ is odd with respect to $S$, i.e. $S(f)=-f$;

therefore, without any calculation, I can conclude that $I=0$.

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On the other hand, if you were to calculate the double integral:

$$ J = \iint\limits_B g(x,\,y)\,\text{d}x\,\text{d}y\,, \quad \text{with} \; g(x,\,y) = \frac{x\,y}{x^2+y^2}\,, \quad B = \{1 \le x^2+y^2 \le 4,\,y \ge x\} $$

through a transformation of coordinates from Cartesian to polar in the plane it is easy to prove that $J = 0$, but I cannot find a way to prove it as above, that is by identifying a symmetry $S$ for $B$ in which $g$ is odd. Ideas?

Answer 1

Let $C:=\{(x,y)\in B:x+y\geqslant 0\}$, and $f(x,y):=(-y,x)$.

Then $B=C\cup f(C)$, $C\cap f(C)$ is a nullset, and $g\circ f=-g$.

Answer 2

Here is a Simpler Solution, similar to the wording in your Question:

The 2 Dark Gray Domains have :

• equal Area,
• enjoy Symmetry,
• have odd function g(x,y) = -g(-x,y)

Hence the Double Integral in one Dark Grey Area is the negative of the Double Integral in the other Dark Grey Area, Hence the total is 0.

Likewise, the 2 Light Gray Domains have :

• equal Area,
• enjoy Symmetry,
• have odd function g(x,y) = -g(x,-y)

Hence the Double Integral in one Light Grey Area is the negative of the Double Integral in the other Light Grey Area, Hence the total is 0.

Hence the Double Integral over the whole Domain is 0.