Symmetry of the domain?

Question

For the tiple integral $$\iiint_D (z^2+z) \,dx\,dy\,dz$$ over the domain $D:x^2+y^2+z^2\leq4,\quad z^2\leq x^2+y^2$

The textbook states that by symmetry of the domain the integral simplifies to $$\iiint_D z^2 \,dx\,dy\,dz.$$ How exactly do they arrive at that conclusion?

The domain is symmetric about z, so the odd contribution cancels — Kiryl Pesotski, Mar 06 '17 at 15:34
Thanks for the reply. What exactly do you mean by the odd contribution cancels? What is the odd contribution? — Eiraus, Mar 06 '17 at 15:36

score 3 · Answer 1 · answered Mar 06 '17 at 15:40

This is because $$ \iiint_D z\,dV = 0 $$ Why is that? If $(a,b,c)$ is in $D$, then so is $(a,b,-c)$ (both inequalities have $z^2$ in them). So each point $P$ in $D$ has a mirror-image point $P'$ on the opposite side of the $xy$-plane. If $f(x,y,z) = z$, then $f(P') = -f(P)$. Accumulating all the points $P$ above the $xy$-plane, their $f$-values cancel in pairs with their corresponding points $P'$.

score 3 · Accepted Answer · answered Mar 06 '17 at 15:41

3

The domain is symmetric about $z$, and the function $f(x,y,z)=z$ is odd, meaning that $$ f(x,y,-z)=-f(x,y,z)=-z $$ Therefore $$ \iiint_D z \; dV = 0 $$

To understand why this happens, consider the $1$ dimensional case: do you see why $$ \int_{-a}^{a} x\; dx = 0 \; ? $$

answered Mar 06 '17 at 15:41

Kuifje

9,584

Thanks I think I get it: In your last example the area from 0 to a and from -a to zero cancel each other out? – Eiraus Mar 06 '17 at 15:51
Yes that is exactly it! @Matthew Leingang explains it well in the other answer ;) – Kuifje Mar 06 '17 at 16:09

Symmetry of the domain?

2 Answers2