Symmetry and Odd Integrands in Triple Integrals

Why Symmetry Helps

A triple integral can be difficult to evaluate directly, but a solid may have a reflection symmetry that pairs points with equal and opposite contributions. When that happens, an integrand can cancel across the solid, and the integral may be known without evaluating a lengthy iterated integral.

Symmetry can also simplify an integral that does not vanish. If an integrand is unchanged by a reflection, its integral over the whole symmetric solid can be found by integrating over only one side and multiplying by two. These shortcuts depend on both the function and the region: the relevant reflection must preserve the solid.

Reflection Symmetry and Parity

Reflection across the $yz$-plane sends $(x,y,z)$ to $(-x,y,z)$. Reflection across the $xz$-plane sends it to $(x,-y,z)$, and reflection across the $xy$-plane sends it to $(x,y,-z)$. A solid is symmetric under one of these reflections if reflecting every point in the solid produces the same solid.

Definition: A function $f(x,y,z)$ is odd in $x$ if $f(-x,y,z)=-f(x,y,z)$ wherever both points are in its domain. It is even in $x$ if $f(-x,y,z)=f(x,y,z)$. Oddness or evenness in $y$ and $z$ is defined by changing the sign of the corresponding coordinate.

These definitions concern one coordinate at a time. For example, $f(x,y,z)=xy^2$ is odd in $x$, even in $y$, and even in $z$. Its parity in $x$ is the relevant fact if the solid is symmetric across the $yz$-plane.

The Cancellation and Reduction Rules

Theorem: Suppose $D$ is a bounded solid symmetric across the $yz$-plane, and $f$ is integrable on $D$. If $f$ is odd in $x$, then $$ \iiint_D f(x,y,z)\,dV=0. $$ If $f$ is even in $x$, then $$ \iiint_D f(x,y,z)\,dV =2\iiint_{D\cap\{x\ge 0\}} f(x,y,z)\,dV. $$ The same statements apply to $y$ or $z$, using symmetry across the corresponding coordinate plane.

For continuous functions on a bounded solid, the required integrability is automatic. The result follows from the change of variables for a reflection. The reflection $S(x,y,z)=(-x,y,z)$ has absolute Jacobian determinant $1$, so it preserves volume. Since $S$ maps $D$ onto itself, changing variables gives

\iiint_D f(x,y,z)\,dV = \iiint_D f(-x,y,z)\,dV.

If $f$ is odd in $x$, the right-hand side is the negative of the original integral. Thus the integral equals its own negative, so it is zero. If $f$ is even in $x$, its values at reflected points agree. The two halves of $D$, on either side of the $yz$-plane, therefore have equal integrals. The plane itself has zero volume and does not affect the triple integral, giving the factor of two in the theorem.

The reasoning can be applied successively. If $D$ is symmetric under reflection across all three coordinate planes and $f$ is even in each coordinate, then all eight octants contribute equally. Consequently, the integral over $D$ is eight times the integral over its portion in the first octant. The solid must have each of these symmetries, and the function must be even in each coordinate for this full reduction to apply.

Worked Examples

Worked Example: An Odd Integrand on a Cube

Evaluate $\iiint_D x(1+y^2+z^2)\,dV$, where $D=[-1,1]\times[-1,1]\times[-1,1]$.

The cube is symmetric across the $yz$-plane. Replacing $x$ with $-x$ changes the sign of the integrand:

$$ (-x)(1+y^2+z^2)=-x(1+y^2+z^2). $$

The integrand is odd in $x$, so the theorem gives an integral of zero. This can also be checked by integrating first with respect to $x$:

\begin{aligned} \iiint_D x(1+y^2+z^2)\,dV &=\int_{-1}^{1}\int_{-1}^{1} \left[\int_{-1}^{1}x(1+y^2+z^2)\,dx\right]dy\,dz\\ &=\int_{-1}^{1}\int_{-1}^{1} (1+y^2+z^2)\left[\frac{x^2}{2}\right]_{-1}^{1}dy\,dz\\ &=\int_{-1}^{1}\int_{-1}^{1}(1+y^2+z^2)(0)\,dy\,dz\\ &=0. \end{aligned}

Indeed, the inner endpoint values are $1/2$ and $1/2$, so their difference is zero. This agrees with the reflection argument.

Worked Example: Reducing an Even Integral to One Octant

Evaluate $\iiint_D (x^2+y^2+z^2)\,dV$ over $D=[-1,1]^3$ by using symmetry.

The cube is symmetric across all three coordinate planes. Each squared coordinate is unchanged by changing the sign of any coordinate, so the integrand is even in each coordinate. The integral is eight times its value on $0\le x,y,z\le1$:

\begin{aligned} \iiint_D(x^2+y^2+z^2)\,dV &=8\int_0^1\int_0^1\int_0^1(x^2+y^2+z^2)\,dz\,dy\,dx\\ &=8\left( \int_0^1\int_0^1\int_0^1x^2\,dz\,dy\,dx +\int_0^1\int_0^1\int_0^1y^2\,dz\,dy\,dx +\int_0^1\int_0^1\int_0^1z^2\,dz\,dy\,dx \right). \end{aligned}

Each term is $1/3$. For example, the first is $(\int_0^1x^2\,dx)(\int_0^1dy)(\int_0^1dz)=(1/3)(1)(1)=1/3$. The second and third have the same value, with the squared variable integrated over $[0,1]$. Therefore,

\iiint_D(x^2+y^2+z^2)\,dV =8\left(\frac13+\frac13+\frac13\right)=8.

As a check, integrate over the full cube term by term. The $x^2$ term is $(2/3)(2)(2)=8/3$; the $y^2$ and $z^2$ terms are each $8/3$ as well. Their sum is $8$, matching the result from one octant.

Worked Example: Symmetry in One Coordinate Only

Let $D$ be given by $-1\le x\le1$, $0\le y\le1$, and $0\le z\le2-x^2$. Evaluate $\iiint_D x(1+y+z)\,dV$.

The upper bound for $z$ depends on $x^2$, so changing $x$ to $-x$ leaves the bounds unchanged. The solid is symmetric across the $yz$-plane. The integrand changes sign under that reflection, so the theorem gives zero. To check by direct integration, set $h=2-x^2$. Integrating first in $z$, then $y$, gives

\begin{aligned} \int_0^1\int_0^h x(1+y+z)\,dz\,dy &=\int_0^1 x\left((1+y)h+\frac{h^2}{2}\right)dy\\ &=x\left(\frac{3h}{2}+\frac{h^2}{2}\right)\\ &=x\left(5-\frac{7}{2}x^2+\frac{1}{2}x^4\right). \end{aligned}

The last expression is odd in $x$. Thus the remaining integral is

\begin{aligned} \int_{-1}^{1}x\left(5-\frac{7}{2}x^2+\frac{1}{2}x^4\right)dx &=\left[\frac{5}{2}x^2-\frac{7}{8}x^4+\frac{1}{12}x^6\right]_{-1}^{1}\\ &=\left(\frac{41}{24}\right)-\left(\frac{41}{24}\right)=0. \end{aligned}

This solid is not symmetric across the other two coordinate planes: its $y$- and $z$-bounds are not centered on zero. Only symmetry in $x$ was needed, because the integrand is odd in $x$.

Check the Solid as Well as the Integrand

A function’s parity by itself does not determine the value of its integral. The solid must be preserved by the same reflection. For instance, an odd-in-$x$ function need not integrate to zero over a region that lies entirely on one side of the $yz$-plane.

Worked Example: An Odd Formula on a Nonsymmetric Region

Consider $f(x,y,z)=x$ on $D=[0,1]\times[-1,1]\times[-1,1]$. It would be incorrect to set the integral to zero merely because $x$ is odd as a function of $x$. The solid is not symmetric across the $yz$-plane: reflection sends its $x$-interval $[0,1]$ to $[-1,0]$.

Direct evaluation shows the actual value:

\begin{aligned} \iiint_D x\,dV &=\int_0^1\int_{-1}^{1}\int_{-1}^{1}x\,dz\,dy\,dx\\ &=\int_0^1x(2)(2)\,dx\\ &=4\left[\frac{x^2}{2}\right]_0^1\\ &=2. \end{aligned}

The result is positive because the region contains only nonnegative $x$-values. The mistaken zero would follow only if the domain were also symmetric under $x\mapsto -x$.

Using Symmetry Reliably

Before applying a symmetry shortcut, check the geometry and the function separately. A useful sequence is:

Name the reflection.
For example, reflection across the $yz$-plane changes $x$ to $-x$ and leaves $y,z$ unchanged.

Check the solid.
Verify that the reflected bounds or constraints describe the same region.

Check the integrand.
Determine whether it changes sign or stays the same under that reflection.

Apply the matching conclusion.
An odd integrand gives zero; an even integrand gives twice the integral over one half of the solid.

Symmetry matters because it can replace a long calculation with a short argument, or reduce the amount of integration by a factor of two or more. It also provides a check on a computed answer: an integral over a symmetric solid of a function odd in the reflected coordinate must be zero. But symmetry does not excuse checking the bounds. A region that is symmetric in one coordinate may be asymmetric in another, and each proposed reduction must match the particular reflection being used.

Check Your Understanding

For each question, identify the relevant reflection and verify that the region and integrand satisfy the required conditions.

Why does an integrable function odd in $y$ have integral zero over a solid symmetric across the $xz$-plane?
If a solid is symmetric across the $yz$-plane and $f$ is even in $x$, how can its integral be written using only the part with $x\ge0$?
For $f(x,y,z)=x^2z$, identify its parity in each coordinate. Which coordinate reflection would make its integral vanish, if the solid has that symmetry?
What symmetry conditions are needed to replace an integral of a function even in all three coordinates by eight times an integral over the first octant?
Why does oddness in $x$ not imply a zero integral over $[0,1]\times[-1,1]\times[-1,1]$?