Separable Integrands on Product Regions

Why Separation Matters

A triple integral over a rectangular box can involve three variables at once, but some integrands have a simpler structure: the $x$-, $y$-, and $z$-dependence appears in separate factors. When the integration region also separates into three independent intervals, the triple integral becomes a product of ordinary one-variable integrals.

This is more than a convenient way to shorten a calculation. It follows from the structure of the iterated integral: each factor can be taken outside the integrations with respect to the other variables. The shortcut is valid only when both the region and the integrand have the required form.

Product Regions and Separable Integrands

Let $I=[a,b]$, $J=[c,d]$, and $K=[e,f]$ be finite intervals, and let $R=I\times J\times K$. Thus every combination of coordinates with $x\in I$, $y\in J$, and $z\in K$ lies in $R$. Suppose the integrand is a product of three one-variable functions: $F(x,y,z)=f(x)g(y)h(z)$.

Definition: The integrand $F(x,y,z)$ is separable on a product region $R=I\times J\times K$ if it can be written as $F(x,y,z)=f(x)g(y)h(z)$, where each factor depends on just its indicated variable.

For example, $x^2 e^y\sin z$ is separable, with factors $x^2$, $e^y$, and $\sin z$. By contrast, $x+y$ is not a product of an $x$-only factor, a $y$-only factor, and a $z$-only factor.

Theorem: If $f$, $g$, and $h$ are continuous on the finite intervals $I$, $J$, and $K$, respectively, then $$ \iiint_R f(x)g(y)h(z)\,dV = \left(\int_a^b f(x)\,dx\right) \left(\int_c^d g(y)\,dy\right) \left(\int_e^f h(z)\,dz\right). $$ Here $dV$ denotes $dx\,dy\,dz$ in any iterated order over the box. Continuity is a sufficient condition for the iterated integrals and the stated factorization to apply.

The theorem does not say that every triple integral can be factored. It gives a direct product formula when the region is a product of intervals and the integrand is a product of one-variable functions. The factors may be zero or negative; the formula still applies.

Why the Product Formula Works

Use the iterated integral in the order $dz\,dy\,dx$, which is valid for a continuous integrand on a rectangular box. Since $f(x)$ and $g(y)$ do not depend on $z$, they are constants with respect to the inner integration. Integrating the $h(z)$ factor therefore gives the first separation. The remaining two factors separate in the same way.

\begin{aligned} \iiint_R f(x)g(y)h(z)\,dV &=\int_a^b\int_c^d\int_e^f f(x)g(y)h(z)\,dz\,dy\,dx\\ &=\int_a^b\int_c^d f(x)g(y) \left(\int_e^f h(z)\,dz\right)\,dy\,dx\\ &=\int_a^b f(x) \left(\int_c^d g(y)\,dy\right) \left(\int_e^f h(z)\,dz\right)\,dx\\ &=\left(\int_a^b f(x)\,dx\right) \left(\int_c^d g(y)\,dy\right) \left(\int_e^f h(z)\,dz\right). \end{aligned}

Each step uses the fact that a factor independent of the current variable can be taken outside that integral. The rectangular product region matters too: the limits for each variable are constant and independent of the other coordinates, so the three one-variable integrals are over separate intervals.

Using the Formula in Practice

Check the region.
Confirm that it is a product of three intervals with fixed endpoints.

Check the integrand.
Rewrite it as a product with one factor depending only on $x$, one only on $y$, and one only on $z$.

Integrate each factor.
Evaluate the three ordinary definite integrals separately, then multiply their values.

Check the result.
Substitute the computed one-variable integrals into the product formula and verify the arithmetic.

The order of the factors in the final product does not affect its value. Still, keeping the $x$-, $y$-, and $z$-integrals in the same order as the factors makes the calculation easier to audit.

Worked Example: An Exponential and Trigonometric Product

Evaluate $$ \int_0^2\int_{-1}^1\int_0^\pi x^2e^y\sin z\,dz\,dy\,dx. $$

The region is $[0,2]\times[-1,1]\times[0,\pi]$, a product of intervals. The integrand separates into $x^2$, $e^y$, and $\sin z$, so the product formula applies:

\begin{aligned} \int_0^2 x^2\,dx &=\left[\frac{x^3}{3}\right]_0^2 =\frac{8}{3}-0 =\frac{8}{3},\\ \int_{-1}^1 e^y\,dy &=[e^y]_{-1}^1 =e-e^{-1},\\ \int_0^\pi \sin z\,dz &=[-\cos z]_0^\pi =1-(-1) =2. \end{aligned}

Multiplying these values gives $$ \iiint_R x^2e^y\sin z\,dV =\frac{8}{3}(e-e^{-1})(2) =\frac{16}{3}(e-e^{-1}). $$

Check by carrying out the stated iterated order: the inner integral in $z$ is $x^2e^y(2)$; integrating that in $y$ gives $2x^2(e-e^{-1})$; integrating in $x$ gives $\frac{16}{3}(e-e^{-1})$, as obtained from the product formula.

Worked Example: Polynomial Factors on Three Intervals

Evaluate $$ \int_0^2\int_1^3\int_{-1}^1 (x+1)y^2(1-z)\,dz\,dy\,dx. $$

The region is a product of intervals, and the integrand has the factors $x+1$, $y^2$, and $1-z$. Evaluate them individually:

\begin{aligned} \int_0^2 (x+1)\,dx &=\left[\frac{x^2}{2}+x\right]_0^2 =(2+2)-0 =4,\\ \int_1^3 y^2\,dy &=\left[\frac{y^3}{3}\right]_1^3 =\frac{27}{3}-\frac{1}{3} =\frac{26}{3},\\ \int_{-1}^1(1-z)\,dz &=\left[z-\frac{z^2}{2}\right]_{-1}^1\\ &=\left(1-\frac12\right) -\left(-1-\frac12\right) =\frac12-\left(-\frac32\right) =2. \end{aligned}

Therefore, $$ \iiint_R (x+1)y^2(1-z)\,dV =4\left(\frac{26}{3}\right)(2) =\frac{208}{3}. $$

As a check, integration first in $z$ contributes the constant factor $2$. Integrating next in $y$ contributes $\frac{26}{3}$, and integrating last in $x$ contributes $4$. Their product is again $\frac{208}{3}$.

When Direct Factorization Does Not Apply

There are two separate conditions to check. First, a region that is not a product of independent intervals does not permit the direct three-factor formula over that region. Second, even on a rectangular box, a coupled integrand such as $x+y$ is not a product of one-variable factors. Neither failure implies that the triple integral cannot be evaluated; it means only that this particular shortcut is not justified.

Some expressions that look coupled can be rewritten as products. For instance, $e^{x+y}=e^x e^y$, so $e^{x+y}h(z)$ is separable. It is the actual algebraic form of the function, not just the way it first appears, that determines whether the product formula can be used.

Worked Example: A Coupled Sum on a Box

Evaluate $$ \int_0^1\int_0^2\int_0^3(x+y)\,dz\,dy\,dx. $$

The region is a rectangular box, but $x+y$ is a sum, not a product of one-variable factors. So it cannot be replaced by a single product of three one-variable integrals. Instead, integrate directly. The inner integral is $$ \int_0^3(x+y)\,dz=3(x+y), $$ because $x$ and $y$ are constant with respect to $z$. Next, $$ \int_0^2 3(x+y)\,dy =3\left[xy+\frac{y^2}{2}\right]_0^2 =3(2x+2) =6x+6. $$ Finally, $$ \int_0^1(6x+6)\,dx =\left[3x^2+6x\right]_0^1 =3+6 =9. $$

Check by using linearity to split the integrand into $x+y$. The contribution from $x$ is $(\int_0^1 x\,dx)(\int_0^2 1\,dy)(\int_0^3 1\,dz)=\frac12(2)(3)=3$. The contribution from $y$ is $(\int_0^1 1\,dx)(\int_0^2 y\,dy)(\int_0^3 1\,dz)=(1)(2)(3)=6$. Their sum is $3+6=9$, confirming the direct computation.

A Common Mistake

A common error is to treat a sum of terms as if it were a product of factors. For the last example, the incorrect calculation $$ \left(\int_0^1 x\,dx\right) \left(\int_0^2 y\,dy\right) \left(\int_0^3 1\,dz\right) =\frac12(2)(3)=3 $$ does not compute the integral of $x+y$. It combines the $x$-part of one term with the $y$-part of another, and omits the $x$-term's required $y$- and $z$-factors as well as the separate $y$-term. The correct use of linearity gives the sum of two products, $3+6=9$. Factorization applies to a product-form integrand, not to an arbitrary sum.

Likewise, do not multiply three one-variable integrals over a box if the actual region is not that whole product of intervals. The formula is about the integral over $I\times J\times K$; changing the region changes the integral, even if the integrand itself separates.

Why the Shortcut Is Useful

When both conditions hold, separating the integral can reduce a three-variable calculation to three familiar one-variable calculations. This is especially useful when the factors have different forms—for example, a polynomial in $x$, an exponential in $y$, and a trigonometric function in $z$. Each factor can be evaluated with the appropriate one-variable method, and the resulting numbers are multiplied.

The conditions are also a practical safeguard: check the shape of the region and the structure of the integrand before using the formula. If either condition fails, do not force a product. Use an appropriate direct evaluation or, when the integrand is a sum, apply linearity term by term if that produces valid separated terms.

Check Your Understanding

For each question, decide whether direct factorization applies before calculating.

On $[0,1]\times[1,2]\times[0,\pi]$, write the product formula for the integral of $x^3\ln y\cos z$.
Evaluate each one-variable integral in Question 1 and give the resulting triple integral.
On $[0,2]\times[0,1]\times[-1,1]$, is $x^2+y^2+z^2$ itself a separable product? Explain whether linearity could still help.
Why does the product formula require the integration region to be a product of intervals?
Is $e^{x+y}\sin z$ separable? If so, write it as a product of one-variable factors.