Multivariable calculus

Triple Integral of xy + z^2 over a Box

Q: How do I check 21 quickly?

Bound the integrand. On this box xy tops out at 2 and z^2 at 9, so the maximum is 11 and the volume is 6, giving an upper bound of 66. The answer 21 sits comfortably inside, and its average of 3.5 matches an integrand that is small over much of the box.

Q: Would the answer be the same over a different shaped box of volume 6?

No. Swap the x and z edges so x runs 0 to 3 and z runs 0 to 1, and the z^2 contribution collapses while the xy contribution grows. Volume alone does not determine the integral.

The triple integral of xy plus z squared, over the box where x runs 0 to 1, y runs 0 to 2 and z runs 0 to 3, equals 21. The xy term contributes 3 and the z squared term contributes 18. The box has volume 6, so the average value of the integrand on it is 3.5.

f(x,y,z) = xy + z^2

\int_0^1 \int_0^2 \int_0^3 \left(xy + z^2\right) \,dz\,dy\,dx = 21

Numerically 21.000000, confirmed by quadrature on every build.

Break the sum into two separable integrals

The integrand is a sum of two terms, and neither term uses all three variables. Split by linearity, then treat each piece as a product with edge lengths supplied by the missing variables.

\iiint_B xy \,dV = \left(\int_0^1 x \,dx\right)\left(\int_0^2 y \,dy\right)(3) = \frac{1}{2}\cdot 2\cdot 3 = 3

\iiint_B z^2 \,dV = (1)(2)\left(\int_0^3 z^2 \,dz\right) = 2\cdot 9 = 18

Adding gives $3 + 18 = 21$ . The $z^2$ term dominates because the $z$ edge is longest and it is the term being squared there.

The same answer from the iterated form

If you would rather not split, run the nested integrals straight through. The inner one in $z$ handles both terms at once, with $xy$ acting as a constant.

\int_0^3 \left(xy + z^2\right) \,dz = 3xy + \left[\frac{z^3}{3}\right]_0^3 = 3xy + 9

\int_0^2 \left(3xy + 9\right) \,dy = 6x + 18, \qquad \int_0^1 \left(6x + 18\right) \,dx = 3 + 18 = 21

Notice the constant $9$ that appears after the $z$ stage. It is not idle: the later integrations multiply it by the $y$ and $x$ edge lengths, turning it into $18$ .

The mistake: factoring a sum

The product shortcut only applies to a product. Writing $\iiint_B \left(xy + z^2\right) \,dV$ as $\left(\int x \,dx\right)\left(\int y \,dy\right) + \left(\int z^2 \,dz\right)$ gives $1 + 9 = 10$ , which drops the edge factors from both terms.

The rule is that linearity splits sums and separability splits products, and the two steps are applied in that order. Split the sum first, then check each piece separately for separability.

$xy$ is separable, but it is still missing a $z$ factor, so it collects the edge length $3$ .
$z^2$ is separable, and it collects the edge lengths $1$ and $2$ .
Neither piece can be pulled out of the triple integral without those factors.

Frequently asked questions

How do I check 21 quickly?

Bound the integrand. On this box $xy$ tops out at $2$ and $z^2$ at $9$ , so the maximum is $11$ and the volume is $6$ , giving an upper bound of $66$ . The answer $21$ sits comfortably inside, and its average of $3.5$ matches an integrand that is small over much of the box.

Would the answer be the same over a different shaped box of volume 6?

No. Swap the $x$ and $z$ edges so $x$ runs $0$ to $3$ and $z$ runs $0$ to $1$ , and the $z^2$ contribution collapses while the $xy$ contribution grows. Volume alone does not determine the integral.