Multivariable calculus

Triple Integral of x y^2 z^3 over a Box

Q: Does the answer change if I integrate in x first?

No. The limits are constants, so all six orders give \frac{1}{3}. The three power rule factors are computed independently and multiplication does not care about order.

The triple integral of x times y squared times z cubed, over the box where x runs 0 to 1, y runs 0 to 2 and z runs 0 to 1, equals 1/3. Each variable is handled by the power rule alone: the x factor gives 1/2, the y factor gives 8/3 and the z factor gives 1/4, and their product is 1/3.

f(x,y,z) = x y^2 z^3

\int_0^1 \int_0^2 \int_0^1 x y^2 z^3 \,dz\,dy\,dx = \frac{1}{3}

Numerically 0.333333, confirmed by quadrature on every build.

One power rule per variable

A monomial over a box is the friendliest triple integral there is. Every factor is a power of a single variable, so the integral becomes a product of three power rule computations.

\left(\int_0^1 x \,dx\right)\left(\int_0^2 y^2 \,dy\right)\left(\int_0^1 z^3 \,dz\right) = \frac{1}{2}\cdot\frac{8}{3}\cdot\frac{1}{4}

The middle factor carries the weight: $\int_0^2 y^2 \,dy = \left[\frac{y^3}{3}\right]_0^2 = \frac{8}{3}$ , because the upper limit gets cubed. Multiplying the three gives $\frac{8}{24} = \frac{1}{3}$ .

For a general monomial on a box with all lower limits at zero, the pattern is one clean formula.

\int_0^a \int_0^b \int_0^c x^p y^q z^r \,dz\,dy\,dx = \frac{a^{p+1}}{p+1}\cdot\frac{b^{q+1}}{q+1}\cdot\frac{c^{r+1}}{r+1}

The mistake: adding exponents or moving them

Two errors show up here again and again. The first is treating $x y^2 z^3$ as $\left(xyz\right)^6$ or as a single power of one combined variable. The exponents belong to different variables and never combine.

The second is applying the power rule to the wrong variable during the inner stage. In $\int_0^1 x y^2 z^3 \,dz$ , the factor $x y^2$ is a constant, so the result is $x y^2 \cdot \frac{1}{4}$ . Raising $y$ to a new power there is a pure bookkeeping slip.

Inner stage in $z$ : only $z^3$ moves, giving $\frac{x y^2}{4}$ .
Middle stage in $y$ : only $y^2$ moves, giving $\frac{x}{4}\cdot\frac{8}{3} = \frac{2x}{3}$ .
Outer stage in $x$ : only $x$ moves, giving $\frac{2}{3}\cdot\frac{1}{2} = \frac{1}{3}$ .

A size check before you commit

The box has volume $(1)(2)(1) = 2$ , and the largest value of $x y^2 z^3$ on it is $1 \cdot 4 \cdot 1 = 4$ at the corner $(1, 2, 1)$ . So the answer sits between $0$ and $8$ , and it should be well below $8$ because the integrand is near zero over most of the box.

The value $\frac{1}{3}$ fits that picture. A student answer like $\frac{8}{3}$ or $6$ would be suspiciously close to the crude upper bound and worth rechecking.

Frequently asked questions

Does the answer change if I integrate in x first?

No. The limits are constants, so all six orders give $\frac{1}{3}$ . The three power rule factors are computed independently and multiplication does not care about order.

What if a lower limit is not zero?

The pattern still works, but each factor becomes a difference. For example $\int_1^2 y^2 \,dy = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ , so you subtract the lower limit contribution rather than dropping it.