Triple integrals and volume

Setting up the iterated integral. If the solid $V$ is described by $a\le x\le b$, $g_1(x)\le y\le g_2(x)$, $h_1(x,y)\le z\le h_2(x,y)$, the triple integral is evaluated innermost-first:

where: the innermost integral (over $z$) treats $x,y$ as constants and produces a function of $x,y$; the middle integral (over $y$) treats $x$ as constant and produces a function of $x$; the outer integral (over $x$) finishes the job as an ordinary single integral. The three variables can be integrated in any of six orders, and the point of choosing one order over another is purely practical — the same solid can have simple limits in one order and painful ones in another.

Volume as the special case. Taking $f(x,y,z)=1$ gives

exactly mirroring how $\iint_R 1\,dA$ gives a plane area — the same summing idea, one dimension up.

Reading the solid. The limits almost always come from a picture, not from algebra alone. Fix $x$ and $y$ and ask which $z$-values lie inside $V$ — that is the innermost pair of limits. Then look at the shadow (projection) the solid casts on the $xy$-plane — that 2D region supplies the outer double-integral limits, exactly as in plane double integrals.

Choosing an order. All six orderings of $x,y,z$ give the same numerical answer, but they are not equally easy to set up. A solid bounded by a single slanted plane and the three coordinate planes, like the tetrahedron below, is naturally described with $z$ innermost because its upper surface is a single explicit function $z=h(x,y)$; a solid described instead as "between two cylinders" might integrate far more cleanly with $x$ or $y$ innermost. Before writing any limits, look at which variable's bounding surfaces are given as single explicit functions of the other two — that variable is usually the easiest to make innermost.

Derive, from the definition of a triple integral, the volume of the general first-octant tetrahedron bounded by the coordinate planes $x=0$, $y=0$, $z=0$ and the slanted plane $x/a+y/a+z/a=1$ with mutually perpendicular edges of length $a$ meeting at the origin, and show the result is $V=a^3/6$, the standard formula for such a right-angled corner solid.

Step 1. Identify the region. The plane $x+y+z=a$ cuts the three axes at $(a,0,0)$, $(0,a,0)$, $(0,0,a)$, so the solid sits in the first octant, capped above by $z=a-x-y$ and below by $z=0$. Its shadow on the $xy$-plane is the triangle $x\ge0$, $y\ge0$, $x+y\le a$.

Step 2. Write the iterated integral. With $z$ innermost, then $y$, then $x$:

Step 3. Integrate over $z$. The innermost integral is $\int_0^{a-x-y}dz = a-x-y$, a linear function of $x,y$ that gives the "height" of the solid above each point of the shadow triangle.

Step 4. Integrate over $y$. $$\int_0^{a-x}(a-x-y)\,dy = \left[(a-x)y-\frac{y^2}{2}\right]_0^{a-x} = (a-x)^2 - \frac{(a-x)^2}{2} = \frac{(a-x)^2}{2}.$$

Step 5. Integrate over $x$ and simplify. $$V = \int_0^a \frac{(a-x)^2}{2}\,dx = \frac12\left[-\frac{(a-x)^3}{3}\right]_0^a = \frac12\left(0+\frac{a^3}{3}\right) = \frac{a^3}{6}.$$ Setting $a=1$ metre recovers the Pass-1 result $V=\tfrac16\ \text{m}^3$ exactly, confirming the general formula against the specific case already worked out.

Problem. For the box $V$: $0\le x\le1$, $0\le y\le2$, $0\le z\le3$, evaluate $\displaystyle\iiint_V xyz\,dV$ and hence find the average value of $f(x,y,z)=xyz$ over $V$.

Step 1. Because $V$ is a rectangular box with constant limits and $f=xyz$ factors as a product of single-variable functions, the triple integral separates:

Step 2. Find the volume of the box needed for the average: $\text{Volume}(V) = \iiint_V dV = 1\times2\times3 = 6$ (the same separable-box trick with $f=1$, so each factor is just the side length).

Step 3. The average value of $f$ over a solid region is the integral divided by the volume:

The average value of $xyz$ over the box is $\bar f = 0.75$.

Competitive stretch

Step 1. Check the separable shortcut against a direct nested evaluation for one factor. Integrating $z$ innermost first over the same box: $\int_0^3 xyz\,dz = xy\left[\frac{z^2}{2}\right]_0^3 = \frac{9xy}{2}$.

Step 2. Integrate over $y$: $\int_0^2 \frac{9xy}{2}\,dy = \frac{9x}{2}\left[\frac{y^2}{2}\right]_0^2 = \frac{9x}{2}\times2 = 9x$.

Step 3. Integrate over $x$: $\int_0^1 9x\,dx = 9\left[\frac{x^2}{2}\right]_0^1 = \frac92 = 4.5$. This matches the separable-shortcut answer exactly, confirming the shortcut is just a faster route to the same nested integral, not a different calculation.

• Writing the innermost variable's limits as fixed numbers when they actually depend on the other two variables, e.g. treating $z$'s upper limit as constant when it should be $a-x-y$.
• Skipping the sketch: for a 3D solid, guessing the shadow region or the height function without visualizing the boundary surfaces is the single biggest source of wrong limits.
• Forgetting the separable-box shortcut and grinding through a full nested evaluation when the region is a box and $f$ factors into single-variable pieces.
• Confusing $\iiint_V dV$ (a pure volume) with $\iiint_V f\,dV$ for nontrivial $f$ (a weighted total, such as mass if $f$ is a density) — know which quantity a given setup actually represents.

Ask the tutor: if you are unsure how to read off the z limits or the shadow region for a solid described in words, ask the AI tutor to sketch it with you cross-section by cross-section before committing to the integral.