Iterated Integrals over z-Simple Solids

Why Integrate Along Vertical Fibers?

A triple integral accumulates values throughout a three-dimensional region. A useful way to organize that accumulation is to fix a point $(x,y)$ in the $xy$-plane and integrate vertically through the solid above that point. The points in the plane where such a vertical segment exists form the solid’s projection.

When the solid has one continuous lower height and one continuous upper height over its projection, the integral can be written with $z$ as the inner variable. The height limits may vary with $x$ and $y$; the outer limits then describe the projection, not the full solid.

Definition and Iterated-Integral Formula

Definition: Let $D$ be a planar region in the $xy$-plane, and let $g,h:D\to\mathbb R$ satisfy $g(x,y)\le h(x,y)$ on $D$. The solid $$ E=\{(x,y,z):(x,y)\in D,\ g(x,y)\le z\le h(x,y)\} $$ is z-simple with projection $D$. For each fixed $(x,y)\in D$, its vertical fiber is the interval from $z=g(x,y)$ to $z=h(x,y)$.

To use the standard slicing formula without additional improper-integral issues, assume $D$ is compact and Jordan measurable, $g$ and $h$ are continuous on $D$, and $g\le h$. Then $E$ is a compact solid. If $f$ is continuous on $E$, its triple integral exists, and integrating along each vertical fiber gives

\iiint_E f(x,y,z)\,dV = \iint_D \left( \int_{g(x,y)}^{h(x,y)} f(x,y,z)\,dz \right)dA.

Here $dV$ denotes volume integration in three dimensions, and $dA$ denotes area integration over the projection. The inner integral treats $x$ and $y$ as fixed; after evaluating it, the resulting function of $x$ and $y$ is integrated over $D$. If the projection has a convenient description such as $a\le x\le b$ and $u(x)\le y\le v(x)$, the formula becomes

\int_a^b\int_{u(x)}^{v(x)} \int_{g(x,y)}^{h(x,y)} f(x,y,z)\,dz\,dy\,dx.

The outer two limits describe the projection $D$. The inner two limits describe the vertical fiber. Confusing these roles changes the solid being integrated over.

Why the Slicing Formula Works

The formula is the vertical-slice case of Fubini’s theorem for a continuous function on a compact, Jordan measurable solid. Its geometric content is that the solid is assembled from vertical fibers above points in its projection. For each such point, the contribution along the fiber is the one-dimensional integral of $f$ from the lower surface to the upper surface. Accumulating those contributions over the projection gives the triple integral.

Fix a projected point.
Choose $(x,y)\in D$. The allowed values of $z$ run from $g(x,y)$ to $h(x,y)$.

Integrate along the fiber.
Compute $\int_{g(x,y)}^{h(x,y)} f(x,y,z)\,dz$, treating $x$ and $y$ as constants.

Accumulate over the projection.
Integrate the resulting function over $D$ using two-dimensional area integration.

This also explains the importance of the hypotheses. Compactness and continuity ensure that the solid and integrand are bounded in this setting, while the Jordan-measurability condition supplies the ordinary integrals used in the formula. Continuity of $f$ on the solid alone would not guarantee a finite integral if the region were not compact.

Worked Examples

Worked Example: A Triangular Projection with a Sloping Top

Let $D$ be the triangle $0\le x\le 1,\ 0\le y\le x$, and let the solid lie between $z=0$ and $z=x+y$. Evaluate the integral of $f(x,y,z)=x+y+z$ over this solid.

First, translate the projection and height bounds into an iterated integral:

\iiint_E (x+y+z)\,dV = \int_0^1\int_0^x\int_0^{x+y}(x+y+z)\,dz\,dy\,dx.

For the inner integration, $x+y$ is constant with respect to $z$:

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

Integrate this expression with respect to $y$:

\int_0^x\frac{3}{2}(x+y)^2\,dy = \frac{3}{2}\left[x^2y+xy^2+\frac{y^3}{3}\right]_0^x = \frac{3}{2}\left(x^3+x^3+\frac{x^3}{3}\right) = \frac{7}{2}x^3.

Finally, integrate with respect to $x$:

\int_0^1\frac{7}{2}x^3\,dx = \left[\frac{7}{8}x^4\right]_0^1 = \frac{7}{8}.

Check the inner antiderivative by differentiating $(x+y)z+z^2/2$ with respect to $z$; the result is $x+y+z$, the original integrand. The limits $0$ and $x+y$ are precisely the bottom and top of each vertical fiber.

Worked Example: A Curved Projection and Variable Height

Take $D=\{(x,y):0\le x\le 1,\ x^2\le y\le 1\}$, and let $0\le z\le x+y$. Evaluate the integral of $f(x,y,z)=z$ over the resulting solid.

The projected region supplies the outer limits, while the two height functions supply the inner limits:

\iiint_E z\,dV = \int_0^1\int_{x^2}^1\int_0^{x+y}z\,dz\,dy\,dx.

Integrating first in $z$ gives

\int_0^{x+y}z\,dz = \left[\frac{z^2}{2}\right]_0^{x+y} = \frac{(x+y)^2}{2}.

Now expand before integrating in $y$:

\begin{aligned} \int_{x^2}^1\frac{(x+y)^2}{2}\,dy &=\frac12\int_{x^2}^1(x^2+2xy+y^2)\,dy\\ &=\frac12\left[x^2y+xy^2+\frac{y^3}{3}\right]_{x^2}^1\\ &=\frac12\left(x^2+x+\frac13-x^4-x^5-\frac{x^6}{3}\right). \end{aligned}

Integrate each term over $0\le x\le 1$:

\begin{aligned} \iiint_E z\,dV &=\frac12\left(\frac13+\frac12+\frac13-\frac15-\frac16-\frac1{21}\right)\\ &=\frac12\left(\frac76-\frac{29}{70}\right) =\frac12\left(\frac{79}{105}\right) =\frac{79}{210}. \end{aligned}

For a check, the antiderivative $\frac12x^2y+\frac12xy^2+\frac16y^3$ differentiates with respect to $y$ to $\frac12(x+y)^2$, the result of the $z$-integration. The lower $y$-limit $x^2$ and upper limit $1$ belong to the projection; neither is a height limit.

Worked Example: A Triangular Projection with Both Height Bounds Varying

Let $D$ be the triangle $0\le x\le 1,\ 0\le y\le 1-x$, and let the vertical fiber run from $z=x+y$ to $z=1$. Evaluate the integral of $f(x,y,z)=x+y+z$.

The top surface is at least as high as the bottom throughout $D$, since $x+y\le 1$. Set up the integral with the projection limits outside:

\iiint_E (x+y+z)\,dV = \int_0^1\int_0^{1-x}\int_{x+y}^1(x+y+z)\,dz\,dy\,dx.

Write $a=x+y$ temporarily while integrating in $z$:

\begin{aligned} \int_{x+y}^1(x+y+z)\,dz &=\left[(x+y)z+\frac{z^2}{2}\right]_{x+y}^1\\ &=(x+y)+\frac12-(x+y)^2-\frac{(x+y)^2}{2}\\ &=\frac12+(x+y)-\frac32(x+y)^2. \end{aligned}

For the $y$-integration, use $t=x+y$, so $dt=dy$, and the limits $y=0$ and $y=1-x$ become $t=x$ and $t=1$:

\begin{aligned} \int_0^{1-x}\left(\frac12+x+y-\frac32(x+y)^2\right)dy &=\left[\frac{t}{2}+\frac{t^2}{2}-\frac{t^3}{2}\right]_x^1\\ &=\frac12-\left(\frac{x}{2}+\frac{x^2}{2}-\frac{x^3}{2}\right)\\ &=\frac{1-x-x^2+x^3}{2}. \end{aligned}

Integrating the remaining polynomial gives

\begin{aligned} \int_0^1\frac{1-x-x^2+x^3}{2}\,dx &=\frac12\left[x-\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}\right]_0^1\\ &=\frac12\left(1-\frac12-\frac13+\frac14\right) =\frac{5}{24}. \end{aligned}

As a check, differentiating $(x+y)z+z^2/2$ with respect to $z$ recovers $x+y+z$. Also, the intermediate fiber integral is nonnegative: its integrand is positive and its interval has $x+y\le 1$. The positive final value is consistent with that fact.

A Common Bounds Error

A frequent mistake is to use an $x$- or $y$-limit as though it were a $z$-limit. For the first example, the projection has $0\le y\le x$, but the vertical fiber has $0\le z\le x+y$. Replacing its upper limit by $x$ changes the region.

For instance, the incorrect setup would be

\int_0^1\int_0^x\int_0^x(x+y+z)\,dz\,dy\,dx.

Its value is not the desired $7/8$. Indeed, the inner integral is $x(x+y)+x^2/2=3x^2/2+xy$; integrating this from $y=0$ to $y=x$ gives $2x^3$, and integrating from $x=0$ to $x=1$ gives $1/2$. The error is geometrical, not an antiderivative error. The correct vertical limit is $x+y$, so the correct value is $7/8$.

Key check: At a fixed projected point $(x,y)$, identify the actual lowest and highest $z$-values in the solid. Use those values as the inner limits. Determine the outer limits separately from the projection.

When This Representation Is Useful

A z-simple description is especially convenient when the solid is naturally bounded above and below by surfaces written as $z=h(x,y)$ and $z=g(x,y)$. The bounds can depend on both projected coordinates, while the projection itself can have curved or straight edges. Before integrating, verify that the lower height never exceeds the upper height on $D$, and that the outer limits cover exactly the projection.

The method also clarifies what an integral measures at each stage: first the contribution through a vertical segment, then the accumulated contribution across the base. That structure applies whether $f$ is nonnegative or changes sign; the integral records signed contributions when $f$ takes both positive and negative values.

Check Your Understanding

For each question, distinguish the projection limits from the vertical-fiber limits.

In a z-simple solid, what set of points is described by the projection $D$?
Write the iterated-integral formula for a solid between $z=g(x,y)$ and $z=h(x,y)$ above $D$.
For $0\le x\le 1,\ 0\le y\le x$, with $x+y\le z\le 2$, which bounds describe the projection and which describe a vertical fiber?
Why are compactness and continuity useful hypotheses for applying the stated slicing formula?
If a proposed lower height exceeds the upper height at some point in $D$, what must be checked before using that description of the solid?