A Workshop on Bounds from Mixed Solid Constraints

Reading Mixed Constraints as Bounds

A solid is often described by several inequalities at once. Some may be planar, such as \(z\le 1-y\); others may be curved, such as \(z\le 1-x^2\). A valid iterated integral must satisfy all the inequalities, not just the one that is easiest to integrate.

The main task in this workshop is to turn those simultaneous restrictions into bounds that describe the solid exactly. If two surfaces compete to be an upper or lower boundary, the relevant boundary can change across the projection. In that case, split the projection where the change occurs and use a separate iterated integral on each part.

Definition: For a solid described by \(g(x,y)\le z\le h(x,y)\), its projection onto the \(xy\)-plane is the set of points \((x,y)\) for which \(g(x,y)\le h(x,y)\) and all other constraints on \(x\) and \(y\) hold. Over each point of that projection, \(z\) ranges between the specified lower and upper surfaces.

When there is more than one upper constraint, the actual upper bound is the smallest permitted value. When there is more than one lower constraint, the actual lower bound is the largest permitted value. These minimum and maximum bounds may be piecewise-defined. The examples below show how to detect and handle that change.

Example: Two Competing Upper Surfaces

Worked Example: Splitting Where the Upper Surface Changes

Find the volume of the solid \(S\) defined by \(0\le x\le 1\), \(0\le y\le 1\), \(z\ge 0\), \(z\le 1-x^2\), and \(z\le 1-y\).

For each fixed \((x,y)\), the height runs from \(0\) to the smaller of \(1-x^2\) and \(1-y\). To locate the change, set the two possible upper bounds equal:

1-x^2=1-y \quad\Longleftrightarrow\quad y=x^2.

If \(0\le y\le x^2\), then \(1-y\ge 1-x^2\), so the curved surface \(z=1-x^2\) is the upper boundary. If \(x^2\le y\le 1\), then \(1-x^2\ge 1-y\), so the plane \(z=1-y\) is the upper boundary. Thus the square projection is split along \(y=x^2\):

V= \int_0^1\int_0^{x^2}\int_0^{1-x^2}1\,dz\,dy\,dx + \int_0^1\int_{x^2}^1\int_0^{1-y}1\,dz\,dy\,dx.

Evaluate the first integral from the inside outward:

\int_0^{1-x^2}1\,dz=1-x^2, \qquad \int_0^{x^2}(1-x^2)\,dy=x^2(1-x^2).

For the second integral, the \(z\)-integral is \(1-y\), and then

\int_{x^2}^1(1-y)\,dy = \left[y-\frac{y^2}{2}\right]_{x^2}^{1} = \frac{(1-x^2)^2}{2}.

Adding the two contributions leaves a single integral in \(x\):

\begin{aligned} V &=\int_0^1\left(x^2(1-x^2)+\frac{(1-x^2)^2}{2}\right)\,dx\\ &=\int_0^1\left(x^2-x^4+\frac12-x^2+\frac{x^4}{2}\right)\,dx\\ &=\int_0^1\left(\frac12-\frac{x^4}{2}\right)\,dx\\ &=\left[\frac{x}{2}-\frac{x^5}{10}\right]_0^1 =\frac12-\frac1{10} =\frac25. \end{aligned}

As a check on the integration, the derivative of \(x/2-x^5/10\) is \(1/2-x^4/2\), the final integrand. The split also agrees with the inequalities: below \(y=x^2\) the plane allows a greater height than the curved surface, and above it the curved surface allows a greater height than the plane.

Why the Split Gives the Correct Integral

A split is not a change to the solid. It is a way to write the integral using bounds that are valid throughout each piece. The reason it works is additivity: the integral over a region divided into pieces is the sum of the integrals over those pieces, provided the pieces cover the region and their interiors do not overlap.

Identify the competing bounds.
At each projected point, the allowed vertical interval must satisfy every lower and upper constraint. For two upper bounds \(h_1(x,y)\) and \(h_2(x,y)\), the actual upper endpoint is \(\min(h_1,h_2)\).

Find where their order changes.
Solve \(h_1(x,y)=h_2(x,y)\) within the projection. The resulting curve or line separates points where one upper bound is smaller from points where the other is smaller.

Partition and integrate.
Use one set of bounds on each side of the switching curve. The pieces cover the projection without overlapping interiors, so their integrals add to the integral over the whole solid.

For an integrable function \(f\), the same reasoning applies to \(\iiint_S f(x,y,z)\,dV\), not just to volume. On a piece where \(g(x,y)\le z\le h(x,y)\), the iterated integral is \(\int\int\int_g^h f(x,y,z)\,dz\,dA\), with the two-dimensional bounds chosen to describe that piece of the projection. This uses the usual iterated-integral rule for a bounded solid with integrable integrand and valid bounds.

Example: A Curved Lower Surface and a Planar Upper Surface

Worked Example: Integrating Between a Parabolic Surface and a Plane

Find the volume of the solid given by \(0\le x\le 1\), \(x^2\le y\le 1\), and \(x^2\le z\le y\). The lower surface for \(z\) is curved, \(z=x^2\), and the upper surface is the plane \(z=y\). The projection is already specified by \(0\le x\le 1\) and \(x^2\le y\le 1\).

V=\int_0^1\int_{x^2}^1\int_{x^2}^{y}1\,dz\,dy\,dx.

First integrate with respect to \(z\), obtaining the height \(y-x^2\). Then integrate with respect to \(y\):

\begin{aligned} V &=\int_0^1\int_{x^2}^1(y-x^2)\,dy\,dx\\ &=\int_0^1\left[\frac{y^2}{2}-x^2y\right]_{x^2}^1\,dx\\ &=\int_0^1\left(\frac12-x^2-\frac{x^4}{2}+x^4\right)\,dx\\ &=\int_0^1\left(\frac12-x^2+\frac{x^4}{2}\right)\,dx\\ &=\left[\frac{x}{2}-\frac{x^3}{3}+\frac{x^5}{10}\right]_0^1\\ &=\frac12-\frac13+\frac1{10} =\frac4{15}. \end{aligned}

The integrand after integrating in \(y\) is \(\frac12-x^2+\frac{x^4}{2}\). Differentiating the displayed antiderivative gives exactly this expression, and evaluating it at \(1\) and \(0\) gives \(4/15\). The bounds also ensure that the vertical interval is nonempty: throughout the projection, \(y\ge x^2\), so its height \(y-x^2\) is nonnegative.

Example: A Projection Bounded by a Plane and a Curved Surface

Worked Example: Splitting a Projection at the Intersection

Find the volume of \(S\), where \(x\ge 0\), \(0\le y\le 1\), \(x^2+y^2\le 2\), and \(0\le z\le 1\). The projection must satisfy both the planar restriction \(y\le 1\) and the curved restriction \(y\le\sqrt{2-x^2}\).

Since \(x\ge0\) and \(x^2+y^2\le2\), the largest possible \(x\) is \(\sqrt2\). The two candidate upper bounds for \(y\) switch where

1=\sqrt{2-x^2} \quad\Longleftrightarrow\quad x^2=1.

Within the projection, the switch is at \(x=1\). For \(0\le x\le 1\), the circle permits \(y\ge 1\), so the plane \(y\le1\) is more restrictive. For \(1\le x\le\sqrt2\), the circle gives the smaller upper bound. The \(z\)-height is \(1\), so

V= \int_0^1\int_0^1\int_0^1 1\,dz\,dy\,dx + \int_1^{\sqrt2}\int_0^{\sqrt{2-x^2}}\int_0^1 1\,dz\,dy\,dx.

The first integral is \(1\). Integrating first in \(z\) and then in \(y\) in the second gives

V=1+\int_1^{\sqrt2}\sqrt{2-x^2}\,dx.

An antiderivative is \(\frac{x}{2}\sqrt{2-x^2}+\arcsin(x/\sqrt2)\). Indeed, differentiating its first term gives \((1-x^2)/\sqrt{2-x^2}\), and differentiating its second gives \(1/\sqrt{2-x^2}\); their sum is \(\sqrt{2-x^2}\). Therefore,

\begin{aligned} V &=1+\left[\frac{x}{2}\sqrt{2-x^2}+\arcsin\left(\frac{x}{\sqrt2}\right)\right]_1^{\sqrt2}\\ &=1+\left(0+\frac{\pi}{2}\right)-\left(\frac12+\frac{\pi}{4}\right)\\ &=\frac12+\frac{\pi}{4}. \end{aligned}

The antiderivative differentiates to the integrand, and the endpoint values use \(\sqrt{2-(\sqrt2)^2}=0\), \(\arcsin(1)=\pi/2\), and \(\arcsin(1/\sqrt2)=\pi/4\). The two projection pieces meet at \(x=1\), where both candidate \(y\)-bounds equal \(1\).

A Common Mistake: Using Only One Constraint

In the first example, a tempting but incorrect setup is to use \(z\le1-y\) over the whole square and write \(\int_0^1\int_0^1\int_0^{1-y}1\,dz\,dy\,dx\). That gives \(1/2\), but it ignores the other upper constraint \(z\le1-x^2\). For instance, at \(x=1/2\) and \(y=0\), the proposed upper endpoint is \(1\), while the curved constraint permits only \(1-(1/2)^2=3/4\). The proposed bounds include points outside the solid.

The correction is to impose both inequalities: use the smaller upper endpoint, \(\min(1-x^2,1-y)\), and split where the two expressions are equal. This check is useful whenever several inequalities constrain the same coordinate: test that every proposed bound obeys all of them, not merely the inequality from which the bound was read.

A Reliable Setup Check

Before evaluating a triple integral over a solid with mixed constraints, check the geometric logic of the bounds. In particular:

For each fixed point in the projection, verify that the lower \(z\)-bound does not exceed the upper \(z\)-bound.
When multiple inequalities give upper bounds, use the smallest; when they give lower bounds, use the largest.
Find where competing bounds are equal, then determine which one applies on each side.
Confirm that the projection pieces cover the intended region and have no overlapping interiors.
After integrating, check the antiderivative and evaluate the endpoints carefully.

These checks focus on the part of a mixed-constraint problem most likely to cause errors: whether the bounds describe exactly the solid. Once the bounds are correct, the integral can be evaluated piece by piece using the usual rules for iterated integrals.

Check Your Understanding

For each question, identify how the inequalities determine the bounds before attempting any integration.

For \(0\le x,y\le1\) and \(0\le z\le\min(1-x,1-y^2)\), where in the square projection do the two candidate upper surfaces switch?
If \(0\le x\le1\), \(x^2\le y\le1\), and \(0\le z\le y-x^2\), what is the height of the solid above \((x,y)\)?
For \(x\ge0\), \(0\le y\le1\), and \(x^2+y^2\le2\), which upper bound for \(y\) applies when \(x=1/2\), and which applies when \(x=3/2\)?
Why is it not sufficient to check that an iterated integral respects just one of several upper-surface inequalities?
When is splitting a projected region necessary for a Cartesian iterated integral with competing bounds?