x-Simple and y-Simple Solid Descriptions

Fibers Determine the Shape of a Solid Description

A solid can be described by asking what happens when we fix two coordinates and allow the third to vary. Fixing $y$ and $z$, for example, leaves a line parallel to the $x$-axis. The part of the solid on that line is called an $x$-fiber. If every nonempty $x$-fiber is one interval, the solid can be described by a lower and an upper $x$-coordinate over a region in the $yz$-plane.

This viewpoint is useful because a three-dimensional region can then be recorded as a base region together with two endpoint functions. The base is a projection onto the coordinate plane perpendicular to the fibers. The same idea applies to fibers parallel to the $y$-axis.

Definition: A solid $D$ is x-simple if there is a region $R_{yz}$ in the $yz$-plane and functions $g_1,g_2$ on that region, with $g_1(y,z)\le g_2(y,z)$, such that $$ D=\{(x,y,z):(y,z)\in R_{yz},\ g_1(y,z)\le x\le g_2(y,z)\}. $$ It is y-simple if there is a region $R_{xz}$ in the $xz$-plane and functions $h_1,h_2$, with $h_1(x,z)\le h_2(x,z)$, such that $$ D=\{(x,y,z):(x,z)\in R_{xz},\ h_1(x,z)\le y\le h_2(x,z)\}. $$

The region $R_{yz}$ is the projection of $D$ onto the $yz$-plane: it contains exactly the pairs $(y,z)$ for which at least one point $(x,y,z)$ belongs to $D$. Similarly, $R_{xz}$ is the projection onto the $xz$-plane. The projection records which fibers exist; the two coordinate limits record the endpoints of each fiber.

These descriptions are geometric, not statements about how an integral must be evaluated. They specify the solid itself. To write one correctly, identify the axis of the fibers first, find the projection perpendicular to that axis, and then determine the lower and upper coordinate values on each fiber.

Reading an x-Simple Description

For an x-simple solid, $y$ and $z$ are fixed while $x$ varies. Consequently, the projection must lie in the $yz$-plane, and the bounds on $x$ may depend on $y$ and $z$. The endpoints might be constants or values of functions. In either case, the lower endpoint cannot exceed the upper endpoint anywhere in the projection.

Worked Example: A Tetrahedral Region as an x-Simple Solid

Describe the region in the first octant below the plane $x+y+z=1$ using $x$-fibers. The first octant conditions are $x\ge 0$, $y\ge 0$, and $z\ge 0$. The condition of lying below the plane is $x+y+z\le 1$.

Fix $y$ and $z$. The inequalities involving $x$ give $$ x\ge 0 \qquad\text{and}\qquad x\le 1-y-z. $$ For such an interval to exist, its upper endpoint must be at least its lower endpoint: $$ 1-y-z\ge 0, \qquad\text{so}\qquad y+z\le 1. $$ Together with $y\ge 0$ and $z\ge 0$, this gives the projection.

R_{yz}=\{(y,z):y\ge 0,\ z\ge 0,\ y+z\le 1\}, \qquad 0\le x\le 1-y-z.

Thus the full description is $$ D=\{(x,y,z):(y,z)\in R_{yz},\ 0\le x\le 1-y-z\}. $$ For instance, at $y=\tfrac14$ and $z=\tfrac12$, the upper $x$-limit is $1-\tfrac14-\tfrac12=\tfrac14$. The fiber is $0\le x\le\tfrac14$. At its upper endpoint, $x+y+z=\tfrac14+\tfrac14+\tfrac12=1$; at its lower endpoint, $x+y+z=\tfrac34\le1$. Both endpoints satisfy the original inequalities, as does every point between them.

The projection condition is necessary as well: if $y,z\ge0$ and $y+z\le1$, then $x=0$ satisfies all the inequalities, so that projected pair really does have a fiber. This checks that the stated projection is exact.

Reading a y-Simple Description

For a y-simple solid, fix $x$ and $z$, then let $y$ vary. The projection belongs in the $xz$-plane, and the limits describe the lowest and highest $y$-coordinates in that fiber. To find the projection from inequalities, ask which pairs $(x,z)$ allow at least one $y$ between the proposed endpoints.

Worked Example: A Solid with a Slanted Lower Boundary

Consider the solid defined by $x\ge0$, $z\ge0$, $x+z\le2$, and $x+2z\le y\le4$. Because $x$ and $z$ are fixed while $y$ varies, the proposed fibers are parallel to the $y$-axis. The projection is $$ R_{xz}=\{(x,z):x\ge0,\ z\ge0,\ x+z\le2\}, $$ and the coordinate bounds are $$ x+2z\le y\le4. $$

We must check that these bounds make sense throughout the projection. From $x+z\le2$ and $z\ge0$, we have $x+2z=(x+z)+z\le2+z$. Also $x\ge0$ gives $z\le x+z\le2$, so $x+2z\le4$. Therefore the lower limit never exceeds the upper limit.

For a specific fiber, take $x=\tfrac12$ and $z=1$. The projection conditions hold because both coordinates are nonnegative and $\tfrac12+1=\tfrac32\le2$. The lower $y$-limit is $\tfrac12+2(1)=\tfrac52$, so this fiber is $\tfrac52\le y\le4$. At the lower endpoint $y=\tfrac52$, the inequality $x+2z\le y$ holds with equality. At the upper endpoint $y=4$, it holds because $\tfrac52\le4$. Every intermediate $y$ satisfies the same bounds. This verifies the coordinate limits on the chosen fiber.

The projection is also exact: for every $(x,z)$ in the stated triangle, we have proved $x+2z\le4$, so at least one allowed $y$ exists. For example, $y=4$ is available for every such pair. No pair outside the triangle can belong to a fiber because the inequalities $x\ge0$, $z\ge0$, and $x+z\le2$ are part of the solid's definition.

A Curved Example: The Unit Ball

Curved boundaries lead to function-valued coordinate limits. For the unit ball, the defining inequality is $x^2+y^2+z^2\le1$. To describe it with $x$-fibers, keep $y,z$ fixed and solve this inequality for $x$.

Worked Example: The Unit Ball as an x-Simple Solid

Rearranging the ball inequality gives $$ x^2\le1-y^2-z^2. $$ A real $x$ can satisfy this inequality exactly when $1-y^2-z^2\ge0$. Thus the projection is the closed unit disk in the $yz$-plane: $$ R_{yz}=\{(y,z):y^2+z^2\le1\}. $$ Taking square roots of the inequality for $x^2$ gives $$ -\sqrt{1-y^2-z^2}\le x\le\sqrt{1-y^2-z^2}. $$

To check a fiber, take $y=\tfrac35$ and $z=0$. This pair lies in the projection because $(\tfrac35)^2+0^2=\tfrac9{25}\le1$. The square root in the $x$-limits is $\sqrt{1-\tfrac9{25}}=\sqrt{\tfrac{16}{25}}=\tfrac45$, so the fiber is $-\tfrac45\le x\le\tfrac45$. At either endpoint, $x^2+y^2+z^2=\tfrac{16}{25}+\tfrac9{25}+0=1$; for every $x$ between those endpoints, $x^2\le\tfrac{16}{25}$, so $x^2+y^2+z^2\le1$. This confirms the bounds.

The disk condition is essential: outside it, the quantity under the square root is negative and there is no real $x$-fiber. On the boundary of the disk, the two limits coincide, so the fiber contains just one point. This is allowed: the lower limit is still no greater than the upper limit.

Why the Projection and Limits Give the Whole Solid

The description by a projection and two coordinate limits works because it makes membership in the solid equivalent to two checks: the fixed coordinates must lie in the projection, and the varying coordinate must lie between the corresponding endpoints. For an x-simple solid, this equivalence can be stated and justified directly.

Characterization: If $R_{yz}$ is a region in the $yz$-plane and $g_1,g_2$ are functions on it with $g_1(y,z)\le g_2(y,z)$, then the set of all points whose $(y,z)$-coordinates lie in $R_{yz}$ and whose $x$-coordinate lies between $g_1(y,z)$ and $g_2(y,z)$ is x-simple. For each fixed $(y,z)\in R_{yz}$, its $x$-fiber is exactly the interval $[g_1(y,z),g_2(y,z)]$. The corresponding statement for y-simple solids uses $R_{xz}$ and the interval $[h_1(x,z),h_2(x,z)]$.

To see why, first choose any $(y,z)\in R_{yz}$. By the set description, the points of the solid with these fixed coordinates are precisely those with $g_1(y,z)\le x\le g_2(y,z)$. This is one interval, including its endpoints. Conversely, take any point $(x,y,z)$ in the described set. Its pair $(y,z)$ is in the projection and its $x$-coordinate lies in the stated interval, so it is included in that fiber. Hence the projection and intervals recover every point of the set, and no additional points.

When starting from inequalities instead, the same logic guides the work in reverse. Fix the coordinates perpendicular to the fibers, solve the inequalities for the remaining coordinate, and then impose the conditions under which the resulting interval is nonempty. Those conditions describe the projection. Checking that the limits are ordered over the entire projection is part of verifying the description.

When One Pair of Bounds Is Not Enough

A solid is not x-simple in a single description of this form if some $x$-fiber consists of two separated intervals. Two endpoint functions would include the gap between them, incorrectly adding points to the solid. The same issue can occur with $y$-fibers. In such a situation, a different fiber direction may give a single interval, or the solid may need to be split into pieces before each piece can be described this way.

Practical test: Fix the two coordinates that are meant to belong to the projection. If the remaining coordinate runs continuously from one endpoint to another for every projected point, a single pair of bounds can describe those fibers. If a fiber is empty, identify that by excluding its base point from the projection. If a fiber has a gap, one interval of bounds cannot represent it without including points outside the solid.

Common Mistake: Projecting onto the Wrong Plane

Consider again the first-octant region $x+y+z\le1$. A tempting but incorrect x-simple description is to use the triangular $xy$-projection and write $0\le z\le1-x-y$. That is a valid way to describe the region using $z$-fibers, but it is not an x-simple description: it varies $z$ while $x,y$ are fixed.

For x-fibers, the fixed pair is $(y,z)$, so the projection must be onto the $yz$-plane. Solving the same inequality for $x$ gives $0\le x\le1-y-z$, with $y\ge0$, $z\ge0$, and $y+z\le1$. The correction is to choose the fiber direction before choosing the projection. The projection plane is perpendicular to the fibers, not a plane chosen merely because its outline is familiar.

A Reliable Procedure

Choose the fiber direction.
For x-simple bounds, vary $x$ and fix $y,z$. For y-simple bounds, vary $y$ and fix $x,z$.

Project perpendicular to the fibers.
Use the $yz$-plane for x-fibers and the $xz$-plane for y-fibers. The projection consists of the fixed-coordinate pairs for which the fiber is nonempty.

Solve for the varying coordinate.
Find its lower and upper limits from the solid's inequalities, and make sure the lower limit does not exceed the upper limit on the projection.

Verify the description.
Check that points between the limits satisfy the original conditions, that the endpoints are included when appropriate, and that the projection contains exactly the base points with nonempty fibers.

These steps turn a three-dimensional description into a family of one-dimensional intervals indexed by a two-dimensional region. Keeping the projection and the fiber limits distinct is the key to writing bounds that describe exactly the intended solid.

Check Your Understanding

For each question, identify the fiber direction and justify the proposed projection or limits.

For an x-simple description, which two coordinates remain fixed, and which coordinate-plane projection is used?
Describe the solid $x^2+y^2+z^2\le4$ using y-fibers. What is its projection, and what are the lower and upper $y$-limits?
For the region $x\ge0,\ z\ge0,\ x+z\le1,\ 0\le y\le x+z$, write a y-simple description and check that the interval is nonempty throughout the projection.
Why must the square-root limits for an x-fiber of the unit ball be accompanied by a disk condition on $y,z$?
What geometric difficulty arises if one fiber consists of two separated intervals?