Projections and Fibers of Solid Regions

Why Projections and Fibers Matter

A solid in three-dimensional space can be difficult to understand from a list of inequalities alone. A useful geometric question is: if we look down toward a coordinate plane, which points can the solid reach? Those points form a projection. At each projected point, we can then ask which points of the solid lie directly above it. Those points form a fiber.

This viewpoint separates a three-dimensional region into a two-dimensional footprint and the one-dimensional pieces attached to that footprint. When each fiber is a single line segment, its endpoints describe the limits for the coordinate perpendicular to the projection plane. The projection identifies where those limits apply. This is the geometric information needed to set up bounds in an iterated integral.

Central idea: A projection records where fibers are nonempty; a fiber records the solid’s extent along a chosen coordinate direction above one point of the projection.

Definitions

Let \(S\) be a solid in \(\mathbb{R}^3\), with coordinates \((x,y,z)\). Its projection onto the \(xy\)-plane is the set of points \((x,y)\) for which at least one value of \(z\) places \((x,y,z)\) in \(S\). The fiber parallel to the \(z\)-axis above \((x,y)\) is the set of all such \(z\)-values.

D_{xy}=\{(x,y)\in\mathbb{R}^2:\text{ there exists }z\in\mathbb{R} \text{ such that }(x,y,z)\in S\},

S_{(x,y)}=\{z\in\mathbb{R}:(x,y,z)\in S\}.

Here \(D_{xy}\) is the projection, and \(S_{(x,y)}\) is the fiber at the projected point. The corresponding definitions for the other coordinate planes are obtained by leaving out the coordinate perpendicular to the plane:

D_{xz}=\{(x,z):\text{ some }y\text{ has }(x,y,z)\in S\}, \qquad S_{(x,z)}=\{y:(x,y,z)\in S\},

D_{yz}=\{(y,z):\text{ some }x\text{ has }(x,y,z)\in S\}, \qquad S_{(y,z)}=\{x:(x,y,z)\in S\}.

A fiber may be empty when its base point is outside the projection. Inside the projection it may be one interval, a single point, several disjoint intervals, or a more general set. It is not automatically a single segment.

Definition: A coordinate-plane projection consists of all base points for which the corresponding fiber is nonempty. The fiber over a base point consists of all values of the omitted coordinate that place a point in the solid.

Finding a Projection from the Fiber Condition

To find a projection, keep the two projected coordinates and ask whether there is at least one value of the omitted coordinate satisfying all the solid’s inequalities. The projection is exactly the set of base points for which the answer is yes. This existence condition matters: merely deleting an inequality or copying some visible bounds does not necessarily give the projection.

If the fiber is a single interval, its lower and upper endpoints describe the solid along the omitted coordinate. Those endpoint functions apply only over the projection. If the fiber splits into pieces, one pair of lower and upper limits does not describe it without splitting the fiber into separate intervals.

Worked Example: Projecting a Sphere onto the \(xy\)-Plane

Worked Example: A Sphere and Its Vertical Fibers

Consider the solid \(S\) defined by \(x^2+y^2+z^2\le 4\). Find its projection onto the \(xy\)-plane and describe the fibers.

For a fixed \((x,y)\), the inequality for \(z\) is

z^2\le 4-x^2-y^2.

A real \(z\) can satisfy this inequality exactly when the right-hand side is nonnegative. Thus the projection is

D_{xy}=\{(x,y):x^2+y^2\le 4\}.

For every point in this disk, the fiber is the interval

S_{(x,y)} = \left[-\sqrt{4-x^2-y^2},\ \sqrt{4-x^2-y^2}\right].

To check the projection condition, take \((x,y)=(1,1)\). Then \(x^2+y^2=2\), so the fiber is \([-\sqrt{2},\sqrt{2}]\). Its endpoints satisfy \(1^2+1^2+(\sqrt{2})^2=4\), as required. At the boundary point \((2,0)\), the fiber is \([0,0]\), so the sphere touches that vertical line at just one point. Outside the disk, for example at \((2,1)\), \(4-x^2-y^2=-1\), and no real \(z\) exists. This verifies both the disk projection and the fiber description.

Worked Example: A Triangular Projection and Variable Fibers

Worked Example: Finding the Footprint and Vertical Segments

Let \(S\) be defined by \(x\ge 0\), \(y\ge 0\), \(x+y\le 2\), and \(0\le z\le 2-x-y\). Find its projection onto the \(xy\)-plane and describe each fiber.

The inequalities that restrict \(x\) and \(y\) require \(x\ge0\), \(y\ge0\), and \(x+y\le2\). For every such pair, the interval from \(0\) to \(2-x-y\) is nonempty because \(2-x-y\ge0\). Therefore

D_{xy}=\{(x,y):x\ge0,\ y\ge0,\ x+y\le2\}, \qquad S_{(x,y)}=[0,2-x-y].

For example, at \((x,y)=(\tfrac12,\tfrac12)\), the upper endpoint is \(2-\tfrac12-\tfrac12=1\), giving the fiber \([0,1]\). At \((1,1)\), the upper endpoint is \(0\), giving the single point \(\{0\}\). At \((1,2)\), the condition \(x+y\le2\) fails, so the fiber is empty. These checks confirm that the triangle is precisely the projection and that the stated endpoints obey the original inequalities.

The geometric bounds for the \(z\)-coordinate vary with \((x,y)\). They are used over this triangular projection, not over the entire \(xy\)-plane.

Worked Example: Fibers with Two Separate Pieces

Worked Example: A Solid with Disconnected Vertical Fibers

Consider the solid with \(-1\le x\le1\), \(-1\le y\le1\), and \(z\in[-2,-1]\cup[1,2]\). Find its \(xy\)-projection and vertical fibers.

For every \((x,y)\) in the square \([-1,1]\times[-1,1]\), both listed \(z\)-intervals are available. For a point outside the square, no \(z\) can make the point belong to \(S\). Hence

D_{xy}=[-1,1]\times[-1,1], \qquad S_{(x,y)}=[-2,-1]\cup[1,2] \quad\text{for }(x,y)\in D_{xy}.

At \((0,0)\), for instance, the fiber contains \(z=-\tfrac32\) and \(z=\tfrac32\), but it does not contain \(z=0\). Each interval has length \(1\), so the total fiber length is \(1+1=2\). At \((2,0)\), the fiber is empty because \(x=2\) is outside the square. This verifies the projection and shows why the fiber must be described as two intervals rather than one segment joining \(-2\) to \(2\).

How Fibers Produce Integration Bounds

Suppose the projection is onto the \(xy\)-plane and the fiber over each \((x,y)\) in a region \(D_{xy}\) is the single interval \([a(x,y),b(x,y)]\), where \(a(x,y)\le b(x,y)\). Then the omitted coordinate lies between those two endpoint functions over the projected region. This is the geometric interpretation of bounds for an integral taken first along the \(z\)-direction.

If a fiber consists of several intervals, each interval must be accounted for separately when describing the solid by such bounds. The same principle applies to projections onto the \(xz\)-plane or \(yz\)-plane: the fiber limits concern \(y\) or \(x\), respectively. The direction of the fiber determines which coordinate is bounded inside, while the projection determines the base region where those bounds apply.

Why the Volume Formula Requires Measurability

For a solid \(S\), the length of a vertical fiber is its one-dimensional measure. A tempting claim is that volume is obtained by adding these lengths across the projection. The claim needs hypotheses: having measurable fibers one at a time is not enough to guarantee that the solid has a volume or that the fiber-length function is measurable.

A correct general statement uses Fubini’s theorem. Assume \(S\) is bounded and Lebesgue measurable. Its indicator function \(\mathbf{1}_S\), equal to \(1\) on \(S\) and \(0\) elsewhere, is integrable. Fubini’s theorem implies that for almost every \((x,y)\), the vertical section is measurable and its length is the integral of the indicator function along that vertical line. The section-length function is measurable almost everywhere, and integrating it over the plane gives the volume:

\operatorname{Vol}(S) = \int_{\mathbb{R}^2} \left(\int_{\mathbb{R}}\mathbf{1}_S(x,y,z)\,dz\right)\,dx\,dy = \int_{\mathbb{R}^2}\operatorname{length}(S_{(x,y)})\,dx\,dy.

The inner integral is the length of the fiber wherever the section is measurable; values on exceptional base points of measure zero do not affect the integral. If the projection and fiber-length function are measurable in the ordinary sense—as they are, for example, for a compact solid—the outer integral can equivalently be taken over the projection, since fibers outside it are empty. The same theorem applies using either of the other coordinate directions.

Start with the solid’s inequalities.
Keep the two coordinates of the desired projection fixed and regard the third as variable.

Determine when a fiber exists.
Find the base points for which at least one value of the omitted coordinate satisfies every inequality.

Describe the fiber.
List all allowed values of the omitted coordinate, keeping separate intervals separate.

Interpret the bounds geometrically.
The projection is the base domain; the fiber endpoints give the coordinate limits over that domain when the fiber is an interval.

A Common Mistake: Using the Projection as if It Were the Solid

A projection alone does not specify how far the solid extends along each fiber. For the sphere of radius \(2\), the projection lies inside the square \([-2,2]\times[-2,2]\), but it is a disk, and the vertical fiber lengths vary from point to point.

One incorrect calculation would treat the whole square as the projection and assign every point a vertical fiber of length \(4\). That would give \(4\cdot 4\cdot 4=64\). This is wrong for two reasons: corner points such as \((2,2)\) have empty fibers, and even above the disk the fiber length is not constantly \(4\).

The correct description is the disk \(x^2+y^2\le4\), with fiber length \(2\sqrt{4-x^2-y^2}\). The volume relation therefore integrates this varying length over the disk. The correction is not merely a better numerical estimate: it uses the actual projection and the actual fibers, which are the data required by the geometry.

Takeaway: A projection identifies where fibers exist, but does not by itself give their lengths. Determine both the projected domain and the complete fiber above each point.

Check Your Understanding

For each question, identify the projection and fiber information needed to describe the solid.

For \(x^2+y^2+z^2\le9\), what is the projection onto the \(xy\)-plane, and what is the fiber above a point in that projection?
Why does finding the projection require checking whether at least one value of the omitted coordinate exists?
Can a fiber consist of two disjoint intervals? Give the relevant feature of the third worked example.
When do fiber endpoints provide a single pair of bounds for the omitted coordinate?
Why does the volume-by-fibers formula require more than measurability of each individual fiber?