Introduction
A solid can be described by examining its intersections with vertical lines. Fix a point \((x,y)\) in the plane and move along the line parallel to the \(z\)-axis through that point. If the line meets the solid in one continuous segment, the segment has a lowest and a highest \(z\)-value. These values can be expressed as functions of \(x\) and \(y\).
The points \((x,y)\) for which the vertical segment exists form the solid’s projection onto the \(xy\)-plane. Finding this projection is essential: bounds for \(z\) make sense only over points where the lower surface is at or below the upper surface.
Definition and vertical fibers
For a solid \(E\subseteq\mathbb{R}^3\), its vertical fiber above \((x,y)\) is the set of heights \(z\) for which \((x,y,z)\) belongs to \(E\). In set notation, this fiber is \(\{z\in\mathbb{R}:(x,y,z)\in E\}\). The projection of \(E\) onto the \(xy\)-plane is the set of all points whose vertical fibers are nonempty.
The definition includes the bounding surfaces. At a point where \(g(x,y)=h(x,y)\), the fiber consists of just one point. Such a fiber is still nonempty and is allowed. If strict inequalities are used instead, the boundary surfaces are omitted; the choice depends on whether the solid includes its boundary.
Sometimes a solid is first described using additional side conditions, such as \(x\ge 0\) or \(y\le 2\). Those conditions restrict the projection as well. The region \(D\) must contain exactly the points satisfying the side conditions for which a permitted height \(z\) exists.
How to find the projection
Suppose a solid is described by lower and upper restrictions \(g(x,y)\le z\le h(x,y)\), possibly together with restrictions on \(x\) and \(y\). A vertical fiber exists precisely when there is at least one \(z\) satisfying both bounds. That is possible exactly when \(g(x,y)\le h(x,y)\). Thus the projection is found by collecting the lateral restrictions and imposing this inequality.
Write the smallest allowed height as \(z=g(x,y)\) and the largest as \(z=h(x,y)\).
Translate all conditions involving \(x\) and \(y\) into restrictions on the planar region.
Impose \(g(x,y)\le h(x,y)\). The points satisfying all these conditions form \(D\).
Write \((x,y)\in D\) first, followed by the lower and upper bounds for \(z\).
Worked Example: A solid over a rectangular projection
Worked Example: A solid over a rectangular projection
Describe the solid specified by \(0\le x\le1\), \(0\le y\le2\), and \(x+y\le z\le3\). Find its projection and identify its vertical fibers.
The lower and upper functions are \(g(x,y)=x+y\) and \(h(x,y)=3\). The side conditions initially restrict \((x,y)\) to the rectangle \(0\le x\le1\), \(0\le y\le2\). We must also require that the lower bound does not exceed the upper bound: $$ x+y\le 3. $$ On the rectangle, \(x+y\le1+2=3\), so this condition is satisfied everywhere. Therefore the projection is $$ D=\{(x,y):0\le x\le1,\ 0\le y\le2\}. $$ For every point of this rectangle, the fiber is the interval \(x+y\le z\le3\). For example, at \((x,y)=(1,2)\), the lower and upper values are both \(3\), so the fiber is the single point \(z=3\). At \((x,y)=(0,1)\), the bounds are \(1\le z\le3\). These substitutions verify that the stated fibers obey the original inequalities.
Worked Example: A triangular projection from side conditions
Worked Example: A triangular projection from side conditions
Describe the solid bounded below by \(z=0\), above by \(z=2-x-y\), and laterally by \(x\ge0\) and \(y\ge0\). Determine the projection.
The vertical bounds are \(0\le z\le2-x-y\). For the fiber to be nonempty, its lower bound must not exceed its upper bound: $$ 0\le2-x-y, $$ so \(x+y\le2\). Combining this with \(x\ge0\) and \(y\ge0\), the projection is $$ D=\{(x,y):x\ge0,\ y\ge0,\ x+y\le2\}. $$ Equivalently, the region is a triangle with vertices \((0,0)\), \((2,0)\), and \((0,2)\). The full solid description is $$ E=\{(x,y,z):(x,y)\in D,\ 0\le z\le2-x-y\}. $$ At \((0,0)\), the fiber is \(0\le z\le2\). At \((1,1)\), the upper value is \(2-1-1=0\), so the fiber is the single point \(z=0\). At \((2,0)\), it is also \(0\le z\le0\). These boundary checks confirm that the fibers collapse where \(x+y=2\), as required by the bounds.
Why the projection condition works
The condition \(g(x,y)\le h(x,y)\) is not merely a convenient test. It exactly characterizes when a vertical fiber exists. To see this, fix a point \((x,y)\) that satisfies any side conditions. A height in the fiber must be at least \(g(x,y)\) and at most \(h(x,y)\).
This argument proves both directions. It shows that no point failing the inequality can project from the solid, and that every point satisfying it has at least one corresponding height. Together with the side restrictions, it gives precisely the projection region.
Worked Example: The intersection of two curved graphs
Worked Example: The intersection of two curved graphs
Find the projection and describe the fibers of the solid between \(z=x^2+y^2\) and \(z=4-x^2-y^2\).
The lower graph is \(g(x,y)=x^2+y^2\), and the upper graph is \(h(x,y)=4-x^2-y^2\). A vertical fiber exists where $$ x^2+y^2\le4-x^2-y^2. $$ Adding \(x^2+y^2\) to both sides gives $$ 2x^2+2y^2\le4. $$ Dividing by \(2\) gives $$ x^2+y^2\le2. $$ Therefore the projection is the closed disk \(D=\{(x,y):x^2+y^2\le2\}\), and the solid is $$ E=\{(x,y,z):x^2+y^2\le2,\ x^2+y^2\le z\le4-x^2-y^2\}. $$ At the center \((0,0)\), the bounds give \(0\le z\le4\). At \((1,1)\), the lower value is \(1^2+1^2=2\), and the upper value is \(4-1^2-1^2=2\); the fiber is the single height \(z=2\). At \((0,0)\), \(0\le4\), and at every point of the boundary circle \(x^2+y^2=2\), both bounds equal \(2\). These checks verify the bounds and show that the fibers collapse on the boundary of the projection.
Reading geometry from the bounds
A z-simple description separates the geometry into two parts. The planar region \(D\) records where the solid reaches, while \(g\) and \(h\) record its bottom and top over each point of that region. This division makes it possible to test a description systematically rather than relying only on a sketch.
For instance, if the projection is proposed to include a point where \(g(x,y)>h(x,y)\), then the description cannot give a fiber there: no real number \(z\) can be both at least \(g(x,y)\) and at most \(h(x,y)\). Such a point must be excluded from \(D\). Conversely, a point where the two functions agree belongs to \(D\) if the boundary is included, even though its fiber has zero length.
The same solid may have more than one valid description if the bounding surfaces or planar region are expressed differently. What matters is that the set of points specified by the bounds is exactly the intended solid. When describing a given solid, check the vertical direction carefully: the lower function must not exceed the upper function anywhere in the stated projection.
Common mistake: forgetting that the bounds must be compatible
A frequent error is to identify the two bounding graphs but omit the inequality that determines where they are in the correct order. For the curved solid in the previous example, one might incorrectly write \(D=\mathbb{R}^2\), because both graph formulas are defined for every \((x,y)\). But being defined does not mean that the lower graph lies below the upper one.
For example, at \((x,y)=(2,0)\), the proposed lower value is \(2^2+0^2=4\), while the upper value is \(4-2^2-0^2=0\). The supposed bounds would require \(4\le z\le0\), which no \(z\) satisfies. The correction is to solve the compatibility condition: $$ x^2+y^2\le4-x^2-y^2 \quad\Longleftrightarrow\quad x^2+y^2\le2. $$ Thus the projection is the disk \(x^2+y^2\le2\), not the whole plane.
Check Your Understanding
For each question, identify the lower and upper bounds before deciding what the projection is.
- What is a vertical fiber of a solid, and what does it mean for the fiber to be nonempty?
- A solid has bounds \(x\le z\le y\). What condition on \((x,y)\) is required for a vertical fiber to exist?
- For \(0\le z\le 5-x-y\), with \(x\ge0\) and \(y\ge0\), write the projection region.
- For a solid bounded by \(z=x^2+y^2\) below and \(z=6\) above, find the projection.
- Why is a point where \(g(x,y)=h(x,y)\) still part of the projection when the solid includes its boundary?