Coordinate Traces and Intercepts

Introduction

A surface graph \(z=f(x,y)\) lives in three-dimensional space, but its intersections with the coordinate planes are often simpler curves. Those intersections, together with the points where the surface meets the coordinate axes, give concrete constraints for understanding the surface.

The key is to impose the coordinate-plane condition correctly. In the plane \(x=0\), for example, we substitute \(x=0\) into the function, provided that the resulting input is in its domain. In the plane \(z=0\), we instead solve \(f(x,y)=0\). These operations describe different traces.

Definitions: Traces and Intercepts

Let \(f:D\to\mathbb{R}\), where \(D\) is a subset of \(\mathbb{R}^2\). Its graph is the set of points \((x,y,f(x,y))\) for inputs \((x,y)\in D\). A coordinate trace is the intersection of this graph with one of the three coordinate planes.

Definition: The yz-plane trace is the graph's intersection with \(x=0\); the xz-plane trace is its intersection with \(y=0\); and the xy-plane trace is its intersection with \(z=0\). The first two are found by restricting the inputs to \(x=0\) or \(y=0\). The third consists of inputs satisfying \(f(x,y)=0\).

More explicitly, the traces are described by

\begin{aligned} \text{yz-plane: } & (0,y,f(0,y)), && (0,y)\in D,\\ \text{xz-plane: } & (x,0,f(x,0)), && (x,0)\in D,\\ \text{xy-plane: } & (x,y,0), && (x,y)\in D\text{ and }f(x,y)=0. \end{aligned}

An axis intersection is a point of the graph that also lies on one of the coordinate axes. For the \(x\)-axis, both \(y\) and \(z\) must be zero; for the \(y\)-axis, both \(x\) and \(z\) must be zero; and for the \(z\)-axis, both \(x\) and \(y\) must be zero. An axis may have no intersections, one intersection, or many.

Useful conditions: The \(z\)-axis intersection is \((0,0,f(0,0))\) if \((0,0)\in D\). The \(x\)-axis intersections are \((x,0,0)\) for inputs with \((x,0)\in D\) and \(f(x,0)=0\). The \(y\)-axis intersections are \((0,y,0)\) for inputs with \((0,y)\in D\) and \(f(0,y)=0\).

Why the Trace Rules Work

The formulas follow directly from the definition of a graph and the equations of the coordinate planes. A point on the graph has the form \((x,y,f(x,y))\). To lie in the yz-plane it must also satisfy \(x=0\), giving \((0,y,f(0,y))\), when \((0,y)\) belongs to the domain. To lie in the xz-plane it must satisfy \(y=0\), giving \((x,0,f(x,0))\). To lie in the xy-plane it must satisfy \(z=0\); on the graph \(z=f(x,y)\), so this is exactly the equation \(f(x,y)=0\). Thus the three rules are not separate tricks: they are the direct consequences of asking which graph points satisfy each plane's defining equation.

The same reasoning gives the axis conditions. For example, an \(x\)-axis point must be of the form \((x,0,0)\). It belongs to the graph precisely when \((x,0)\in D\) and the graph's height there is \(f(x,0)=0\).

Worked Examples

Worked Example: A Downward-Opening Paraboloid

Find the three coordinate-plane traces and all axis intersections for \(z=4-x^2-y^2\).

The function is defined for all real \(x,y\). For the yz-plane, set \(x=0\):

$$ z=4-0^2-y^2=4-y^2. $$

Its trace is \((0,y,4-y^2)\). For the xz-plane, set \(y=0\):

$$ z=4-x^2-0^2=4-x^2. $$

Its trace is \((x,0,4-x^2)\). For the xy-plane, set \(z=0\), so solve

0=4-x^2-y^2 \quad\Longrightarrow\quad x^2+y^2=4.

The xy-plane trace is the circle of radius \(2\), centered at the origin. For the \(x\)-axis, set \(y=0\) and \(z=0\): \(4-x^2=0\), so \(x^2=4\) and \(x=\pm2\). The intersections are \((2,0,0)\) and \((-2,0,0)\). For the \(y\)-axis, set \(x=0\) and \(z=0\): \(4-y^2=0\), so \(y=\pm2\). The intersections are \((0,2,0)\) and \((0,-2,0)\). On the \(z\)-axis, \(x=y=0\), giving \(z=4\); the intersection is \((0,0,4)\).

Check: substituting either \(x=\pm2,y=0\) or \(x=0,y=\pm2\) into the formula gives \(z=4-4=0\). Substituting \(x=y=0\) gives \(z=4\). Each listed axis point therefore lies on the graph.

Worked Example: A Surface Meeting Two Axes Along Entire Lines

Find the coordinate traces and axis intersections for \(z=xy\).

In the yz-plane, set \(x=0\). Then \(z=0\), so the trace consists of all points \((0,y,0)\). In the xz-plane, set \(y=0\). Again \(z=0\), and the trace consists of all points \((x,0,0)\).

In the xy-plane, solve \(xy=0\). A product is zero exactly when at least one factor is zero. Thus \(x=0\) or \(y=0\), and the trace is the union of the two coordinate axes in the xy-plane:

\{(x,0,0):x\in\mathbb{R}\} \;\cup\; \{(0,y,0):y\in\mathbb{R}\}.

Every point on the \(x\)-axis belongs to the graph because \(y=0\) gives \(z=x\cdot0=0\). Every point on the \(y\)-axis belongs because \(x=0\) gives \(z=0\cdot y=0\). The \(z\)-axis requires \(x=y=0\), which gives \(z=0\); its only intersection is the origin.

Check: each point in either stated axis family satisfies \(z=xy=0\). The origin is also on the graph, since \(0=0\cdot0\). This example shows why axis intersections need not be isolated points.

Worked Example: Traces with a Restricted Domain

Consider \(z=\sqrt{9-x^2-y^2}\). The square root is real only when \(9-x^2-y^2\geq0\), so the domain is \(x^2+y^2\leq9\). Find the coordinate traces and axis intersections.

In the yz-plane, \(x=0\), so the domain condition becomes \(y^2\leq9\), or \(-3\leq y\leq3\). The trace is

(0,y,\sqrt{9-y^2}),\qquad -3\leq y\leq3.

In the xz-plane, \(y=0\). The domain condition becomes \(-3\leq x\leq3\), and the trace is

(x,0,\sqrt{9-x^2}),\qquad -3\leq x\leq3.

In the xy-plane, set \(z=0\). Then \(\sqrt{9-x^2-y^2}=0\), which requires \(9-x^2-y^2=0\). The trace is the circle \(x^2+y^2=9\).

On the \(x\)-axis, \(y=0,z=0\), so \(\sqrt{9-x^2}=0\), giving \(x=\pm3\). The intersections are \((3,0,0)\) and \((-3,0,0)\). On the \(y\)-axis, \(x=0,z=0\), giving \(y=\pm3\), so the intersections are \((0,3,0)\) and \((0,-3,0)\). On the \(z\)-axis, \(x=y=0\), so \(z=\sqrt9=3\), and the intersection is \((0,0,3)\).

Check: the four horizontal axis points have \(x^2+y^2=9\), so their heights are \(\sqrt{9-9}=0\). The vertical axis point has height \(\sqrt{9}=3\). The interval restrictions also matter: the vertical traces stop at \(x=\pm3\) or \(y=\pm3\), where their height is zero.

Using Traces as Constraints

Each trace gives an exact piece of the surface that any sketch must include. For example, if the xz-plane trace is a curve with a particular intercept, the surface must pass through that point. The coordinate-axis intersections provide especially useful shared points: an axis point lies in two coordinate planes, so it must agree with both relevant plane traces.

These constraints do not, by themselves, determine every part of a surface. Many different surfaces can have the same three coordinate-plane traces. The traces are reliable sections of the graph, not a complete description of all its points. They should be used together with other information about the function, such as its domain and its values away from the coordinate planes.

Sketching principle: A proposed surface must contain every point in each computed coordinate trace, and it must obey the domain restrictions. A drawing that contradicts any one of these facts cannot represent the graph.

A Common Mix-Up

A frequent error is to confuse the \(x\)-axis intersection with the xz-plane trace. The xz-plane is defined by \(y=0\), so substituting \(y=0\) finds an entire plane trace; it does not yet require \(z=0\). To find \(x\)-axis intersections, both conditions \(y=0\) and \(z=0\) are necessary.

For \(z=4-x^2-y^2\), setting \(y=0\) gives \(z=4-x^2\). It would be incorrect to call every point \((x,0,4-x^2)\) an \(x\)-axis intersection: for instance, \(x=0\) gives \((0,0,4)\), which lies on the \(z\)-axis, not the \(x\)-axis. Correcting the calculation, impose \(z=0\) as well: \(4-x^2=0\), hence \(x=\pm2\). The actual \(x\)-axis intersections are \((2,0,0)\) and \((-2,0,0)\).

A second important check is whether the function is defined at the inputs being used. If \((0,y)\notin D\), there is no graph point there to include in the yz-plane trace. A restriction may have a limited interval, or the intersection may be empty; do not extend a formula beyond its domain.

Summary

For a graph \(z=f(x,y)\), the yz-plane trace comes from \(x=0\), the xz-plane trace comes from \(y=0\), and the xy-plane trace comes from solving \(f(x,y)=0\). To find an axis intersection, impose both coordinate conditions defining that axis and check that the input belongs to the domain.

Check Your Understanding

Use the definitions to answer each question. Include domain restrictions where they apply.

For \(z=x^2-y\), what equation describes the xz-plane trace?
For \(z=x^2+y^2-1\), find the xy-plane trace.
For \(z=x+y\), find all intersections with the \(x\)-axis and the \(y\)-axis.
For \(z=\sqrt{4-x^2-y^2}\), what interval of \(y\)-values occurs in the yz-plane trace?
Why does setting \(y=0\) alone not find the \(x\)-axis intersections?