Why the Domain Matters
A scalar field assigns a number to each point in its domain. A vector field assigns a vector to each point in its domain. The formula for a field tells us how to calculate its value, but it may not produce a real number or vector at every point in the surrounding space.
For example, division by zero is undefined, and the real square root of a negative number is not defined. The domain of a field formula is the set of all input points where every part of the formula is defined. Knowing that set is necessary before evaluating the field or representing it geometrically.
In this tutorial, the restrictions of interest come from denominators and square roots. A denominator must be nonzero. A square-root radicand must be nonnegative. Both conditions matter: the radicand may equal zero, but a denominator may not.
Two Basic Restrictions
For real inputs, a quotient \(A/B\) is defined only where \(B\ne 0\). Thus, for a formula containing several quotients, each denominator must be checked separately. The points excluded by a denominator are the points where that denominator equals zero.
The real square root \(\sqrt{A}\) is defined when \(A\ge 0\). Equality is allowed: \(\sqrt{0}=0\). Therefore, the boundary where the radicand is zero is included unless another part of the formula excludes it.
For a vector field, the restrictions apply to every component. If \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), then \(\mathbf{F}\) is defined only at points where both \(P\) and \(Q\) are defined. Its domain is the intersection of the domains of its components.
Why the Conditions Are Combined by Intersection
Suppose a formula has restrictions \(C_1,C_2,\ldots,C_k\). An input belongs to the domain precisely when it satisfies every restriction. In set language, if \(S_i\) is the set of inputs satisfying condition \(C_i\), the domain is \(S_1\cap S_2\cap\cdots\cap S_k\).
Find every denominator and every square-root radicand in the formula.
Set each denominator unequal to zero and each radicand greater than or equal to zero.
Keep only inputs satisfying all conditions, including those from every component of a vector field.
Identify the region that remains and mark the forbidden points or sets as excluded.
This procedure works because a formula can be evaluated only if every operation in it can be performed. Satisfying one restriction cannot compensate for violating another.
Worked Example: A Scalar Field on a Disk with a Line Removed
Worked Example: A Scalar Field on a Disk with a Line Removed
Find and describe the domain of \(f(x,y)=\dfrac{\sqrt{9-x^2-y^2}}{x-y}\).
The square root requires its radicand to be nonnegative:
This condition gives the closed disk centered at the origin with radius \(3\). In particular, its circular boundary is allowed by the square root.
The denominator requires:
Combining the conditions gives the domain:
To sketch it, draw the closed disk of radius \(3\), then remove the portion of the line \(y=x\) lying inside that disk. The line meets the circle where \(y=x\) and \(2x^2=9\), so the intersection points are \(\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)\) and \(\left(-\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}}\right)\). Both are excluded because the denominator is zero there.
As a check, at \((3,0)\) the radicand is \(9-9-0=0\), while the denominator is \(3-0=3\). The formula is defined and gives \(f(3,0)=0/3=0\). At \((0,0)\), the radicand is \(9\), but the denominator is \(0\), so the formula is undefined. These checks agree with the stated domain.
Worked Example: Combining Conditions in a Vector Field
Worked Example: Combining Conditions in a Vector Field
Find and sketch the domain of \(\mathbf{F}(x,y)=\left\langle \dfrac{1}{x^2+y^2-4},\sqrt{y-x}\right\rangle\).
The first component requires its denominator to be nonzero:
Thus, points on the circle of radius \(2\) are excluded from the first component. The second component requires:
Both components must be defined at the same point. Therefore:
The sketch begins with the closed half-plane on or above the line \(y=x\). Within that half-plane, remove the part of the circle \(x^2+y^2=4\) that lies there. The circle meets the line \(y=x\) at \((\sqrt{2},\sqrt{2})\) and \((-\sqrt{2},-\sqrt{2})\); those points are excluded too. Points below the line are outside the domain regardless of the circle condition.
Check the point \((0,1)\): the first denominator is \(0^2+1^2-4=-3\), and the radical is \(\sqrt{1-0}=1\). Thus \(\mathbf{F}(0,1)=\langle-1/3,1\rangle\), so the point is in the domain. At \((0,-1)\), the denominator is \(-3\), but the radical would be \(\sqrt{-1}\), which is not real; this point is excluded. At \((\sqrt{2},\sqrt{2})\), the radicand is zero, but the first denominator is \(2+2-4=0\), so that point is also excluded. Each check matches the combined conditions.
Worked Example: A Three-Dimensional Domain
Worked Example: A Three-Dimensional Domain
Find and describe the domain of \(g(x,y,z)=\dfrac{\sqrt{z-x^2-y^2}}{z-1}\).
The radicand condition is:
This describes the points on or above the paraboloid \(z=x^2+y^2\). The denominator condition is:
Consequently, the domain is:
For a sketch, indicate the solid region on or above the paraboloid, then remove the part of the plane \(z=1\) that belongs to that region. In that plane the inequality becomes \(1\ge x^2+y^2\), so the removed set is the disk \(x^2+y^2\le 1\) at height \(z=1\), including its circular edge.
Check \((0,0,2)\): the radicand is \(2\), and the denominator is \(1\), so \(g(0,0,2)=\sqrt{2}\). At \((0,0,1)\), the radicand is \(1\), but the denominator is zero, so the point is excluded. At \((1,0,1)\), the radicand is \(1-1-0=0\), but again the denominator is zero, so it is excluded. These substitutions confirm that the plane restriction removes points even when the square root itself is defined.
Common Mistake: Excluding a Radical Boundary
A frequent error is to require a square-root radicand to be strictly positive. That would wrongly discard points where the radicand is zero. For example, consider the scalar formula \(h(x,y)=\sqrt{4-x^2-y^2}\). Its condition is:
not \(x^2+y^2<4\). At \((2,0)\), \(h(2,0)=\sqrt{4-4-0}=\sqrt{0}=0\), so the point is included. The correction is to use a non-strict inequality for a real square root, then check separately whether some other part of the formula excludes that point.
The reverse error is also possible: including a point where a denominator vanishes just because the radical is valid. In the three-dimensional example, \((0,0,1)\) satisfies the radical condition but not the denominator condition. Always check each restriction independently before combining them.
Reading a Domain Sketch
A domain sketch should distinguish the allowed region from the forbidden set. A closed inequality such as \(x^2+y^2\le 9\) includes its boundary; a condition such as \(x-y\ne0\) removes the zero set of that expression. For a vector field, the final sketch must show the intersection of the component domains, not just the domain of one component.
These sketches describe where the formulas can be evaluated. They do not describe the field’s values or how it behaves near an excluded set. The essential task here is to state exactly which inputs are permitted and to make that set visible geometrically.
Check Your Understanding
For each question, identify every restriction before describing the domain.
- What condition does the denominator impose in \(p(x,y)=1/(x^2+y^2-9)\)? Describe the excluded set.
- What is the domain of \(q(x,y)=\sqrt{5-x-y}\)? Is its boundary included?
- For \(\mathbf{G}(x,y)=\langle\sqrt{x},1/(y-2)\rangle\), state the conditions that must hold simultaneously.
- Why is a point with a zero square-root radicand not automatically excluded from a field’s domain?
- For \(r(x,y,z)=\sqrt{z-x^2-y^2}/(z-2)\), describe the region allowed by the radical and the set removed by the denominator.