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Vector Calculus An Introduction

Daniel Daners

Section 9.2 Implicit and Explicit Representations of Surfaces

Quite often, surfaces are given as the set of points \(\vect x=(x_1,x_2,x_3)\in\mathbb R^3\) such that \(f(\vect x)=0\) for some given function \(\vect f\colon\mathbb R^3\to\mathbb R\text{.}\) We call this an implicit representation of a surface. In some cases it is possible to solve \(f(\vect x)=0\) for one of the variables. For instance we can solve for \(x_3\text{.}\) If we can solve with one single function then \(S\) is the graph of a function. Graphs of a function \(h\) defined on a subset \(D\subset\mathbb R^3\) are the simplest possible surfaces. We call \(x_3=h(x_1,x_2)\text{,}\) \((x_1,x_2)\in D\) an explicit representation of the surface. In many cases, however, solving for one variable will only be possible locally, even for very simple surfaces like the sphere.
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Example 9.5. Implicit representation of a sphere.

Suppose that \(S\) is a sphere of radius \(R\text{.}\) By definition \(S\) is the set of points in \(\mathbb R^3\) with distance \(R\) from the origin. Hence \(S\) is the set of all points \(\vect x=(x_1,x_2,x_3)\in\mathbb R^3\) such that
\begin{equation*} f(\vect x):=x_1^2+x_2^2+x_3^2-R^2=0, \end{equation*}
providing an implicit representation of the sphere. We can solve the above equation for \(x_3\) to get
\begin{equation*} x_3=\pm\sqrt{R^2-x_1^2-x_2^2}\text{.} \end{equation*}
Note that we need two different functions to describe the upper and the lower half of the sphere. The above provides an explicit representation of \(S\text{.}\)
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Not for every function \(f\) does the equation \(f(\vect x)=0\) describe a smooth surface. For instance the only solution to \(x_1^2+x_2^2+x_3^2=0\) is \((0,0,0)\text{,}\) that is, a single point. Hence we need to find criteria which guarantee that the solutions of \(f(\vect x)=0\) form a surface. We have the following theorem, which is a generalised form of the implicit function Theorem 4.33, where also the ideas for a more general result are indicated.
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In the above example of the sphere we have \(\grad f(\vect x)=(2x_1,2x_2,2x_3)\text{,}\) which is clearly nonzero if \(R\gt 0\) and \(x_1^2+x_2^2+x_3^2=R\text{.}\)
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As mentioned already smooth surfaces, in principle, can be represented by all three means. We consider some examples.
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Example 9.7. Representations of a Sphere.

We already saw three different representations of the sphere with radius \(R\) centred at \(\vect 0\text{:}\)
  • parametric representation by spherical coordinates:
    \begin{equation*} \vect g(\theta,\varphi) :=\bigl(R\cos\varphi\sin\theta,R\sin\varphi\sin\theta,R\cos\theta\bigr), \end{equation*}
    where \((\theta,\varphi)\in D:=[0,\pi]\times[0,2\pi)\text{;}\)
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  • implicit representation by \(x_1^2+x_2^2+x_3^2-R^2=0\text{;}\)
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  • explicit representation by \(x_3=\sqrt{R^2-x_1^2-x_2^2}\) and \(x_3=-\sqrt{R^2-x_1^2-x_2^2}\text{.}\)
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Example 9.8. Graph of a function.

Suppose that \(h\colon D\to\mathbb R\) is continuously differentiable on the open set \(D\subset\mathbb R^2\text{.}\) Then the graph
\begin{equation*} S:=\{(x_1,x_2,h(x_1,x_2))\colon (x_1,x_2)\in D\} \end{equation*}
forms a smooth surface in \(\mathbb R^3\text{.}\) Then the three possible representations are as follows:
  • explicit representation is \(x_3=h(x_1,x_2)\text{,}\) \((x_1,x_2)\in D\text{;}\)
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  • implicit representation is \(f(\vect x):=x_3-h(x_1,x_2)=0\text{;}\)
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  • parametric representation is \(\vect g(x_1,x_2)=(x_1,x_2,h(x_1,x_2))\) for \((x_1,x_2)\in D\text{.}\)
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To check that \(S\) is smooth just note that \(\grad f(\vect x)=(-\grad h(x_1,x_2), 1)\text{,}\) which is always nonzero. Hence by the above theorem the surface is smooth.
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Example 9.9. Implicit representation of a plane.

Most commonly, a plane in \(\mathbb R^3\) is given in implicit form by the equation
\begin{equation*} ax+by+cz+d=0\text{.} \end{equation*}
At least one of \(a,b,c\) is nonzero. If for instance \(a\neq 0\) we get the explicit representation
\begin{equation*} x=-\frac{1}{a}\Bigl(by+cz+d\Bigr)\text{.} \end{equation*}
A parametric representation can be obtained as follows. Suppose that \(\vect v_1\) and \(\vect v_2\) are linearly independent vectors, lying on the plane \(ax+by+cz=0\) (the original plane translated to the origin). If \(\vect a\) is a point on the original plane, then
\begin{equation*} \vect g(s,t):=\vect a+s\vect v_1+t\vect v_2,\qquad (s,t)\in\mathbb R^2 \end{equation*}
is a parametric representation of the plane.
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