The Comuptation of Unit Normals

Section 9.4 The Comuptation of Unit Normals

We now want to see how to compute the unit normals depending on how the surface is given.

Subsection 9.4.1 Normals for surfaces defined implicitly

Theorem 9.15. Normal to implicitly given surface.

Suppose that \(S\) is a surface given implicitly by \(g(\vect x)=0\text{,}\) and that \(\grad g(\vect x)\neq\vect 0\) for all \(\vect x\in S\text{.}\) Then

\begin{equation} \vect n(\vect x) =\frac{\grad g(\vect x)}{\|\grad g(\vect x)\|}\tag{9.1} \end{equation}

defines a continuous field of unit normal vectors on \(S\text{.}\)

As an important special case we might look at a surface which is given explicitly as the graph of a function \(h\colon D\to\mathbb R\text{.}\) Then \(x_3-h(x_1,x_2)=0\) is an implicit representation and therefore

\begin{equation} \vect n=\frac{\bigl(-\grad h(\vect x),1\bigr)}{\sqrt{1+\|\grad h(\vect x)\|^2}}\text{.}\tag{9.2} \end{equation}

defines a unit normal vector field to the graph of \(h\text{.}\) The above normal field is pointing upwards, that is, has a positive \(x_3\)-component.

Subsection 9.4.2 Computation of unit normals using parametric representations

Suppose that \(S\) is an orientable curve, and that \(\vect g\colon D\to\mathbb R^3\) is a regular parametrisation of \(S\text{.}\) If we fix \(\vect a=(a_1,a_2)\in D\text{,}\) then the maps \(x_1\to\vect g(x_1,a_2)\) and \(x_2\to\vect g(a_1,x_2)\) define two curves on \(S\) crossing at \(\vect g(\vect a)\text{.}\) The vectors

\begin{equation*} \vect v_i:=\frac{\partial\vect g}{\partial x_i}(\vect a)\text{,} \end{equation*}

\(i=1,2\text{,}\) are tangent vectors to \(S\) at \(\vect g(\vect a)\text{.}\) As \(\vect g\) is a regular parametrisation (see Definition 7.2) we have that \(\vect v_1\times\vect v_2\neq\vect 0\text{.}\) Recall that the vector product of two vectors is perpendicular to the original vectors (see Theorem 1.24). Hence we have the following theorem.

Theorem 9.16. Normal to a parametrised surface.

Suppose that \(S\) is a smooth orientable surface with regular parametrisation \(\vect g\text{.}\) Then

\begin{equation} \vect n(\vect x) =\frac{\dfrac{\partial\vect g}{\partial x_1}(\vect x) \times \dfrac{\partial\vect g}{\partial x_2}(\vect x)} {\Bigl\|\dfrac{\partial\vect g}{\partial x_1}(\vect x) \times \dfrac{\partial\vect g}{\partial x_2}(\vect x)\Bigr\|}\tag{9.3} \end{equation}

defines a continuous field of unit normal vectors on \(S\text{.}\)

Let us write down the components of the above vector product product in components. Using (1.9) and the definition of the Jacobian determinant in Remark 5.18 we can write

\begin{equation} \frac{\partial}{\partial x_1}\vect g(\vect x) \times \frac{\partial}{\partial x_2}\vect g(\vect x) = \begin{bmatrix} \dfrac{\partial(g_2,g_3)}{\partial(x_1,x_2)} \\ \dfrac{\partial(g_3,g_1)}{\partial(x_1,x_2)} \\ \dfrac{\partial(g_1,g_2)}{\partial(x_1,x_2)} \end{bmatrix}\tag{9.4} \end{equation}

and

\begin{equation*} \Bigl\|\frac{\partial\vect g}{\partial x_1}(\vect x) \times \frac{\partial\vect g}{\partial x_2}(\vect x)\Bigr\| = \sqrt{ \Bigl|\dfrac{\partial(g_2,g_3)}{\partial(x_1,x_2)}\Bigr|^2+ \Bigl|\dfrac{\partial(g_3,g_1)}{\partial(x_1,x_2)}\Bigr|^2+ \Bigl|\dfrac{\partial(g_1,g_2)}{\partial(x_1,x_2)}\Bigr|^2}\text{.} \end{equation*}

There is another way to express the norm of the above vector product. We know from Theorem 1.24 that the norm of the vector product is the area of the parallelogram spanned by the two vectors. In Theorem 1.16 we derived another formula for the area of a parallelogram spanned by two vectors. We have to form the matrix with the two vectors as columns. In our case this turns out to be the Jacobian matrix of \(\vect g\text{:}\)

\begin{equation*} J_{\vect g}(\vect x) = \begin{bmatrix} \dfrac{\partial\vect g}{\partial x_1}(\vect x) & \dfrac{\partial\vect g}{\partial x_2}(\vect x) \end{bmatrix}\text{.} \end{equation*}

As a consequence of Theorem 1.16 we can express the norm of the above vector product in terms of the Jacobian of \(\vect g\text{.}\)

Proposition 9.17. Jacobian via the vector product.

If \(D\subset\mathbb R^2\) and \(\vect g\colon D\to\mathbb R^3\) is a differentiable function, then

\begin{equation} \Bigl\|\frac{\partial\vect g}{\partial x_1}(\vect x) \times \frac{\partial\vect g}{\partial x_2}(\vect x)\Bigr\| =\sqrt{\det\bigl(\bigl(J_{\vect g}(\vect x)\bigr)^T J_{\vect g}(\vect x)\bigr)}\text{.}\tag{9.5} \end{equation}

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