Vector Calculus An Introduction

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

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.

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

\(i=1,2\text{,}\) 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.

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

and

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 \(\vect g\text{:}\)

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