## 9.4 Writing a Surface Integral as an Iterated Integral

The task here is similar to that in the last section, but now the vector w sought is normal to the surface rather than tangent to a line. Here there is a very easy way to find w from an equation, and a slightly more complicated way to find it from a parametric representation of the surface.
Suppose first we want to write our integral as an integral over the variables x and y; this happens when x and y are the parameters of the surface which is defined by z - g(x, y) = 0. It is useful sometimes when the surface is defined by an equation, say, h(x, y, z) = 0, rather than by a parametrization.
In this case the vector w must be normal to the surface and its z component should have magnitude 1.
(This is obviously true when the surface is parallel to the x - y plane; as one tilts the surface, its area relative to dxdy increases precisely as the length of this vector does.) If we seek the normal that points "upward" we choose the z component to be 1, otherwise we make it -1.
We can find the upward pointing normal vector w by taking the gradient of the  defining equation:

w =(z - g(x, y));

in the more general case, h(x, y, z) = 0,  we get

If we parametrize the surface by parameters u and v and write the surface integral as an integral dudv, then the z component of w (oriented upward) should be the Jacobian of the two dimensional transformation from x, y to u, v, namely .

Proof

Similarly (as follows by symmetry among the coordinates), the magnitude of the x and y components of w are corresponding Jacobians. To get the relative sign of the x and y coefficients you must cyclically permute x y and z in forming the objects above and use the relative signs obtained among them. The overall sign of w depends on which normal you want (upward or downward).

Examples