Surface integral
Surface integrals of scalar fields

Assume that f is a scalar, vector, or tensor field defined on a surface S.To find an explicit formula for the surface integral of f over S, we need to parameterize S by defining a system of curvilinear coordinates on S, like the latitude and longitude on a sphere. Let such a parameterization be r (s, t), where (s, t) varies in some region T in the plane. Then, the surface integral is given by

where the expression between bars on the right-hand side is the magnitude of the cross product of the partial derivatives of r (s, t), and is known as the surface element (which would, for example, yield a smaller value near the poles of a sphere, where the lines of longitude converge more dramatically, and latitudinal coordinates are more compactly spaced). The surface integral can also be expressed in the equivalent form

where g is the determinant of the first fundamental form of the surface mapping r (s, t).

For example, if we want to find the surface area of the graph of some scalar function, say z = f (x, y), we have

where r = (x, y, z) = (x, y, f (x, y)). So that ∂ r ∂ x = (1, 0, f x (x, y)) {\displaystyle {\partial \mathbf {r} \over \partial x}=(1,0,f_{x}(x,y))}, and ∂ r ∂ y = (0, 1, f y (x, y)) {\displaystyle {\partial \mathbf {r} \over \partial y}=(0,1,f_{y}(x,y))}. So,

which is the standard formula for the area of a surface described this way. One can recognize the vector in the second-last line above as the normal vector to the surface.

Because of the presence of the cross product, the above formulas only work for surfaces embedded in three-dimensional space.

This can be seen as integrating a Riemannian volume form on the parameterized surface, where the metric tensor is given by the first fundamental form of the surface.
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