Editing First fundamental form (section)

==Calculating lengths and areas==

The first fundamental form completely describes the metric properties of a surface.  Thus, it enables one to calculate the lengths of curves on the surface and the areas of regions on the surface.  The [[line element]] {{math|''ds''}} may be expressed in terms of the coefficients of the first fundamental form as
<math display="block">ds^2 = E\,du^2+2F\,du\,dv+G\,dv^2 \,.</math> 

The classical area element given by {{math|1=''dA'' = {{abs|''X<sub>u</sub>'' × ''X<sub>v</sub>''}} ''du'' ''dv''}} can be expressed in terms of the first fundamental form with the assistance of [[Lagrange's identity]],
<math display="block">dA = |X_u \times X_v| \ du\, dv= \sqrt{ \langle X_u,X_u \rangle \langle X_v,X_v \rangle - \left\langle X_u,X_v \right\rangle^2 } \, du\, dv = \sqrt{EG-F^2} \, du\, dv.</math>

===Example: curve on a sphere===
A [[spherical curve]] on the [[unit sphere]] in {{math|'''R'''<sup>3</sup>}} may be parametrized as
<math display="block">X(u,v) = \begin{bmatrix} \cos u \sin v \\ \sin u \sin v \\ \cos v \end{bmatrix},\ (u,v) \in [0,2\pi) \times [0,\pi].</math>
Differentiating {{math|''X''(''u'',''v'')}} with respect to {{mvar|u}} and {{mvar|v}} yields
<math display="block">\begin{align}
X_u &= \begin{bmatrix} -\sin u \sin v \\ \cos u \sin v \\ 0 \end{bmatrix},\\[5pt]
X_v &= \begin{bmatrix} \cos u \cos v \\ \sin u \cos v \\ -\sin v \end{bmatrix}.
\end{align}</math>
The coefficients of the first fundamental form may be found by taking the dot product of the [[partial derivatives]].

<math display="block">\begin{align}
E &= X_u \cdot X_u = \sin^2 v \\
F &= X_u \cdot X_v = 0 \\
G &= X_v \cdot X_v = 1
\end{align}</math>
so:
<math display="block"> \begin{bmatrix}E & F \\F & G\end{bmatrix} =\begin{bmatrix} \sin^2 v & 0 \\0 & 1\end{bmatrix}.</math>

====Length of a curve on the sphere====

The [[equator]] of the unit sphere is a parametrized curve given by
<math display="block">(u(t),v(t))=(t,\tfrac{\pi}{2})</math>
with {{mvar|t}} ranging from 0 to 2{{pi}}.  The line element may be used to calculate the length of this curve.

<math display="block">\int_0^{2\pi} \sqrt{ E\left(\frac{du}{dt}\right)^2 + 2F \frac{du}{dt} \frac{dv}{dt} + G\left(\frac{dv}{dt}\right)^2 } \,dt = \int_0^{2\pi} \left|\sin v\right| \, dt = 2\pi \sin \tfrac{\pi}{2} = 2\pi</math>

====Area of a region on the sphere====

The area element may be used to calculate the area of the unit sphere.

<math display="block">\int_0^\pi \int_0^{2\pi} \sqrt{ EG-F^2 } \ du\, dv = \int_0^\pi \int_0^{2\pi} \sin v \, du\, dv = 2\pi \Big[ {-\cos v} \Big]_0^{\pi} = 4\pi</math>