In this article, we will examine the nature and utility of the first fundamental form, a quadratic form on a surface. Given a curve parametrized by \(u=u(t)\) and \(v=v(t)\) which lies on a surface \(X\) parametrized by \(X=X(u,v)\) we have that
$$ ds = |\frac{dX}{dt}| dt = |X_u \frac{du}{dt} + X_v \frac{dv}{dt}| dt = \sqrt{(X_u \dot{u} + X_v \dot{v}) \cdot (X_u \dot{u} + X_v \dot{v})} $$
Now defining the coefficients, called the First Fundamental Form coefficients E, F and G, we have
$$ E=X_u \cdot X_u, F=X_u \cdot X_v, G=X_v \cdot X_v $$
and moreover,
$$ \sqrt{(X_u \dot{u} + X_v \dot{v}) \cdot (X_u \dot{u} + X_v \dot{v})} = \sqrt{Edu^2 + 2F du dv + Gdv^2} $$
by direct computation. Now the First Fundamental Form, denoted I, is defined as
$$ I=ds^2=dX \cdot dX = Edu^2 + 2F du dv + Gdv^2 $$
This quadratic form allows us to derive the curvature and metric properties of any surface, namely arc length and area. We see that
$$ (X_{u} \times X_{v})^2=(X_{u} \times X_{v}) \cdot (X_{u} \times X_{v}) = (X_{u} \cdot X_{u})(X_{v} \cdot X_{v})-(X_{u} \cdot X_{v})^2=EG - F^2 \geq 0 $$
and because \(E>0\) we have that I is positive definite, i.e. \(I \geq 0\) and \(I=0 \iff du=dv=0\) It is worth nothing that I also has the symmetric matrix representation
$$ I=x^T \begin{bmatrix} E & F \\ F & G \end{bmatrix} y $$
where \(x,y\) are vectors.
Example 1: Arc Length of a Line on a Paraboloid
Let \(u(t)=t\) and \(v(t)=t\) be the equation of a line which lies on the paraboloid parametrized by
$$ X(u,v) = (u,v,u^2+v^2) $$
and let \(0 \geq t \geq 1\) and \(0 \geq u,v \geq 1\)
We will use the first fundamental form to evaluate this arc length. First we see that
$$ E=X_u \cdot X_u = 1+4u^2 , F=X_u \cdot X_v = 4uv , G=X_v \cdot X_v = 1+4v^2 $$
substituting \(u(t)\) and \(v(t)\) in, we have that $$ E= 1+4t^2 , F= 4t^2 , G= 1+4t^2 $$ Now, we have our equation
$$ ds=\sqrt{Edu^2 + 2F du dv + Gdv^2} dt=\sqrt{E \dot{u}^2 + 2F \dot{u} \dot{v} + G \dot{v}^2} dt = \sqrt{2+16t^2} dt= 4 \sqrt{\frac{1}{8} + t^2} dt $$ Finally, we have that our arc length is given by $$ \ell = \int_{0}^{1} ds = 4 \int_{0}^{1} \sqrt{\frac{1}{8} + t^2} dt $$ This is quite a difficult integral, which is usually the case when computing arc length, but we do obtain an analytical antiderivative, so our definite integral has the solution $$ \ell = 4 (\frac{t}{8} \sqrt{16t^2} + \frac{1}{16} ln (\sqrt{8t^2+1}+2\sqrt{2}t) |_{0}^{1}) = \frac{1}{2} (3\sqrt{2}+\frac{1}{2} ln(2+2 \sqrt{2})) \approx 2.562 $$
Example 2: Area of a Region on a Surface
Here, we can utilize the First Fundamental Form to compute a given region on a surface. Note that for the area of a small parallelogram with verticies \(X(u_0,v_0) , X(u_0, \delta v + v_0), X(\delta u + u_0, v_0), X(\delta u + u_0, \delta v + v_0)\) we can compute the area of this parallelogram using the cross product: $$ \delta A = |X_u \delta u \times X_v \delta v| = \sqrt{EG - F^2} \delta u \delta v \implies dA = \sqrt{EG - F^2} du dv $$ Now let us look at an example using the same paraboloid parametrization we used in Example one, but let \(-1 \leq u,v \leq 1\) and let \(u^2+v^2=1\) be a circle on the paraboloid. Our first fundamental form coefficients are the same, so we have that $$ dA=\sqrt{EG - F^2} du dv= \sqrt{(1+4u^2)(1+4v^2)-(4uv)^2} du dv =\sqrt{1+4u^2+4v^2} du dv $$ Thus, letting \(D=\{(u_0,v_0) : u_0^2+v_0^2 = 1\}\) we have that $$ A_{S} = \int_{D} dA = \int_{D} \sqrt{1+4u^2+4v^2} du dv $$ Performing a change to polar coordinates with \(u=r\cos\theta\) and \(v=r\sin\theta\) and integrating the full unit circle with radius 1, we have that $$ A_{S}=\int_{0}^{2\pi} \int_{0}^{1} \sqrt{1+4(r\cos\theta)^2+4(r\sin\theta)^2} r dr d\theta = \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1+4r^2} r dr d\theta $$ Now using substitution with \(p=1+4r^2\) we have that \(\frac{dp}{8}=r dr\) and thus $$ A_{S}=\int_{0}^{2\pi} \int_{0}^{1} \sqrt{1+4r^2} r dr d\theta = \int_{0}^{2\pi} \int_{1}^{5} \frac{\sqrt{p}}{8} dp d\theta = \int_{0}^{2\pi} \frac{x^{3/2}}{12}|_{1}^{5} d\theta = \int_{0}^{2\pi} \frac{5^{3/2}-1}{12} d\theta = \frac{\pi (5^{3/2}-1)}{6} \approx 5.3304 $$