Notes on Line and Surface Integrals

This note examines a few properties of line and surface integrals. In particular, it indicates how they can be computed using differential forms.

Classical Definitions

First, let's look at the classical definitions of line and surface integrals. The goal in this note is to show how to reproduce these definitions and various calculations based on these definitions using differential forms, so we will just briefly state the definitions.

Line Integrals

First, we'll examine the classical definitions of line integrals of scalar and vector fields.

Scalar Fields

For a scalar field \(f : U \rightarrow \mathbb{R}\) defined on a subset \(U \subseteq \mathbb{R}^n\), the line integral along a piecewise smooth curve \(C \subseteq U\) with a bijective parameterization \(r : [a, b] \rightarrow C\) on the interval \([a, b] \subseteq \mathbb{R}\) is defined as follows:

\[\int_C f(r)~ds = \int_a^b f(r(t)) ~\lVert r'(t) \rVert ~dt.\]

Vector Fields

For a vector field \(F : U \rightarrow \mathbb{R}^n\) defined on a subset \(U \subseteq \mathbb{R}^n\), the line integral along a piecewise smooth curve \(C \subseteq U\) with bijective parameterization \(r : [a, b] \rightarrow C\) on the interval \([a, b] \subseteq \mathbb{R}\) is defined as follows:

\[\int_C F(r) \cdot ~dr = \int_a^b F(r(t)) \cdot r'(t) ~dt.\]

Surface Integrals

Next, we'll examine the classical definitions of surface integrals of scalar and vector fields.

Scalar Fields

For a scalar field \(f : U \rightarrow \mathbb{R}\) defined on a subset \(U \subseteq \mathbb{R}^n\), the surface integral over a surface \(S \subseteq U\) with parameterization \(r : T \rightarrow S\) over a planar region \(T\) is defined as follows:

\[\iint_S f dS = \iint_T f(r(s, t)) ~\left\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t} \right\rVert\ ~ds~dt.\]

Vector Fields

For a vector field \(F : U \rightarrow \mathbb{R}^n\) defined on a subset \(u \subseteq \mathbb{R}^n\), the surface integral over a surface \(S \subseteq U\) with parameterization \(r : T \rightarrow S\) over a planar region \(T\) relative to to the surface normal \(\mathbf{n}\) is defined as follows:

\[\iint_S F \cdot ~d\mathbf{s} = \iint_S (F \cdot \mathbf{n}) ~ds = \iint_T F(r(s, t)) \cdot \left(\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t} \right) ~ds~dt.\]

Note that this computes the normal component of the vector field, also called the flux through the surface.

Pullbacks of Differential Forms

Next, let's determine how to express line and surface integrals using pullbacks of differential forms.

Line Integrals

For a smooth curve \(\gamma : [a, b] \rightarrow \mathbb{R}^n\) and smooth differential 1-form (covector field) \(\omega\) on \(\mathbb{R}^n\), the line integral of \(\omega\) over \(\gamma\) is defined as follows:

\[\int_{\gamma}\omega = \int_{[a,b]} \gamma^*\omega = \int_a^b (\gamma^*\omega)_t ~dt.\]

Thus, line integrals can be computed by pulling back covector fields along smooth curves. In this notation, the line integral of a vector field would be expressed as follows:

\[\int_a^b \omega_{\gamma(t)}\left(\gamma'(t)\right)~dt.\]

Note that, according to the musical isomorphism (tangent-cotangent bundle isomorphism), the covector field \(\omega\) corresponds to a unique vector field \(\omega^{\sharp}\) such that

\[\omega_{\gamma(t)}\left(\gamma'(t)\right) = \langle \omega^{\sharp}_{\gamma(t)}, \gamma'(t) \rangle = \omega^{\sharp}_{\gamma(t)} \cdot \gamma'(t).\]

Writing \(F(r(t))\) for \(\omega^{\sharp}_{\gamma(t)}\) and \(r'(t)\) for \(\gamma'(t)\), this reproduces the classical equation for the line integral of a vector field:

\[\int_a^b F(r(t)) \cdot r'(t) ~dt.\]

Thus, we just need to establish the following equation:

\[\int_{[a,b]}\gamma^*\omega = \int_a^b \omega_{\gamma(t)}\left(\gamma'(t)\right)~dt.\]

Let's write the coordinate representations of \(\gamma\) as \(\gamma^i(t)\) and the coordinate representations of \(\omega\) as \(\omega_i dx^i\). We then compute:

\begin{align}\omega_{\gamma(t)}\left(\gamma'(t)\right) &= (\omega_i ~dx^i)_{\gamma(t)}\left(\gamma'(t)\right) \\&= \omega_i\left(\gamma(t)\right)~dx^i_{\gamma(t)}\left(\gamma'(t)\right) \\&=\omega_i\left(\gamma(t)\right)\left(\dot{\gamma}^i(t)\right) \\&= (\omega_i \circ \gamma)(t)d(\gamma^i)_t \\&= (\gamma^*\omega)_t.\end{align}

Thus, we have established that

\[\int_{[a,b]}\gamma^*\omega = \int_a^b (\gamma^*\omega)_t ~dt = \int_a^b \omega_{\gamma(t)}\left(\gamma'(t)\right)~dt.\]

Thus, the pullback of a covector field (differential 1-form) along a smooth curve computes a line integral (which corresponds to the line integral of a vector field).

Surface Integrals

Each differential 2-form on \(\mathbb{R}^3\) has the form

\[f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy.\]

Given a surface \(S\) in \(\mathbb{R}^3\), a differential 2-form \(\omega\) on \(\mathbb{R}^3\), and a parameterization \(F : D \rightarrow S\) defined on a domain of integration \(D \subseteq \mathbb{R}^2\), the surface integral of \(\omega\) over \(S\) is defined as follows:

\[\int_S \omega = \int_D F^*\omega.\]

Thus, line and surface integrals are defined in essentially the same manner: via pullbacks of differential forms. The unifying nature of differential forms makes them especially useful for both theoretical and computational purposes.

Consider a parameterization \(F(s, t) = (x(s, t), y(s, t), z(s, t))\) of a surface \(S\).

The formula for a covector field in coordinates indicates the following:

\[dx = \frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\]

\[dy = \frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\]

\[dz = \frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\]

Then, under pullback:

\begin{align}F^*(dx \wedge dy) &= \left(\frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\right) \wedge \left(\frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\right)\\&= \left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(x,y)}{\partial(s,t)}ds \wedge dt\end{align}

\begin{align}F^*(dy \wedge dz) &= \left(\frac{\partial y}{\partial s}ds + \frac{\partial y}{\partial t}dt\right) \wedge \left(\frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\right)\\&= \left(\frac{\partial y}{\partial s} \frac{\partial z}{\partial t} - \frac{\partial y}{\partial t} \frac{\partial z}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(y,z)}{\partial(s,t)}ds \wedge dt\end{align}

\begin{align}F^*(dz \wedge dx) &= \left(\frac{\partial z}{\partial s}ds + \frac{\partial z}{\partial t}dt\right) \wedge \left(\frac{\partial x}{\partial s}ds + \frac{\partial x}{\partial t}dt\right)\\&= \left(\frac{\partial z}{\partial s} \frac{\partial x}{\partial t} - \frac{\partial z}{\partial t} \frac{\partial x}{\partial s}\right)ds \wedge dt \\&= \frac{\partial(z,x)}{\partial(s,t)}ds \wedge dt\end{align}

Thus, the differential form

\[f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy\]

transforms as

\begin{align}F^*(f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy) &= F^*(f_x ~ dy \wedge dz) \\&+ F^*(f_y ~ dz \wedge dx) \\&+ F^*(f_z ~ dx \wedge dy) \\&= (f_x \circ F)(\frac{\partial(y,z)}{\partial(s,t)}ds \wedge dt) \\&+ (f_y \circ F)(\frac{\partial(z,x)}{\partial(s,t)}ds \wedge dt) \\&+ (f_z \circ F)(\frac{\partial(x,y)}{\partial(s,t)}ds \wedge dt).\end{align}

This is a general pattern of differential forms: the pullback has a coefficient of the determinant of the respective Jacobian matrix.

The derivatives of the parameterization are as follows:

\[\frac{\partial F}{\partial s} = \left(\frac{\partial x}{\partial s}, \frac{\partial y}{\partial s}, \frac{\partial z}{\partial s} \right)\]

\[\frac{\partial F}{\partial t} = \left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}, \frac{\partial z}{\partial t} \right)\]

Then

\begin{align}\frac{\partial F}{\partial s} \times \frac{\partial F}{\partial t} &= \left(\left(\frac{\partial y}{\partial s} \frac{\partial z}{\partial t} - \frac{\partial y}{\partial t} \frac{\partial z}{\partial s}\right), \left(\frac{\partial z}{\partial s} \frac{\partial x}{\partial t} - \frac{\partial z}{\partial t} \frac{\partial x}{\partial s}\right), \left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t} - \frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right)\right) \\&= \left(\frac{\partial(y,z)}{\partial(s,t)}, \frac{\partial(z,x)}{\partial(s,t)}, \frac{\partial(x,y)}{\partial(s,t)}\right).\end{align}

Thus

\begin{align}\int_D F^*(f_x ~ dy \wedge dz + f_y ~ dz \wedge dx + f_z ~ dx \wedge dy) &= \int_D f(F(s,t)) \cdot \left(\frac{\partial(y,z)}{\partial(s,t)}, \frac{\partial(z,x)}{\partial(s,t)}, \frac{\partial(x,y)}{\partial(s,t)}\right) ~ds ~dt.\end{align}

Replacing \(F\) with \(r\) and \(f\) with \(F\) and \(D\) with \(T\) and using the double integral symbol, this reproduces the classical expression for a surface integral:

\[\iint_T F(r(s, t)) \cdot \left(\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t} \right) ~ds~dt.\]

Pullbacks and Scalar Fields

We previously indicated how pullbacks of certain differential forms correspond to line and surface integrals of vector fields. We now indicate how similar pullbacks corresponds to line and surface integrals of scalar fields.

Line Integrals

Given any scalar field (i.e. function) \(f : C \rightarrow \mathbb{R}\) on a curve \(C\) with parameterization \(r : [a,b] \rightarrow C\), we can define a corresponding vector field as follows:

\[F(p) = f(p)\frac{r'(r^{-1}(p))}{\lVert r'(r^{-1}(p)) \rVert}.\]

That is, we simply multiply the scalar by the normalized derivative evaluated at the parameter \(t \in [a,b]\) corresponding to the point \(p\). Then, when we compute the corresponding integral of this vector field, we obtain

\begin{align}\int_C F(r) \cdot ~dr &= \int_a^b F(r(t)) \cdot r'(t) ~dt \\&= \int_a^b f(r(t))\frac{r'(t)}{\lVert r'(t) \rVert} \cdot r'(t) ~dt \\&= \int_a^b f(r(t))\frac{r'(t)}{\lVert r'(t) \rVert} \cdot \frac{r'(t)}{\lVert r'(t) \rVert} \lVert r'(t) \rVert ~dt \\&= \int_a^b f(r(t)) \lVert r'(t) \rVert ~dt .\end{align}

The ultimate step uses the fact that the dot product of any unit vector with itself has value \(1\). Thus, we can reproduce the integral of a scalar field in this manner. In general, using any unit length tangent vector will suffice.

In particular, if \(f=1\), i.e. the constant function \(1\) (which is equivalent to omitting \(f\)), we obtain the integral for the length of the curve:

\[\int_a^b \lVert r'(t) \rVert ~dt .\]

Surface Integrals

Given any scalar field \(f : S \rightarrow \mathbb{R}\) on a surface \(S\) with parameterization \(r : D \rightarrow S\) over a domain of integration \(D\), we can define a corresponding vector field as follows:

\[F(p) = f(p) \frac{\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}}{\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\rVert}.\]

We then compute

\begin{align}\iint_S F \cdot ~d\mathbf{s} &= \iint_S (F \cdot \mathbf{n}) ~ds \\&= \iint_T f(r(s,t)) \frac{\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}}{\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\rVert}\cdot \left(\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t} \right) ~ds~dt \\&= \iint_T f(r(s,t)) \frac{\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}}{\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\rVert}\cdot \frac{\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}}{\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\rVert} \left\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right\rVert~ds~dt \\&= \iint_T f(r(s,t)) \left\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right\rVert~ds~dt .\end{align}

The final step uses the fact that the dot product of any two unit vectors has value \(1\). Thus, surface integrals of vector fields are strictly more general than surface integrals of scalar fields. The key was to use a unit length normal vector (oriented in the right direction).

In particular, if \(f=1\), i.e. the constant function \(1\) (which is equivalent to omitting \(f\)), we obtain the integral that computes the area of the surface:

\[\iint_T \left\lVert \frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right\rVert~ds~dt .\]

Examples

Now we will consider a few examples.

The Circle

The circle with radius \(r \in \mathbb{R}\) in \(\mathbb{R}^2\) can be defined as an implicit surface:

\[S^1 = \{(x, y) \mid x^2 + y^2 = r\}.\]

We will use the parametric curve \(\gamma : [0, 2\pi] \rightarrow S^1\) defined using polar coordinates:

\[\gamma(t) = (r\cos t, r\sin t).\]

To compute the length of the circle, we need to construct a differential 1-form in \(\mathbb{R}^2\) whose components are the components of a unit-length vector tangent to the circle. Given a point \((x,y)\) on the circle, the unit length tangent vector at this point is

\[\frac{1}{\sqrt{x^2 + y^2}}(-y, x).\]

The denominator can be simplified to \(r\).

Thus, we define the following differential form:

\[\omega = \frac{-y~dx + x~dy}{\sqrt{x^2 + y^2}}.\]

We compute the following by substituting the coordinate functions of \(\gamma\) for the component functions of \(\omega\):

\begin{align}\int_{\gamma}\omega &= \int_{[0,2\pi]}\gamma^*\omega \\&= \int_0^{2\pi} (\gamma^*\omega)_t \\&= \int_0^{2\pi} \frac{-r\sin t ~d(r\cos t) + r\cos t ~d(r\sin t)}{\sqrt{r^2\cos^2 t + r^2\sin^2 t}} \\&= \int_0^{2\pi} \frac{(-r\sin t)(-r\sin t) ~dt + (r\cos t) (r\cos t) ~dt}{r} \\&= \int_0^{2\pi} r ~dt \\&= 2\pi r.\end{align}

This differential form can be modified to compute the line integral of a scalar field \(f\):

\[\omega = f\frac{-y~dx + x~dy}{\sqrt{x^2 + y^2}}.\]

The pullback computes \(\gamma^* (f\omega) = (f \circ \gamma)\gamma^*(\omega)\), so the corresponding integral is

\[\int_0^{2\pi} f(\gamma(t)) r~ dt.\]

The Sphere

The sphere with radius \(r\) in \(\mathbb{R}^3\) can be defined as an implicit surface:

\[S^2 = \{(x, y, z) \mid x^2 + y^2 + z^2 = r\}.\]

We use the parametric curve \(\gamma : [0,\pi] \times [0,2\pi] \rightarrow S^2\) defined using spherical coordinates as follows:

\[\gamma(\varphi, \theta) = (r\sin \varphi \cos \theta, r\sin \varphi \sin \theta, r\cos \varphi).\]

To compute the area of the sphere, we need to construct a differential \(2\)-form in \(\mathbb{R}^3\) whose components at a point are the components of the unit normal vector at the point. Given a point \((x,y,z)\) on the sphere, its unit normal vector is simply the vector \((x/r, y/r, z/r)\).

We thus define the following differential form:

\[\omega = \frac{x}{r}dy \wedge dz + \frac{y}{r} dz \wedge dx + \frac{z}{r} dx \wedge dy.\]

We first compute the following by substitution:

• \(\gamma^*dx = d(r\sin \varphi \cos \theta) = r\cos \varphi \cos \theta ~d\varphi - r\sin \varphi \sin \theta ~d\theta,\)
• \(\gamma^*dy = d(r\sin \varphi \sin \theta) = r\cos \varphi \sin \theta ~d\varphi + r\sin \varphi \cos \theta ~d\theta,\)
• \(\gamma^*dz = d(r\cos \varphi) = -r\sin \varphi ~d\varphi.\)

Next, we compute each of the wedge products:

• \(\gamma^*(dy \wedge dz) = \gamma^*(dy) \wedge \gamma^*(dz) = -r^2\sin^2 \varphi \cos \theta ~d\theta \wedge d\varphi,\)
• \(\gamma^*(dz \wedge dx) = \gamma^*(dz) \wedge \gamma^*(dx) = r^2\sin^2 \varphi \sin \theta ~d\varphi \wedge d\theta,\)
• \(\gamma^*(dx \wedge dy) = \gamma^*(dx) \wedge \gamma^*(dy) = r^2\cos \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta - r^2\sin \varphi \cos \varphi \sin^2 \theta ~d\theta \wedge d\varphi.\)

We then normalize the wedge products, using \(d\varphi \wedge d\theta\) consistently, and exploiting the fact that \(d\theta \wedge d\varphi = - d\varphi \wedge d\theta\):

• \(\gamma^*(dy \wedge dz) = r^2\sin^2 \varphi \cos \theta ~d\varphi \wedge d\theta\)
• \(\gamma^*(dz \wedge dx) = \gamma^*(dz) \wedge \gamma^*(dx) = r^2\sin^2 \varphi \sin \theta ~d\varphi \wedge d\theta,\)
• \(\gamma^*(dx \wedge dy) = \gamma^*(dx) \wedge \gamma^*(dy) = r^2\cos \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta + r^2\sin \varphi \cos \varphi \sin^2 \theta ~d\varphi \wedge d\theta.\)

Next, we pullback the components of the differential form:

• \(\gamma^*((x/r) dy \wedge dz) = r\sin^3 \varphi \cos^2 \theta ~d\varphi \wedge d\theta\)
• \(\gamma^*((y/r) dz \wedge dx) = r\sin^3 \varphi \sin^2 \theta ~d\varphi \wedge d\theta,\)
• \(\gamma^*((z/r) dx \wedge dy) = r\cos^2 \varphi \sin \varphi \cos^2 \theta ~d\varphi \wedge d\theta + r\sin \varphi \cos^2 \varphi \sin^2 \theta ~d\varphi \wedge d\theta.\)

Finally, we add and group using trigonometric identities to simplify:

\begin{align}\gamma^*\omega &= (r\sin^3 \varphi \cos^2 \theta + r\sin^3 \varphi \sin^2 \theta + r\cos^2 \varphi \sin \varphi \cos^2 \theta + r\sin \varphi \cos^2 \varphi \sin^2 \theta)~d\varphi \wedge d\theta \\&= r\sin^3 \varphi (\cos^2 \theta + \sin^2 \theta) + r\sin \varphi \cos^2 \theta (\cos^2 \theta + \sin^2 \theta) \\&= r\sin^3 \varphi + r\sin \varphi \cos^2 \theta \\&= r\sin \varphi (\cos^2 \varphi + \sin^2 \varphi) \\&= r \sin \varphi.\end{align}

Thus, we compute the integral as follows:

\begin{align}\int_{\gamma}\omega &= \int_0^{2 \pi}\int_0^{\pi} r \sin \varphi ~d\varphi~d\theta \\&= \int_0^{2 \pi} -r \cos(\pi) - (-r \cos(0))~d\theta \\&= \int_0^{2 \pi} 2r ~d\theta \\&= 4\pi r.\end{align}

If we include a function (scalar field) \(f\) and pull back \(f\omega\) along \(\gamma\), we obtain

\[\int_0^{2 \pi}\int_0^{\pi} f(\gamma(\varphi,\theta)) r \sin \varphi ~d\varphi~d\theta,\]

which is the formula for integration using spherical coordinates.

Other Classical Expressions of Line and Surface Integrals

In this section, we will relate another, classical expression for line and area integrals.

Line Integrals

One classical expression for the line integral is as follows: given a parameterization \(r(t) = (x^1(t),\dots,x^n(t))\) of a curve \(C\) in \(\mathbb{R}^n\) over an interval \([a,b]\), the line integral of a function \(f\) along \(C\) is

\[\int_C f dx^1 \dots dx^n = \int_a^b f(x^1(t),\dots,x^n(t)) \sqrt{\left(\frac{d x^1}{d t}\right)^2 + \dots + \left(\frac{d x^n}{d t}\right)^2 }~dt.\]

However, noting that

\[\lVert r'(t) \rVert = \left\lVert \left(\frac{d x^1}{d t}, \dots, \frac{d x^n}{d t}\right ) \right\rVert,\]

which, using the standard Euclidean norm is equal to

\[\sqrt{\left(\frac{d x^1}{d t}\right)^2 + \dots + \left(\frac{d x^n}{d t}\right)^2 },\]

then we see that these are the same expression.

Surface Integrals

For surface integrals, there is a classical formula for the surface integral of a surface expressed as a graph of a function \(f : \mathbb{R}^2 \rightarrow \mathbb{R}\). The surface is comprised of points of the form \((x,y,f(x,y))\). We can parameterize the surface by \(r(x,y) = (x,y,f(x,y))\). Then, we compute

\begin{align}\left\lVert \frac{\partial r}{\partial x} \times \frac{\partial r}{\partial y} \right\rVert &= \left\lVert \left(1, 0, \frac{\partial f}{\partial x}\right) \times \left(0, 1, \frac{\partial f}{\partial y}\right) \right\rVert \\&= \left\lVert \left(-\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1\right)\right\rVert \\&= \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 +\left(\frac{\partial f}{\partial y}\right)^2 + 1}.\end{align}

This provides an alternative expression for a surface integral:

\[\iint_S f dS = \iint_T f(x, y) \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 +\left(\frac{\partial f}{\partial y}\right)^2 + 1} ~dx ~dy. \]

Riemannian Volume Forms

In addition to the classical formulas and formulas involving pullbacks of differential forms, since \(\mathbb{R}^n\) is a Riemannian manifold, we can also express line and surface integrals using the Riemannian volume form.

Given an injective smooth map \(F : M \rightarrow N\) and a Riemannian metric \(g\) (also written \(\langle -,-\rangle\)) on \(N\), the (point-wise) pullback metric \(F^*g\) is defined as follows for any \(v,w \in T_pM\):

\[\left(F^*g\right)_p(v,w) = g_p\left(dF_p(v), dF_p(w)\right).\]

The map must be injective, otherwise, \(dF_p(v) = 0\) for some \(v \neq 0\), and thus \((F^*g)_p(v, v) = 0\), violating the positivity requirement.

This can be applied to vector fields \(X,Y\) on \(M\) to yield a function whose value at any point \(p \in M\) is the respective point-wise pullback:

\[F^*g(X,Y) = g\left(dF(X), dF(Y)\right),\]

\[\left(F^*g(X,Y)\right)_p = g_p\left(dF_p(X_p), dF_p(Y_p)\right).\]

Given smooth coordinate chart \((U,(x^i))\) on \(M\), we can compute the representation of \(F^*g\) in these coordinates:

\[(F^*g)_{ij} = \left\langle dF\left(\frac{\partial}{\partial x^i}\right), dF\left(\frac{\partial}{\partial x^j}\right) \right\rangle.\]

This is an \(n \times n\) (where \(n\) is the dimension of \(M\)) matrix of functions, which, when evaluated at a point \(p\), yields

\[((F^*g)_{ij})_p = \left\langle dF_p\left(\frac{\partial}{\partial x^i}\bigg\rvert_p\right), dF_p\left(\frac{\partial}{\partial x^j}\bigg\rvert_p\right) \right\rangle,\]

or, equivalently, a matrix-valued function.

Line Integrals

Given a parameterization \(r(t) = (r^1(t),\dots,r^n(t))\) of a curve \(C\) in \(\mathbb{R}^n\) on an interval \([a,b]\), the representation of the pullback metric tensor \(r^*g\) is

\[(r^*g)_{ij} = \left\langle dr\left(\frac{\partial}{dx^i}\right), dr\left(\frac{\partial}{dx^j}\right) \right\rangle.\]

However, \(i = j = 1\) in this case, and we use the label \(t\) for \(x^1\) so this is simply a \(1 \times 1\) matrix-valued function, i.e. a scalar-valued function:

\[(r^*g) = \left\langle dr\left(\frac{d }{dt}\right), dr\left(\frac{d}{dt}\right) \right\rangle.\]

The standard Euclidean metric on \(\mathbb{R}^n\) is defined as follows:

\[g = (dx^1)^2 + \dots + (dx^n)^2.\]

This metric yields the dot product of vector fields \(\langle X, Y \rangle\), i.e. \(\langle X, Y \rangle_p = \langle X_p,Y_p \rangle\). Thus, we compute the representation of \(r^*g\) in coordinates as follows:

\begin{align}(r^*g) &= \left((dx^1)^2 + \dots + (dx^n)^2\right)\left(dr\left(\frac{d}{dt}\right),dr\left(\frac{d}{dt}\right)\right)\\&= dx^1\left(dr\left(\frac{d}{dt}\right)\right) dx^1\left(dr\left(\frac{d}{dt}\right)\right) + \dots + dx^n\left(dr\left(\frac{d}{dt}\right)\right) dx^n\left(dr\left(\frac{d}{dt}\right)\right) \\&= \frac{dr^1}{dt}\frac{dr^1}{dt} + \dots + \frac{dr^n}{dt}\frac{dr^n}{dt} \\&= \left\langle \frac{dr}{dt}, \frac{dr}{dt} \right\rangle \\&= \left(r'(t)\right)^2.\end{align}

The volume form is

\[\sqrt{\mathrm{det}(g)}dt = \lVert r'(t)\rVert.\]

Thus, this reproduces the same result. Thus, we can also compute

\[\int_C f \sqrt{\mathrm{det(g)}}dt.\]

Surface Integrals

Likewise, given a parameterization \(r(s,t) = (r_x(s,t), r_y(s,t), r_z(s,t))\) of a surface, the representation of the pullback metric tensor \(r^*g\) in these coordinates is

\[(r^*g)_{ij} = \left\langle dr\left(\frac{\partial}{dx^i}\right), dr\left(\frac{\partial}{dx^j}\right) \right\rangle.\]

In this case, \(i = j = 2\) and we use the labels \(s\) and \(t\) for \(x^1\) and \(x^2\), respectively, which yields

\[g = \begin{bmatrix} \left\langle \frac{\partial r}{\partial s}, \frac{\partial r}{\partial s} \right\rangle & \left\langle \frac{\partial r}{\partial s}, \frac{\partial r}{\partial t} \right\rangle \\ \left\langle \frac{\partial r}{\partial s}, \frac{\partial r}{\partial t} \right\rangle & \left\langle \frac{\partial r}{\partial t}, \frac{\partial r}{\partial t} \right\rangle \end{bmatrix}\]

Then, \(\mathrm{det}(g)\) evaluates to

\[\left(\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right) \cdot \left(\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right) = \left\lVert\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right\rVert^2,\]

and so \(\sqrt{\mathrm{det}(g)}\) is

\[\left\lVert\frac{\partial r}{\partial s} \times \frac{\partial r}{\partial t}\right\rVert.\]

Thus, this again reproduces the classical surface integral.

Summary

There are thus several equivalent methods for calculating line and surface integrals:

• The classical expressions
• Pullbacks of appropriate differential forms
• Riemannian volume forms induced by pullback metrics