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Section 3.4 Line and Surface Integrals

To integrate over a curve or a surface, we need a precise way to describe the domain of integration and its orientation. Chains provide that language. They are built from parameterized cubes, and the boundary operator keeps track of endpoints, edges, and faces with the correct signs.

Subsection 3.4.1 Chains

Definition 3.4.1. Singular Cubes.

Let \(A \subseteq \R^m\text{.}\) A singular \(n\)-cube in \(A\) is a continuous map
\begin{equation*} c \colon [0,1]^n \to A. \end{equation*}
A singular \(0\)-cube amounts to a point of \(A\text{.}\) A singular \(1\)-cube is a parameterized curve, and a singular \(2\)-cube is a parameterized surface.
The standard \(n\)-cube in \(\R^n\) is the identity map \(\iota_n \colon [0,1]^n \to \R^n\text{.}\)

Definition 3.4.2. Chains.

An \(n\)-chain in \(A\) is a finite formal integer linear combination of singular \(n\)-cubes:
\begin{equation*} \Gamma = a_1c_1+\cdots+a_Nc_N, \qquad a_1,\ldots,a_N \in \Z. \end{equation*}
We add and subtract chains by combining coefficients of the same cubes. For example,
\begin{equation*} 3c_1-2c_2+7c_3 \end{equation*}
is a \(n\)-chain.
The coefficients record orientation and multiplicity. A negative coefficient means the cube is taken with the opposite orientation, while a coefficient of magnitude greater than \(1\) means the same cube is counted several times.

Definition 3.4.3. Boundary of a Singular Cube.

Let \(c \colon [0,1]^n \to A\) be a singular \(n\)-cube. For each \(1 \le i \le n\) and \(\epsilon \in \{0,1\}\text{,}\) define the face map
\begin{equation*} c_{i,\epsilon}(t_1,\ldots,t_{n-1}) := c(t_1,\ldots,t_{i-1},\epsilon,t_i,\ldots,t_{n-1}). \end{equation*}
Then each \(c_{i,\epsilon}\) is a singular \((n-1)\)-cube, and the boundary of \(c\) is the \((n-1)\)-chain
\begin{equation*} \partial c := \sum_{i=1}^n (-1)^i(c_{i,0}-c_{i,1}). \end{equation*}
We extend the boundary operator to chains by linearity:
\begin{equation*} \partial(a_1c_1+\cdots+a_Nc_N) = a_1\partial c_1+\cdots+a_N\partial c_N. \end{equation*}
The signs in the boundary formula encode orientation. With this choice, the boundary of an interval is terminal point minus initial point, and opposite faces of adjacent cubes appear with opposite signs, so shared faces cancel when cubes are added together.

Example 3.4.4.

Compute the boundary of a singular \(1\)-cube.
Solution.
Let \(c(t)=\langle t,t^2 \rangle\) for \(0 \le t \le 1\text{.}\) Since \(n=1\text{,}\) the boundary formula gives
\begin{equation*} \partial c = (-1)^1(c_{1,0}-c_{1,1}) = c(1)-c(0). \end{equation*}
Therefore
\begin{equation*} \partial c = \langle 1,1 \rangle - \langle 0,0 \rangle. \end{equation*}
So the boundary of a parameterized curve consists of its endpoint minus its starting point.

Example 3.4.5.

Compute the boundary of the standard square \(\iota_2(s,t)=\langle s,t \rangle\text{.}\)
Solution.
Define the four edge paths
\begin{gather*} b(t)=\langle t,0 \rangle,\\ r(t)=\langle 1,t \rangle,\\ u(t)=\langle t,1 \rangle,\\ \ell(t)=\langle 0,t \rangle. \end{gather*}
Then
\begin{gather*} (\iota_2)_{1,0}=\ell,\\ (\iota_2)_{1,1}=r,\\ (\iota_2)_{2,0}=b,\\ (\iota_2)_{2,1}=u. \end{gather*}
Therefore
\begin{equation*} \partial \iota_2 = -\ell+r+b-u = b+r-u-\ell. \end{equation*}
This means the square is traversed counterclockwise: along the bottom from left to right, then up the right edge, then back along the top, and finally down the left edge.
A shaded unit square in the plane. Arrows run right along the bottom edge, up the right.
Figure 3.4.6. The standard square with its oriented boundary \(b+r-u-\ell\text{.}\)

Example 3.4.7.

Compute the boundary of a chain formed by two adjacent squares.
Solution.
Let
\begin{gather*} c_1(s,t)=\langle s,t \rangle,\\ c_2(s,t)=\langle 1+s,t \rangle. \end{gather*}
These parameterize the left and right halves of the rectangle \([0,2]\times[0,1]\text{.}\) If \(\Gamma=c_1+c_2\text{,}\) then \(\partial \Gamma=\partial c_1+\partial c_2\text{.}\) The common vertical edge appears once with positive orientation and once with negative orientation, so it cancels. The result is the boundary of the outer rectangle.
Two shaded adjacent unit squares forming a 2-by-1 rectangle..
Figure 3.4.8. The chain \(c_1+c_2\) with outer boundary orientation; the interior edge cancels.

Example 3.4.9.

Verify in an example that the boundary of a boundary is zero.
Solution.
Starting from the standard square,
\begin{equation*} \partial \iota_2 = b+r-u-\ell. \end{equation*}
Taking the boundary again gives
\begin{equation*} \partial(\partial \iota_2) = (\langle 1,0 \rangle-\langle 0,0 \rangle) + (\langle 1,1 \rangle-\langle 1,0 \rangle) - (\langle 0,1 \rangle-\langle 1,1 \rangle) - (\langle 0,1 \rangle-\langle 0,0 \rangle). \end{equation*}
Every point appears twice with opposite signs, so all terms cancel. Hence
\begin{equation*} \partial(\partial \iota_2)=0. \end{equation*}
This algebraic fact is fundamental. Later, when we integrate forms over chains, it will be the geometric reason behind the familiar pattern that the integral over a boundary is related to the derivative of what is being integrated.

Subsection 3.4.2 Integrating Forms over Chains

We now define the integral of a \(p\)-form over a singular \(p\)-cube, and hence over a \(p\)-chain. The key idea is to pull the form back to the standard cube \([0,1]^p\text{,}\) where ordinary multiple integration is already available.

Definition 3.4.11. Integral over the Standard Cube.

Let \(\omega\) be a smooth \(p\)-form on \([0,1]^p\text{.}\) Since \(\Omega^p([0,1]^p)\) is generated by the single form \(dx_1 \wedge \cdots \wedge dx_p\text{,}\) there is a unique smooth function \(f\) such that
\begin{equation*} \omega = f\,dx_1 \wedge \cdots \wedge dx_p. \end{equation*}
We define the integral of \(\omega\) over the standard \(p\)-cube by
\begin{equation*} \int_{[0,1]^p}\omega := \int_{[0,1]^p} f(x_1,\ldots,x_p)\,dx_1\cdots dx_p. \end{equation*}

Definition 3.4.12. Integral of a Form over a Singular Cube.

Let \(A \subseteq \R^m\text{,}\) let \(\omega\) be a smooth \(p\)-form on \(A\text{,}\) and let \(c \colon [0,1]^p \to A\) be a singular \(p\)-cube. The integral of \(\omega\) over \(c\) is defined by
\begin{equation*} \int_c \omega := \int_{[0,1]^p} c^*\omega. \end{equation*}
This definition is natural: the pullback \(c^*\omega\) is a \(p\)-form on the standard cube, so the previous definition applies. The parameterization \(c\) carries both the geometry of the cube and its orientation.
When \(p=0\text{,}\) the cube \([0,1]^0\) consists of a single point. A singular \(0\)-cube is therefore just a point \(c(0) \in A\text{,}\) and a \(0\)-form is an ordinary function. In this case,
\begin{equation*} \int_c \omega = \omega(c(0)). \end{equation*}

Definition 3.4.13. Integral of a Form over a Chain.

Let \(\Gamma = a_1c_1+\cdots+a_Nc_N\) be a \(p\)-chain in \(A\text{,}\) and let \(\omega\) be a smooth \(p\)-form on \(A\text{.}\) We define
\begin{equation*} \int_{\Gamma}\omega := a_1\int_{c_1}\omega+\cdots+a_N\int_{c_N}\omega. \end{equation*}
So the integral over a chain is linear in both the chain and the form. In particular, reversing orientation changes the sign of the integral, and repeated cubes contribute with their multiplicity.
For \(p=1\) this is a line integral, and for \(p=2\) it is a surface integral.

Example 3.4.14.

Compute the integral of a \(0\)-form over a singular \(0\)-cube.
Solution.
Let \(\omega(x,y)=x^2+y\) on \(\R^2\text{,}\) and let \(c\) be the singular \(0\)-cube determined by \(c(0)=\langle 1,2 \rangle\text{.}\) Then
\begin{equation*} \int_c \omega = \omega(c(0)) = \omega(1,2) = 1^2+2 = 3. \end{equation*}

Example 3.4.15.

Compute a line integral by pulling back a \(1\)-form.
Solution.
Let \(\omega = x\,dy+y\,dx\) on \(\R^2\text{,}\) and let \(c(t)=\langle t,t^2 \rangle\) for \(0 \le t \le 1\text{.}\) Then
\begin{gather*} c^*(x)=t,\\ c^*(y)=t^2,\\ c^*(dx)=dt,\\ c^*(dy)=2t\,dt. \end{gather*}
Therefore
\begin{equation*} c^*\omega = t(2t\,dt)+t^2(dt) = 3t^2\,dt. \end{equation*}
Hence
\begin{equation*} \int_c \omega = \int_0^1 3t^2\,dt = 1. \end{equation*}

Example 3.4.16.

Compute the integral of a \(2\)-form over the standard square.
Solution.
Let \(c=\iota_2\) be the standard square in \(\R^2\text{,}\) and let \(\omega = x\,dx \wedge dy\text{.}\) Since \(c=\iota_2\text{,}\) we have \(c^*\omega=\omega\text{.}\) Thus
\begin{equation*} \int_c \omega = \int_{[0,1]^2} x\,dx\,dy = \int_0^1\int_0^1 x\,dx\,dy = \int_0^1 \frac12\,dy = \frac12. \end{equation*}

Subsection 3.4.3 Line Integrals

A singular \(1\)-cube is just a parameterized curve, so the general definition above becomes the familiar notion of a line integral. There are two standard kinds of line integrals: integrals of scalar densities with respect to arclength, and integrals of \(1\)-forms or vector fields along an oriented curve.

Definition 3.4.17. Line Integral with Respect to Arclength.

Let \(\gamma \colon [a,b] \to \R^n\) be a piecewise \(C^1\) curve, and let \(g\) be a continuous scalar field along the image of \(\gamma\text{.}\) The line integral of \(g\) with respect to arclength is
\begin{equation*} \int_{\gamma} g\,ds := \int_a^b g(\gamma(t))\norm{\gamma'(t)}\,dt. \end{equation*}
This definition has several basic applications:
\begin{gather*} \text{arclength of }\gamma = \int_{\gamma} 1\,ds,\\ \text{mass of a wire with linear density }\rho = \int_{\gamma} \rho\,ds. \end{gather*}

Example 3.4.18.

Compute the circumference of the circle of radius \(r\) as the length of a parameterized curve.
Solution.
Parameterize the circle by
\begin{equation*} \gamma(t)=\langle r\cos t,\; r\sin t \rangle, \qquad 0 \le t \le 2\pi. \end{equation*}
A circle centered at the origin in the xy-plane with a radius drawn from the origin to.
Figure 3.4.19. A circle of radius \(r\text{.}\)
Then
\begin{equation*} \gamma'(t)=\langle -r\sin t,\; r\cos t \rangle, \qquad \norm{\gamma'(t)}=r. \end{equation*}
Therefore the length of the circle is
\begin{equation*} \int_{\gamma} 1\,ds = \int_0^{2\pi} \norm{\gamma'(t)}\,dt = \int_0^{2\pi} r\,dt = 2\pi r. \end{equation*}

Example 3.4.20.

Compute the length of the helix
\begin{equation*} \gamma(t)=\left\langle \cos t,\; \sin t,\; \frac{t}{2\pi} \right\rangle, \qquad 0 \le t \le 6\pi, \end{equation*}
which makes three full turns.
A three-dimensional helix of radius 1 winding upward around the z-axis for three complete.
Figure 3.4.21. A helix making three full turns.
Solution.
Differentiating gives
\begin{equation*} \gamma'(t) = \left\langle -\sin t,\; \cos t,\; \frac{1}{2\pi} \right\rangle. \end{equation*}
Hence the speed is constant:
\begin{equation*} \norm{\gamma'(t)} = \sqrt{\sin^2 t+\cos^2 t+\frac{1}{4\pi^2}} = \sqrt{1+\frac{1}{4\pi^2}}. \end{equation*}
Therefore the length of the helix is
\begin{equation*} \int_{\gamma} 1\,ds = \int_0^{6\pi} \sqrt{1+\frac{1}{4\pi^2}}\,dt = 6\pi \sqrt{1+\frac{1}{4\pi^2}} = 3\sqrt{4\pi^2+1}. \end{equation*}
Now let \(\mathbf{F}=\langle F_1,\ldots,F_n \rangle\) be a vector field on an open set \(U \subseteq \R^n\text{.}\) The corresponding \(1\)-form is
\begin{equation*} \omega_{\mathbf{F}} := F_1\,dx_1+\cdots+F_n\,dx_n. \end{equation*}

Definition 3.4.22. Line Integral of a Vector Field.

Let \(\gamma \colon [a,b] \to U\) be a piecewise \(C^1\) curve. The line integral of \(\mathbf{F}\) along \(\gamma\) is the integral of the associated \(1\)-form:
\begin{equation*} \int_{\gamma}\omega_{\mathbf{F}} = \int_a^b \mathbf{F}(\gamma(t))\cdot \gamma'(t)\,dt. \end{equation*}
In the usual physics language this is written \(\int_{\gamma}\mathbf{F}\cdot d\mathbf{r}\text{,}\) and it measures the work done by the force field \(\mathbf{F}\) along the motion described by \(\gamma\text{.}\)

Proof.
Write \(\gamma^*\omega=h(t)\,dt\text{.}\) Then
\begin{equation*} (\gamma \circ \phi)^*\omega = \phi^*(\gamma^*\omega) = h(\phi(s))\phi'(s)\,ds. \end{equation*}
Therefore
\begin{equation*} \int_{\gamma \circ \phi}\omega = \int_{\alpha}^{\beta} h(\phi(s))\phi'(s)\,ds. \end{equation*}
If \(\phi' \gt 0\text{,}\) the substitution \(t=\phi(s)\) gives \(\int_a^b h(t)\,dt\text{.}\) If \(\phi' \lt 0\text{,}\) the same substitution reverses the limits and contributes a minus sign.
So a line integral of a \(1\)-form depends only on the oriented path, not on the particular parameterization. However, it may still depend on which path is chosen between the same endpoints.

Example 3.4.24.

Show that a line integral can depend on the path, not just the endpoints.
Solution.
On the punctured plane \(\R^2 \setminus \{(0,0)\}\text{,}\) consider the \(1\)-form
\begin{equation*} \omega = \frac{-y}{x^2+y^2}\,dx + \frac{x}{x^2+y^2}\,dy. \end{equation*}
In Subsection 3.3.4, we saw that this form is closed but not exact. Let
\begin{gather*} \gamma_+(t)=\langle \cos(\pi t),\sin(\pi t) \rangle,\\ \gamma_-(t)=\langle \cos(\pi t),-\sin(\pi t) \rangle, \end{gather*}
for \(0 \le t \le 1\text{.}\) Both curves start at \(\langle 1,0 \rangle\) and end at \(\langle -1,0 \rangle\text{.}\) On the upper semicircle \(\gamma_+\text{,}\) we have
\begin{equation*} \gamma_+^*\omega = \pi\,dt, \end{equation*}
so
\begin{equation*} \int_{\gamma_+}\omega = \int_0^1 \pi\,dt = \pi. \end{equation*}
On the lower semicircle \(\gamma_-\text{,}\) we get
\begin{equation*} \gamma_-^*\omega = -\pi\,dt, \end{equation*}
hence
\begin{equation*} \int_{\gamma_-}\omega = \int_0^1 (-\pi)\,dt = -\pi. \end{equation*}
The endpoints are the same, but the integrals are different. Thus the integral depends on the path. This happens precisely because \(\omega\) is closed but not exact.

Definition 3.4.25. Conservative Field.

A vector field \(\mathbf{F}\) on an open set \(U \subseteq \R^n\) is conservative if there exists a differentiable scalar field \(f \colon U \to \R\) such that
\begin{equation*} \mathbf{F}=\nabla f. \end{equation*}
The function \(f\) is called a potential function for \(\mathbf{F}\text{.}\)

Proof.
The \(1\)-form corresponding to \(\mathbf{F}\) is \(\omega_{\mathbf{F}}=df\text{.}\) Therefore
\begin{equation*} \int_{\gamma}\mathbf{F}\cdot d\mathbf{r} = \int_{\gamma} df = \int_a^b \gamma^*(df) = \int_a^b d(f \circ \gamma). \end{equation*}
By the one-variable Fundamental Theorem of Calculus,
\begin{equation*} \int_a^b d(f \circ \gamma) = (f \circ \gamma)(b)-(f \circ \gamma)(a) = f(\vb)-f(\va). \end{equation*}
In the plane, if \(\mathbf{F}=\langle P,Q \rangle\) on a star-shaped region and \(Q_x=P_y\text{,}\) then the associated \(1\)-form \(P\,dx+Q\,dy\) is closed. By Poincare’s Lemma from Theorem 3.3.22, it is exact, so \(\mathbf{F}\) is conservative and its line integrals are path independent.

Subsection 3.4.4 Surface Integrals

A singular \(2\)-cube is a parameterized surface. As with curves, there are two standard kinds of integrals: scalar surface integrals, which measure area or mass, and integrals of vector fields through an oriented surface, which measure flux.

Definition 3.4.27. Scalar Surface Integral.

Let \(\sigma \colon [0,1]^2 \to \R^3\) be a \(C^1\) parameterized surface, and let \(g\) be a continuous scalar field on the image of \(\sigma\text{.}\) The surface integral of \(g\) over \(\sigma\) is
\begin{equation*} \int_{\sigma} g\,dS := \int_0^1\int_0^1 g(\sigma(u,v)) \norm{\frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v}}\,du\,dv. \end{equation*}
Two important applications are
\begin{gather*} \text{surface area of }\sigma = \int_{\sigma} 1\,dS,\\ \text{mass of a surface with density }\rho = \int_{\sigma} \rho\,dS. \end{gather*}
In practice, one often parameterizes a surface by a rectangle \([a,b]\times[c,d]\) rather than by \([0,1]^2\text{.}\) The same formula applies, with the limits of integration changed accordingly.

Example 3.4.28.

Compute the surface area of the sphere of radius \(R\text{.}\)
Solution.
Parameterize the sphere by
\begin{equation*} \sigma(\theta,\phi) = \langle R\sin\phi\cos\theta,\; R\sin\phi\sin\theta,\; R\cos\phi \rangle, \end{equation*}
where \(0 \le \theta \le 2\pi\) and \(0 \le \phi \le \pi\text{.}\)
A sphere centered at the origin with an outward-pointing normal arrow drawn from a point.
Figure 3.4.29. A sphere of radius \(R\) with outward normal direction.
Differentiating gives
\begin{align*} \frac{\partial \sigma}{\partial \theta} \amp = \langle -R\sin\phi\sin\theta,\; R\sin\phi\cos\theta,\; 0 \rangle,\\ \frac{\partial \sigma}{\partial \phi} \amp = \langle R\cos\phi\cos\theta,\; R\cos\phi\sin\theta,\; -R\sin\phi \rangle. \end{align*}
A direct computation shows that
\begin{equation*} \norm{ \frac{\partial \sigma}{\partial \theta} \times \frac{\partial \sigma}{\partial \phi}} = R^2\sin\phi. \end{equation*}
Therefore the surface area is
\begin{equation*} \int_{\sigma} 1\,dS = \int_0^{2\pi}\int_0^{\pi} R^2\sin\phi\,d\phi\,d\theta = R^2\left(\int_0^{2\pi} d\theta\right) \left(\int_0^{\pi}\sin\phi\,d\phi\right) = 4\pi R^2. \end{equation*}
If \(\mathbf{F}=\langle P,Q,R \rangle\) is a vector field on \(\R^3\text{,}\) the corresponding \(2\)-form is
\begin{equation*} \eta_{\mathbf{F}} := P\,dy \wedge dz + Q\,dz \wedge dx + R\,dx \wedge dy. \end{equation*}

Definition 3.4.30. Flux Integral.

Let \(\sigma \colon [0,1]^2 \to \R^3\) be a \(C^1\) parameterized surface. The integral of \(\eta_{\mathbf{F}}\) over \(\sigma\) is
\begin{equation*} \int_{\sigma}\eta_{\mathbf{F}} = \int_0^1\int_0^1 \mathbf{F}(\sigma(u,v))\cdot \left( \frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v} \right) du\,dv. \end{equation*}
This is called the flux of \(\mathbf{F}\) across the oriented surface \(\sigma\text{.}\)
The ordered parameters \((u,v)\) determine the orientation of the surface through the normal vector \(\frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v}\text{.}\) Reversing the orientation changes the sign of a flux integral, while scalar surface integrals are unaffected.

Example 3.4.31.

Compute the flux of the vector field \(\mathbf{F}=\langle 0,0,z \rangle\) through the parameterized surface
\begin{equation*} \sigma(u,v)=\langle u,\; v,\; u+v \rangle, \qquad 0 \le u,v \le 1. \end{equation*}
Then reverse the orientation by changing the parameterization.
Solution.
First compute the tangent vectors:
\begin{equation*} \frac{\partial \sigma}{\partial u} = \langle 1,0,1 \rangle, \qquad \frac{\partial \sigma}{\partial v} = \langle 0,1,1 \rangle. \end{equation*}
Hence
\begin{equation*} \frac{\partial \sigma}{\partial u} \times \frac{\partial \sigma}{\partial v} = \langle -1,-1,1 \rangle. \end{equation*}
Two panels show the same slanted square surface..
Figure 3.4.32. The same parametric surface with two opposite orientations.
Since \(\mathbf{F}(\sigma(u,v))=\langle 0,0,u+v \rangle\text{,}\) the flux is
\begin{equation*} \int_{\sigma}\eta_{\mathbf{F}} = \int_0^1\int_0^1 \langle 0,0,u+v \rangle \cdot \langle -1,-1,1 \rangle\,du\,dv = \int_0^1\int_0^1 (u+v)\,du\,dv = 1. \end{equation*}
Now reverse the orientation by swapping the parameters:
\begin{equation*} \tilde{\sigma}(u,v)=\sigma(v,u)=\langle v,\; u,\; u+v \rangle. \end{equation*}
Then
\begin{equation*} \frac{\partial \tilde{\sigma}}{\partial u} \times \frac{\partial \tilde{\sigma}}{\partial v} = \langle 1,1,-1 \rangle = - \left( \frac{\partial \sigma}{\partial u} \times \frac{\partial \sigma}{\partial v} \right). \end{equation*}
Therefore
\begin{equation*} \int_{\tilde{\sigma}}\eta_{\mathbf{F}} = \int_0^1\int_0^1 \langle 0,0,u+v \rangle \cdot \langle 1,1,-1 \rangle\,du\,dv = -1. \end{equation*}
So the flux depends on the orientation of the surface.

Example 3.4.33.

Compute the area of the unit square in the \(xy\)-plane and the flux of the constant field \(\mathbf{F}=\langle 0,0,1 \rangle\) through it.
Solution.
Let \(\sigma(u,v)=\langle u,v,0 \rangle\) for \(0 \le u,v \le 1\text{.}\) Then
\begin{equation*} \frac{\partial \sigma}{\partial u}=\langle 1,0,0 \rangle, \qquad \frac{\partial \sigma}{\partial v}=\langle 0,1,0 \rangle, \qquad \frac{\partial \sigma}{\partial u}\times \frac{\partial \sigma}{\partial v} = \langle 0,0,1 \rangle. \end{equation*}
Therefore
\begin{gather*} \int_{\sigma} 1\,dS = \int_0^1\int_0^1 1\,du\,dv = 1,\\ \int_{\sigma}\eta_{\mathbf{F}} = \int_0^1\int_0^1 1\,du\,dv = 1. \end{gather*}
So the square has area \(1\text{,}\) and the upward flux of \(\mathbf{F}\) across it is also \(1\text{.}\)
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