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Section 2.6 Surface integrals

Subsection 2.6.1 Surface integrals of scalar functions

Definition 2.6.1. Surface integral of scalar function.

Let \(\boldr\colon \mathcal{R}\rightarrow \R^3\) be a smooth parametrization of the surface \(\mathcal{S}\text{,}\) and let \(f\) be a continuous function defined on \(\mathcal{S}\text{.}\) The surface integral of \(f\) over \(\mathcal{S}\) is defined as

\begin{equation*} \iint\limits_\mathcal{S}f(x,y,z)\, d\sigma=\iint\limits_{\mathcal{R}}f(\boldr(u,v))\abs{\boldr_u\times \boldr_v}\, dA\text{.} \end{equation*}

Remark 2.6.2. Taxonomy of scalar integrals.

It might be useful to list (and organize) all the different forms of scalar integrals we have met thus far.

\begin{equation*} \begin{array}{c|c|c} \text{Region} \amp \text{Type}\amp \text{Notation}\\ \hline \text{interval } [a,b] \amp \text{integral} \amp \int_a^b f(x)\, dx\\ \text{planar region } \mathcal{R}\subseteq\R^2 \amp \text{double integral} \amp \iint\limits_\mathcal{R} f(x,y)\, dA\\ \text{solid region } \mathcal{R}\subseteq\R^3 \amp \text{triple integral} \amp \iiint\limits_\mathcal{R}f(x,y,z)\, dV\\ \text{parametrized curve } \mathcal{C}\subseteq\R^n \amp \text{line integral} \amp \int\limits_\mathcal{C}f\, ds \\ \text{parametrized surface } \mathcal{S}\subseteq\R^3 \amp \text{surface integral} \amp \iint\limits_\mathcal{S}f(x,y,z)\, d\sigma \end{array} \end{equation*}
Figure 2.6.3. Taxonomy of scalar integrals

Remark 2.6.4. Independence of parametrization.

As with scalar line integrals, it is possible to show that our definition of the surface integral of \(f\) over a smoothly parametrized surface \(\mathcal{S}\) is independent of the parametrization chosen. (Recall that by definition a surface parametrization is one-to-one on the interior of the parameter domain.)

Interpretation 2.6.5. Surface integral interpretation.

As always, the key to understanding the meaning of a surface integral is to make sense of the right-hand side of the approximation formula

\begin{equation} \iint\limits_\mathcal{S}f\, d\sigma\approx \sum_{k=1}^{n}f(\boldr(u_k,v_k))\underset{\operatorname{approx. area of} \Delta\sigma_k}{\underbrace{\abs{\boldr_u\times \boldr_v}\Delta u\Delta v}}\text{.}\tag{2.6.1} \end{equation}

Understanding \(\abs{\boldr_u\times \boldr_v}\Delta u\Delta v\) as an approximation of the area of one of the subpatches of \(\mathcal{S}\text{,}\) the interpretation then depends on what the function \(f\) is. Here are two typical examples.

  1. Geometric: surface area.

    If \(f=1\) (the constant function), then the right-hand side of (2.6.1) simply adds up \(\abs{\boldr_u\times \boldr_v}\Delta u\Delta v\text{,}\) yielding an approximation of the area of \(S\text{.}\) The integral of \(f=1\) computes this area exactly.

  2. Physical: density function.

    Assume \(f(x,y,z)\geq 0\) gives us the density of quantity \(Q\) per unit area at position \((x,y,z)\text{.}\) In this case the right-hand side of (2.6.1) can be understood as an approximation of the total quantity \(Q\) over \(\mathcal{S}\text{,}\) and thus the integral computes this total quantity exactly.

Example 2.6.6.

Let \(\mathcal{S}\) be the surface obtained by rotating the curve

\begin{equation*} \mathcal{C}\colon x=\cos z, y=0, -\pi/2\leq z\leq \pi/2 \end{equation*}

around the \(z\)-axis. Assume the mass density at a point \((x,y,z)\) on \(\mathcal{S}\) is given by \(f(x,y,z)=\sqrt{1-x^2-y^2}\text{.}\) Compute the total mass of \(\mathcal{S}\) assuming the variables \(x,y,z\) are measured in cm, and \(f(x,y,z)\) is measured in g per cm\(^2\text{.}\)

Solution.

  1. First we parametrize \(\mathcal{S}\text{.}\) Observe that intersecting \(\mathcal{S}\) with the plane \(z=z_0\) yields a circle centered at \((0,0,z_0)\) of radius \(R=\cos z_0 \text{,}\) which we can parametrize as

    \begin{equation*} (R\cos\theta, R\sin\theta, z_0)=(\cos z_0\cos\theta, \cos z_0\sin\theta, z_0)\text{.} \end{equation*}

    Letting \(z\) vary between \(-\pi/2\) and \(\pi/2\text{,}\) we get the parametrization

    \begin{align*} \boldr(\theta, z) \amp = (\cos z\cos\theta, \cos z\sin\theta, z) \\ \mathcal{R} \colon 0\amp \leq \theta\leq 2\pi, -\pi/2\leq z\leq \pi/2 \text{.} \end{align*}

  2. Assemble the necessary ingredients:

    \begin{align*} \boldr_\theta \amp = \angvec{-\cos z\sin\theta, \cos z\cos\theta, 0}\\ \boldr_z\amp =\angvec{-\sin z\cos\theta, -\sin z\sin\theta, 1} \\ \boldr_\theta\times \boldr_z \amp=\angvec{\cos z\cos\theta,\cos z\sin\theta, \sin z\cos z\sin^2\theta+\sin z\cos z\cos^2\theta} \\ \amp =\angvec{\cos z\cos\theta,\cos z\sin\theta, \sin z\cos z}\\ \abs{\boldr_\theta\times \boldr_z} \amp = \sqrt{\cos^2(1+\sin^2 z)} \\ \amp =\abs{\cos z}\sqrt{1+\sin^2 z}\\ \amp =\cos z\sqrt{1+\sin^2z} \text{,} \end{align*}

    where the last equality follows since \(\cos z\geq 0\) for \(z\in [-\pi/2, \pi/2]\text{.}\)

  3. According to Interpretation 2.6.5, integrating a mass density function over \(\mathcal{S}\) gives us the total mass of \(\mathcal{S}\text{.}\) Thus

    \begin{align*} \operatorname{mass}\mathcal{S} \amp = \iint\limits_{\mathcal{S}}f(x,y,z)\, d\sigma \\ \amp=\iint\limits_{\mathcal{R}}f(\boldr(\theta, z))\, \abs{\boldr_\theta\times \boldr_z}\, dA \\ \amp = \int_0^{2\pi}\int_{-\pi/2}^{\pi/2}\cos z\sqrt{1-\cos^2 z\cos^2\theta-\cos^2z\sin^2\theta}\sqrt{1+\sin^2z}\, dz\, d\theta\\ \amp =2\pi\int_{-\pi/2}^{\pi/2}\cos z\sqrt{\sin^2z(1+\sin^2z)}\, dz\\ \amp = 2\pi \int_{-\pi/2}^{\pi/2}\abs{\sin z}\cos z\sqrt{1+\sin^2z}\, dz\\ \amp =4\pi\int_0^{\pi/2}\sin z\cos z\sqrt{1+\sin^2 z}\, dz , \hspace{5pt} \ (\abs{\sin z} \text{ even})\\ \amp =4\pi\left( \frac{1}{3}(1+\sin^2 z)^{3/2}\right)\Bigr\vert_0^{\pi/2}\\ \amp =\frac{4\pi}{3}(2^{3/2}-1)\ \operatorname{grams} \text{.} \end{align*}

Definition 2.6.7. Surface integral for piecewise smooth surfaces.

A piecewise smooth surface is the union of smoothly parametrized surfaces whose interiors are non-overlapping, and whose intersections form piecewise smooth curves. If \(\mathcal{S}=\mathcal{S}_1\cup\mathcal{S}_2\cdots\cup\mathcal{S}_n\) is piecewise smooth, and if \(f\) is continuous on \(\mathcal{S}\text{,}\) we define the surface integral of \(f\) over \(S\) as

\begin{equation*} \iint\limits_\mathcal{S}f\, d\sigma=\sum_{k=1}^n \iint\limits_{S_k}f\, d\sigma\text{.} \end{equation*}

Example 2.6.8. Piecewise smooth surface.

Compute the surface integral of \(f(x,y,z)=xy\) over the tetrahedron \(\mathcal{T}\) with vertices \((0,0,0),(1,0,0),(0,1,0),(0,0,1)\text{.}\)

Solution.

We have \(\mathcal{T}=T_1\cup T_2\cup T_3\cup T_4\text{,}\) where the \(T_k\) are the four triangular faces of \(\mathcal{T}\text{,}\) and thus

\begin{equation*} \iint\limits_\mathcal{T}f\, d\sigma=\iint\limits_{T_1}f\, d\sigma+\iint\limits_{T_2}f\, d\sigma+\iint\limits_{T_3}f\, d\sigma+\iint\limits_{T_4}f\, d\sigma\text{.} \end{equation*}

Let \(T_1, T_2\) be the triangular faces lying in the coordinate planes \(x=0\) and \(y=0\text{,}\) respectively. Since \(f(x,y,z)=0\) for all points in these two surfaces, the corresponding surface integrals \(\iint_{T_k}f\, d\sigma\text{,}\) \(k=1,2\text{,}\) are equal to zero. Thus we need only compute the integrals of \(f\) over \(T_3\) and \(T_4\text{.}\)

  • Integral over \(T_3\).

    Let \(T_3\) be the face of the tetrahedron in the \(xy\)-plane. We parametrize as

    \begin{align*} \boldr(x,y) \amp =(x,y,0)\\ \mathcal{R}_1\colon 0\amp\leq x\leq 1, 0\leq y\leq 1-x \text{.} \end{align*}

    A simple computation shows \(\abs{\boldr_x\times \boldr_y}=1\text{.}\) Thus

    \begin{align*} \iint\limits_{T_1}f\, d\sigma\amp =\int_0^1\int_0^{1-x}xy\, dy\, dx\\ \amp =\frac{1}{2}\int_0^1x(1-x)^2\, dx \\ \amp =-\frac{1}{2}\int_1^0 (1-u)u^2\, du \amp (u=1-x, du=-dx)\\ \amp =\frac{1}{2}(1/3-1/4)=\frac{1}{24} \end{align*}
  • Integral over \(T_4\).

    Let \(T_4\) be triangle with vertices \((1,0,0), (1,0,0),(0,1,0)\text{.}\) This is the region of the plane \(z=x+y\) lying over the planar triangle with vertices \((0,0,0), (1,0,0), (0,1,0)\text{.}\) Since \(z\) is expressed as a function of \(x,y\) we parametrize easily as

    \begin{align*} \bolds(x,y) \amp =(x,y,x+y)\\ \mathcal{R}_{2}\colon 0 \amp\leq x\leq 1, 0\leq y\leq 1-x \text{.} \end{align*}

    Again, a straightforward computation shows \(\abs{\bolds_x\times \bolds_y}=\sqrt{3}\text{,}\) and thus

    \begin{align*} \iint\limits_{T_4}f\, d\sigma\amp =\iint\limits_{\mathcal{R}_2}f(\bolds(x,y))\sqrt{3}\, dA \amp \\ \amp= \sqrt{3}\int_0^1\int_0^{1-x}xy\, dy\, dx \\ \amp =\frac{\sqrt{3}}{24}\text{,} \end{align*}

    where we use our work from above for the last equality.

In conclusion we have
\begin{equation*} \iint\limits_{\mathcal{T}}=\iint\limits_{T_1}f\, d\sigma+\iint\limits_{T_2}f\, d\sigma =\frac{\sqrt{3}+1}{24}\text{.} \end{equation*}

Subsection 2.6.2 Surface integrals of vector fields

Definition 2.6.9. Surface orientation.

Let \(\mathcal{S}\subseteq\R^3\) be a smoothly parametrized surface. Informally, an orientation of \(\mathcal{S}\) is a continuous choice of unit normal vectors to points on the surface. More precisely, an orientation is a continuous function

\begin{align*} \boldn \colon \mathcal{S}\amp\rightarrow \R^3 \\ (x,y,z)\amp\mapsto \boldn(x,y,z) \text{,} \end{align*}

where for each \(P=(x,y,z)\in \mathcal{S}\text{,}\) \(\boldn(P)\) is a unit normal vector to \(\mathcal{S}\) at \(P\text{.}\) The surface \(\mathcal{S}\) is orientable if an orientation \(\boldn\colon \mathcal{S}\rightarrow \R^3\) exists for \(\mathcal{S}\text{.}\) If no orientation function exists, the surface is nonorientable.

Example 2.6.10. Orientations of the sphere.

Let \(\mathcal{S}\) be the unit sphere \(x^2+y^2+z^2=1\text{.}\) We have seen previously that for a point \(P=(x,y,z)\) on \(\mathcal{S}\) the vector \(\overrightarrow{OP}=\angvec{x,y,z}\) is a normal vector to \(\mathcal{S}\) at \(P\text{;}\) furthermore, since \(P\in \mathcal{S}\text{,}\) its magnitude is \(\sqrt{x^2+y^2+z^2}=1\text{.}\) Thus the function \(\boldn_1(x,y,z)=\angvec{x,y,z}\) assigns a unit normal vector to each point \(P=(x,y,z)\in \mathcal{S}\text{.}\) Since \(\boldn_1\) is clearly continuous, it defines an orientation on \(\mathcal{S}\text{,}\) called the outward orientation. The function \(\boldn_2(x,y,z)=-\angvec{x,y,z}\) also defines an orientation on the sphere, called the inward orientation.

More generally, if \(\mathcal{S}\) is a smooth surface that is the boundary of a bounded solid region \(D\subseteq\R^3\text{,}\) then \(\mathcal{S}\) has both an outward and inward orientation. In particular, all such surfaces are orientable. The sphere discussed above is one such surface: it is the boundary of the solid ball \(x^2+y^2+z^2\leq 1\text{.}\) Another example is the torus: the boundary surface of a solid doughnut.

Example 2.6.11. Graph of function.

Suppose \(\mathcal{S}\) is a smooth surface defined as the graph of the function \(z=f(x,y)\) for inputs \((x,y)\) lying in the region \(\mathcal{R}\subseteq\R^2\text{.}\) As usual this gives rise to the parametrization

\begin{align*} \boldr\colon \mathcal{R} \amp \rightarrow \R^3\\ (x,y) \amp\mapsto (x,y,f(x,y)) \text{.} \end{align*}

We have seen that in this case \(\boldr_x\times \boldr_y=\angvec{-f_x, -f_y, 1}\text{.}\) It follows that the function

\begin{equation*} \boldn(x,y,z)=\frac{\boldr_x\times \boldr_y}{\abs{\boldr_x\times \boldr_y}}=\frac{\angvec{-f_x,-f_y,1}}{\sqrt{f_x^2+f_y^2+1}} \end{equation*}

is an orientation of \(\mathcal{S}\text{.}\) Since the \(z\)-component of \(\boldn(x,y,z)\) is positive, this is called the upward orientation of \(\mathcal{S}\text{.}\)

Remark 2.6.12. Orientations of orientable surfaces.

Assume \(\mathcal{S}\) is a smooth orientable surface with orientation \(\boldn\colon \mathcal{S}\rightarrow\R^3\) and parametrization \(\boldr(u,v)\text{.}\) We have the following facts:

  1. There are exactly two orientations on \(\mathcal{S}\text{:}\) namely, the given orientation \(P\mapsto \boldn(P)\) and its opposite \(P\mapsto -\boldn(P)\text{.}\)

  2. The function

    \begin{equation*} P=\boldr(u_0,v_0)\mapsto \frac{\boldr_u\times \boldr_v(u_0,v_0)}{\abs{\boldr_u\times\boldr_v(u_0,v_0)}} \end{equation*}

    is one orientation on \(\mathcal{S}\text{;}\) the function

    \begin{equation*} P=\boldr(u_0,v_0)\mapsto \frac{\boldr_v\times \boldr_u(u_0,v_0)}{\abs{\boldr_u\times\boldr_v(u_0,v_0)}} \end{equation*}

    is the other.

Fix a constant \(R>1/2\text{,}\) and let \(\mathcal{S}\) be the Möbius strip with parametrization

\begin{equation*} \boldr(\theta, t)=\left((R+t\cos(\theta/2))\cos\theta, (R+t\cos(\theta/2))\sin\theta, t\sin(\theta/2\right)\text{,} \end{equation*}

where \(\ 0\leq \theta\leq 2\pi, -\frac{1}{2}\leq t\leq\frac{1}{2}\text{.}\) The Sage cell below depicts \(\mathcal{S}\) along with the labeled line segments we used in its original description.

Fact: \(\mathcal{S}\) is a nonorientable surface! If this is so, why doesn't the function

\begin{equation*} P=\boldr(\theta_0, t_0)\mapsto \frac{\boldr_t\times \boldr_\theta(\theta_0,t_0)}{\abs{\boldr_\theta\times\boldr_t(\theta_0,t_0)}} \end{equation*}

provide us with an orientation? The answer: this is a well-defined function on the parameter domain \([0,2\pi]\times[-1/2,1/2]\text{,}\) but not on the actual surface \(\mathcal{S}\text{!}\) Indeed, the point \(P=(R,0,0)\) can be expressed as both \(\boldr(0,0)\) and \(\boldr(2\pi, 0)\text{.}\) As the Sage cell below illustrates the two normal vectors corresponding to \((0,0)\) and \((2\pi,0)\) are \(\angvec{1,0,0}\) and \(\angvec{-1,0,0}\text{!}\)

We can see what's going on by plotting some of the normal vectors given by \(\boldr_t\times \boldr_\theta\) for points of the form \(P=\boldr(\theta, 0)\text{.}\)

Definition 2.6.13. Surface integral of vector field.

Let \(\mathcal{S}\) be a smooth orientable surface, and let \(\boldF\) be a vector field that is continuous on \(\mathcal{S}\text{.}\) We define the surface integral of \(\boldF\) over \(\mathcal{S}\) with respect to the orientation \(\boldn\colon \mathcal{S}\rightarrow \R^3\) as the scalar surface integral

\begin{equation*} \iint_\mathcal{S}\boldF\cdot \boldn\, d\sigma\text{.} \end{equation*}

This integral is also called the flux of \(\boldF\) across \(\mathcal{S}\) with respect to the orientation \(\boldn\text{.}\)

Example 2.6.14. Flux across a sphere.

Let \(\mathcal{S}\) be the sphere \(x^2+y^2+z^2=4\text{.}\) Compute the flux of \(\boldF(x,y,z)=\angvec{z,y,x}\) out of the sphere: i.e., with respect to the outward orientation on \(\mathcal{S}\text{.}\)

Solution.

We use the usual parametrization of the sphere of radius 2:

\begin{align*} \boldr(\phi,\theta) \amp = (2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)\\ \mathcal{R}\colon 0 \amp\leq \phi\leq \pi, 0\leq \theta\leq 2\pi \text{.} \end{align*}

Given a point \(P=(x,y,z)\in \mathcal{S}\) the vector \(\angvec{x,y,z}\) is an outward pointing normal vector of magnitude \(\sqrt{x^2+y^2+z^2}=2\text{.}\) Thus the function \(\boldn(x,y,z)=\frac{1}{2}\angvec{x,y,z}\) is the outward orientation on \(\mathcal{S}\text{.}\) It follows that

\begin{equation*} \boldF(x,y,z)\cdot \boldn(x,y,z)=\frac{1}{2}\angvec{z,y,x}\cdot \angvec{x,y,z}=\frac{1}{2}(2xz+y^2)\text{,} \end{equation*}

and thus that

\begin{align*} \iint\limits_\mathcal{S}\boldF\cdot\boldn\, d\sigma\amp = \frac{1}{2}\iint\limits_\mathcal{S}2xz+y^2\, d\sigma \\ \amp = \frac{1}{2}\iint\limits_{\mathcal{R}}(8\sin\phi\cos\phi\cos\theta+4\sin^2\phi\sin^2\theta)(4\sin\phi)\, dA\\ \amp =8\int_0^{2\pi}\int_0^\pi2\sin^2\phi\cos\phi\cos\theta+\sin^3\phi\sin^2\theta\, d\phi\, d\theta\\\ \amp = 8\left(\cancelto{0}{\int_0^{2\pi}\cos\theta\, d\theta}\int_0^{\pi}\sin^2\phi\cos\phi\, d\phi+\int_0^{2\pi}\sin^2\theta\, d\theta\int_0^{\pi}\sin^3\phi\, d\phi\right)\\ \amp =8\int_0^{2\pi}\sin^2\theta\, d\theta\int_0^{\pi}\sin^3\phi\, d\phi\\ \amp =\frac{32\pi}{3}\text{.} \end{align*}

The last step above is done using the identity \(\sin^2\theta=\frac{1}{2}(1-\cos 2\theta)\) and the fact that \(\sin^3\phi=\sin\phi(1-\cos^2\phi)=\sin\phi-\sin\phi\cos^2\phi\) has antiderivative \(-\cos\phi+\frac{1}{3}\cos^3\phi\text{.}\)

Note that in this case we computed the surface integral by (a) identifying the outward orientation \(\boldn\) on \(\mathcal{S}\) geometrically (as opposed to using \(\boldr_\phi\times \boldr_\theta\) or its opposite), (b) computing the scalar function \(\boldF(x,y,z)\cdot \boldn(x,y,z)\text{,}\) and (c) integrating with respect to \(d\sigma=\vert \boldr_\phi\times\boldr_\theta\vert=4\sin\phi\text{.}\) This method works in this case because we could easily identify the orientation function \(\boldn(x,y,z)\text{.}\) When this is not easily done, resort to the procedure indicated in Theorem 2.6.15. See Example 2.6.16 for an illustration of this technique.

Example 2.6.16. Flux across a cone.

Let \(\mathcal{S}\) be the portion of the cone \(z=\sqrt{x^2+y^2}\) lying below the plane \(z=4\text{,}\) and let \(\boldF(x,y,z)=\angvec{xy,0,-z}\text{.}\) Compute the flux of \(\boldF\) across \(\mathcal{S}\) with respect to the outward (away from the \(z\)-axis) orientation of \(\mathcal{S}\text{.}\)

Solution.

  • Parametrize. Using cylindrical coordinates, we see that \(\mathcal{S}\) has parametrization

    \begin{align*} \bolds(r,\theta) \amp = (r\cos\theta, r\sin\theta, r)\\ \mathcal{R}\colon 0\amp\leq \theta\leq 2\pi, 0\leq r\leq 4 \text{.} \end{align*}

  • Ingredients.

    \begin{align*} \bolds_r \amp = \angvec{\cos\theta, \sin\theta, 1}\\ \bolds_\theta \amp= \angvec{-r\sin\theta, r\cos\theta, 0} \\ \bolds_r\times\bolds_\theta \amp =\angvec{-r\cos\theta, -r\sin\theta, r} \text{.} \end{align*}

    Observe that since \(r\geq 0\text{,}\) the normal vector \(\bolds_r\times\bolds_\theta\) points upward, and hence toward the \(z\)-axis. Thus the outward orientation \(\boldn\) comes from \(\bolds_\theta\times\bolds_r\text{.}\)

  • Now compute

    \begin{align*} \iint_\mathcal{S}\boldF\cdot\boldn\, d\sigma \amp = \iint_\mathcal{R}\boldF(r\cos\theta, r\sin\theta, r)\cdot \angvec{r\cos\theta,r\sin\theta, -r}\, dA \\ \amp =\iint_\mathcal{R}\angvec{r^2\cos\theta\sin\theta,0,-r}\cdot\angvec{r\cos\theta,r\sin\theta, -r}\, dA \\ \amp =\int_0^{2\pi}\int_0^4r^3\cos^2\theta\sin\theta+r^2\, dr\, d\theta\\ \amp = \frac{128}{3}\pi\text{.} \end{align*}

The Sage cell below assembles the many necessary ingredients to compute a surface integral of a vector field \(\boldF\text{,}\) then returns the value of the integral, as well as a diagram of the surface, some normal vectors (in green), and the vector field. You can use this to check your surface integral computations by adjusting the parametrization r(u,v) and its parameter domain description. You may also want to adjust the domain specified for the vector field plot Field.