3.7: Surface Integral

Learning Objectives

• Describe the surface integral of a scalar-valued function over a parametric surface.
• Use a surface integral to calculate the area of a given surface.
• Explain the meaning of an oriented surface, giving an example.
• Describe the surface integral of a vector field.
• Use surface integrals to solve applied problems.

Surface Integral of a Scalar-Valued Function

Now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. First, let’s look at the surface integral of a scalar-valued function. Informally, the surface integral of a scalar-valued function is an analog of a scalar line integral in one higher dimension. The domain of integration of a scalar line integral is a parameterized curve (a one-dimensional object); the domain of integration of a scalar surface integral is a parameterized surface (a two-dimensional object). Therefore, the definition of a surface integral follows the definition of a line integral quite closely. For scalar line integrals, we chopped the domain curve into tiny pieces, chose a point in each piece, computed the function at that point, and took a limit of the corresponding Riemann sum. For scalar surface integrals, we chop the domain region (no longer a curve) into tiny pieces and proceed in the same fashion.

Let $S$ be a piecewise smooth surface with parameterization $\mathbf{r}(u,v) = \langle x(u,v), \, y(u,v), \, z(u,v) \rangle$ with parameter domain $D$ and let $f(x,y,z)$ be a function with a domain that contains $S$. For now, assume the parameter domain $D$ is a rectangle, but we can extend the basic logic of how we proceed to any parameter domain (the choice of a rectangle is simply to make the notation more manageable). Divide rectangle $D$ into subrectangles $D_{ij}$ with horizontal width $\Delta u$ and vertical length $\Delta v$. Suppose that $i$ ranges from $1$ to $m$ and $j$ ranges from $1$ to $n$ so that $D$ is subdivided into $mn$ rectangles. This division of $D$ into subrectangles gives a corresponding division of $S$ into pieces $S_{ij}$. Choose point $P_{ij}$ in each piece $S_{ij}$ evaluate $P_{ij}$ at $f$, and multiply by area $S_{ij}$ to form the Riemann sum

$$\sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \, \Delta S_{ij}. \nonumber$$

To define a surface integral of a scalar-valued function, we let the areas of the pieces of $S$ shrink to zero by taking a limit.

Definition: Surface Integral of a Scalar-Valued Function

The surface integral of a scalar-valued function of $f$ over a piecewise smooth surface $S$ is

$$\iint_S f(x,y,z) dA = \lim_{m,n\rightarrow \infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}.$$

Again, notice the similarities between this definition and the definition of a scalar line integral. In the definition of a line integral we chop a curve into pieces, evaluate a function at a point in each piece, and let the length of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. In the definition of a surface integral, we chop a surface into pieces, evaluate a function at a point in each piece, and let the area of the pieces shrink to zero by taking the limit of the corresponding Riemann sum. Thus, a surface integral is similar to a line integral but in one higher dimension.

The definition of a scalar line integral can be extended to parameter domains that are not rectangles by using the same logic used earlier. The basic idea is to chop the parameter domain into small pieces, choose a sample point in each piece, and so on. The exact shape of each piece in the sample domain becomes irrelevant as the areas of the pieces shrink to zero.

Scalar surface integrals are difficult to compute from the definition, just as scalar line integrals are. To develop a method that makes surface integrals easier to compute, we approximate surface areas $\Delta S_{ij}$ with small pieces of a tangent plane, just as we did in the previous subsection. Recall the definition of vectors $\mathbf t_u$ and $\mathbf t_v$:

$$\mathbf t_u = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle\, \text{and} \, \mathbf t_v = \left\langle \dfrac{\partial x}{\partial u},\, \dfrac{\partial y}{\partial u},\, \dfrac{\partial z}{\partial u} \right\rangle. \nonumber$$

From the material we have already studied, we know that

$$\Delta S_{ij} \approx ||\mathbf t_u (P_{ij}) \times \mathbf t_v (P_{ij})|| \,\Delta u \,\Delta v. \nonumber$$

Therefore,

$$\iint_S f(x,y,z) \,dS \approx \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij})|| \mathbf t_u(P_{ij}) \times \mathbf t_v(P_{ij}) ||\,\Delta u \,\Delta v. \nonumber$$

This approximation becomes arbitrarily close to $\displaystyle \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n f(P_{ij}) \Delta S_{ij}$ as we increase the number of pieces $S_{ij}$ by letting $m$ and $n$ go to infinity. Therefore, we have the following equation to calculate scalar surface integrals:

$$\iint_S f(x,y,z)\,dS = \iint_D f(\mathbf r(u,v)) ||\mathbf t_u \times \mathbf t_v||\,dA. \label{scalar surface integrals}$$

Equation $\ref{scalar surface integrals}$ allows us to calculate a surface integral by transforming it into a double integral. This equation for surface integrals is analogous to the equation for line integrals:

$$\iint_C f(x,y,z)\,ds = \int_a^b f(\mathbf r(t))||\mathbf r'(t)||\,dt. \nonumber$$

In this case, vector $\mathbf t_u \times \mathbf t_v$ is perpendicular to the surface, whereas vector $\mathbf r'(t)$ is tangent to the curve.

Example $\PageIndex{1}$: Calculating a Surface Integral

Calculate surface integral

$$\iint_S 5 \, dS, \nonumber$$

where $S$ is the surface with parameterization $\mathbf r(u,v) = \langle u, \, u^2, \, v \rangle$ for $0 \leq u \leq 2$ and $0 \leq v \leq u$.

Solution

Notice that this parameter domain$D$ is a triangle, and therefore the parameter domain is not rectangular. This is not an issue though, because Equation $\ref{scalar surface integrals}$ does not place any restrictions on the shape of the parameter domain.

To use Equation $\ref{scalar surface integrals}$ to calculate the surface integral, we first find vectors $\mathbf t_u$ and $\mathbf t_v$. Note that $\mathbf t_u = \langle 1, 2u, 0 \rangle$ and $\mathbf t_v = \langle 0,0,1 \rangle$. Therefore,

$$\mathbf t_u \times \mathbf t_v = \begin{vmatrix} \mathbf{ i} & \mathbf{ j} & \mathbf{ k} \nonumber \\ 1 & 2u & 0 \nonumber \\ 0 & 0 & 1 \end{vmatrix} = \langle 2u, \, -1, \, 0 \rangle\ \nonumber$$

and

$$||\mathbf t_u \times \mathbf t_v|| = \sqrt{1 + 4u^2}.$$

By Equation $\ref{scalar surface integrals}$,

$$\begin{align*} \iint_S 5 \, dS &= 5 \iint_D u \sqrt{1 + 4u^2} \, dA \\ &= 5 \int_0^2 \int_0^u \sqrt{1 + 4u^2} \, dv \, du = 5 \int_0^2 u \sqrt{1 + 4u^2}\, du \\ &= 5 \left[\dfrac{(1+4u^2)^{3/2}}{3} \right]_0^2 \\ &= \dfrac{5(17^{3/2}-1)}{3} \approx 115.15. \end{align*}$$

Example $\PageIndex{2}$: Calculating the Surface Integral of a Cylinder

Calculate surface integral $$\iint_S (x + y^2) \, dS,$$ where $S$ is cylinder $x^2 + y^2 = 4, \, 0 \leq z \leq 3$ (Figure $\PageIndex{1}$).

Figure $\PageIndex{1}$: $S$ is cylinder $x^2 + y^2 = 4, \, 0 \leq z \leq 3$

Solution

To calculate the surface integral, we first need a parameterization of the cylinder. A parameterization is $\mathbf r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, 0 \leq u \leq 2\pi, \, 0 \leq v \leq 3.$

The tangent vectors are $\mathbf t_u = \langle \sin u, \, \cos u, \, 0 \rangle$ and $\mathbf t_v = \langle 0,0,1 \rangle$. Then,

$$\mathbf t_u \times \mathbf t_v = \begin{vmatrix} \mathbf{ i} & \mathbf{ j} & \mathbf{ k} \\ -\sin u & \cos u & 0 \\ 0 & 0 & 1 \end{vmatrix} = \langle \cos u, \, \sin u, \, 0 \rangle$$

and $||\mathbf t_u \times \mathbf t_v || = \sqrt{\cos^2 u + \sin^2 u} = 1$. By Equation $\ref{scalar surface integrals}$,

$$\begin{align*} \iint_S f(x,y,z)dS &= \iint_D f (\mathbf r(u,v)) ||\mathbf t_u \times \mathbf t_v|| \, dA \\ &= \int_0^3 \int_0^{2\pi} (\cos u + \sin^2 u) \, du \,dv \\ &= \int_0^3 \left[\sin u + \dfrac{u}{2} - \dfrac{\sin(2u)}{4} \right]_0^{2\pi} \,dv \\ &= \int_0^3 \pi \, dv = 3 \pi. \end{align*}$$

Exercise $\PageIndex{1}$

Calculate $$\iint_S (x^2 - z) \,dS,$$ where $S$ is the surface with parameterization $\mathbf r(u,v) = \langle v, \, u^2 + v^2, \, 1 \rangle, \, 0 \leq u \leq 2, \, 0 \leq v \leq 3.$

Hint

Use Equation $\ref{scalar surface integrals}$.

Answer

24

Example $\PageIndex{3}$: Calculating the Surface Integral of a Piece of a Sphere

Calculate surface integral $$\iint_S f(x,y,z)\,dS,$$ where $f(x,y,z) = z^2$ and $S$ is the surface that consists of the piece of sphere $x^2 + y^2 + z^2 = 4$ that lies on or above plane $z = 1$ and the disk that is enclosed by intersection plane $z = 1$ and the given sphere (Figure $\PageIndex{2}$).

Figure $\PageIndex{2}$: $S$ is the surface that consists of the piece of sphere $x^2 + y^2 + z^2 = 4$ that lies on or above plane $z = 1$ and the disk that is enclosed by intersection plane $z = 1$ and the given sphere.

Solution

Notice that $S$ is not smooth but is piecewise smooth; $S$ can be written as the union of its base $S_1$ and its spherical top $S_2$, and both $S_1$ and $S_2$ are smooth. Therefore, to calculate

$$\iint_S z^2 dS, \nonumber$$

we write this integral as

$$\iint_{S_1} z^2 \,dS + \iint_{S_2} z^2 \,dS \nonumber$$

and we calculate integrals

$$\iint_{S_1} z^2 \,dS \nonumber$$

and

$$\iint_{S_2}Z^2 \,dS. \nonumber$$

First, we calculate $\displaystyle \iint_{S_1} z^2 \,dS.$ To calculate this integral we need a parameterization of $S_1$. This surface is a disk in plane $z = 1$ centered at $(0,0,1)$. To parameterize this disk, we need to know its radius. Since the disk is formed where plane $z = 1$ intersects sphere $x^2 + y^2 + z^2 = 4$, we can substitute $z = 1$ into equation $x^2 + y^2 + z^2 = 4$:

$$x^2 + y^2 + 1 = 4 \Rightarrow x^2 + y^2 = 3. \nonumber$$

Therefore, the radius of the disk is $\sqrt{3}$ and a parameterization of $S_1$ is $\mathbf r(u,v) = \langle u \, \cos v, \, u \, \sin v, \, 1 \rangle, \, 0 \leq u \leq \sqrt{3}, \, 0 \leq v \leq 2\pi$. The tangent vectors are $\mathbf t_u = \langle \cos v, \, \sin v, \, 0 \rangle$ and $\mathbf t_v = \langle -u \, \sin v, \, u \, \cos v, \, 0 \rangle$, and thus

$$\mathbf t_u \times \mathbf t_v = \begin{vmatrix} \mathbf{ i} & \mathbf{ j} & \mathbf{ k} \\ \cos v & \sin v & 0 \\ -u\sin v & u\cos v& 0 \end{vmatrix} = \langle 0, \, 0, u \, \cos^2 v + u \, \sin^2 v \rangle = \langle 0, 0, u \rangle.$$

The magnitude of this vector is $u$. Therefore,

$$\begin{align*} \iint_{S_1} z^2 \,dS &= \int_0^{\sqrt{3}} \int_0^{2\pi} f(r(u,v))||t_u \times t_v|| \, dv \, du \\ &= \int_0^{\sqrt{3}} \int_0^{2\pi} u \, dv \, du \\ &= 2\pi \int_0^{\sqrt{3}} u \, du \\ &= 2\pi \sqrt{3}. \end{align*}$$

Now we calculate

$$\iint_{S_2} \,dS. \nonumber$$

To calculate this integral, we need a parameterization of $S_2$. The parameterization of full sphere $x^2 + y^2 + z^2 = 4$ is

$$\mathbf r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi.$$

Since we are only taking the piece of the sphere on or above plane $z = 1$, we have to restrict the domain of $\phi$. To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in Figure $\PageIndex{17}$ (the $\sqrt{3}$ comes from the fact that the base of $S$ is a disk with radius $\sqrt{3}$). Therefore, the tangent of $\phi$ is $\sqrt{3}$, which implies that $\phi$ is $\pi / 6$. We now have a parameterization of $S_2$:

$\mathbf r(\phi, \theta) = \langle 2 \, \cos \theta \, \sin \phi, \, 2 \, \sin \theta \, \sin \phi, \, 2 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi / 3.$

Figure $\PageIndex{3}$: The maximum value of $\phi$ has a tangent value of $\sqrt{3}$.

The tangent vectors are $\mathbf t_{\phi} = \langle 2 \, \cos \theta \, \cos \phi, \, 2 \, \sin \theta \,\cos \phi, \, -2 \, \sin \phi \rangle$ and $\mathbf t_{\theta} = \langle - 2 \sin \theta \sin \phi, \, u\cos \theta \sin \phi, \, 0 \rangle$, and thus

$$\begin{align*} \mathbf t_{\phi} \times \mathbf t_{\theta} &= \begin{vmatrix} \mathbf{ i} & \mathbf{ j} & \mathbf{ k} \nonumber \\ 2 \cos \theta \cos \phi & 2 \sin \theta \cos \phi & -2\sin \phi \\ -2\sin \theta\sin\phi & 2\cos \theta \sin\phi & 0 \end{vmatrix} \\[4 pt] &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos^2 \theta \, \cos \phi \, \sin \phi + 4 \, \sin^2 \theta \, \cos \phi \, \sin \phi \rangle \\[4 pt] &= \langle 4 \, \cos \theta \, \sin^2 \phi, \, 4 \, \sin \theta \, \sin^2 \phi, \, 4 \, \cos \phi \, \sin \phi \rangle. \end{align*}$$

The magnitude of this vector is

$$\begin{align*} \mathbf t_{\phi} \times \mathbf t_{\theta} &= \sqrt{16 \, \cos^2\theta \, \sin^4\phi + 16 \, \sin^2\theta \, \sin^4 \phi + 16 \, \cos^2\phi \, \sin^2\phi} \\[4 pt] &= 4 \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi}. \end{align*}$$

Therefore,

$$\begin{align*} \iint_{S_2} z \, dS &= \int_0^{\pi/6} \int_0^{2\pi} f (\mathbf r(\phi, \theta))||\mathbf t_{\phi} \times \mathbf t_{\theta}|| \, d\theta \, d\phi \\ &= \int_0^{\pi/6} \int_0^{2\pi} 16 \, \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2\phi} \, d\theta \, d\phi \\ &= 32 \pi \int_0^{\pi/6} \cos^2\phi \sqrt{\sin^4\phi + \cos^2\phi \, \sin^2 \phi} \, d\phi \\ &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \sqrt{\sin^2\phi + \cos^2\phi} \, d\phi \\ &= 32 \pi \int_0^{\pi/6} \cos^2\phi \, \sin \phi \, d\phi \\ &= 32\pi \left[- \dfrac{\cos^3 \phi}{3} \right]_0^{\pi/6} \\ &= 32 \pi \left[ \dfrac{1}{3} - \dfrac{\sqrt{3}}{8} \right] = \dfrac{32\pi}{3} - 4\sqrt{3}. \end{align*}$$

Since

$$\iint_S z^2 \,dS = \iint_{S_1}z^2 \,dS + \iint_{S_2}z^2 \,dS, \nonumber$$

we have

$$\iint_S z^2 \,dS = (2\pi - 4) \sqrt{3} + \dfrac{32\pi}{3}. \nonumber$$

Analysis

In this example we broke a surface integral over a piecewise surface into the addition of surface integrals over smooth subsurfaces. There were only two smooth subsurfaces in this example, but this technique extends to finitely many smooth subsurfaces.

Exercise $\PageIndex{2}$

Calculate line integral $\displaystyle \iint_S (x - y) \, dS,$ where $S$ is cylinder $x^2 + y^2 = 1, \, 0 \leq z \leq 2$, including the circular top and bottom.

Hint

Break the integral into three separate surface integrals.

Answer

0

Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface $S$ and the density of the sheet at point $(x,y,z)$ is $\rho(x,y,z)$ then mass $m$ of the sheet is

$$\displaystyle m = \iint_S \rho (x,y,z) \,dS. \label{mass}$$

Example $\PageIndex{4}$: Calculating the Mass of a Sheet

A flat sheet of metal has the shape of surface $z = 1 + x + 2y$ that lies above rectangle $0 \leq x \leq 4$ and $0 \leq y \leq 2$. If the density of the sheet is given by $\rho (x,y,z) = x^2 yz$, what is the mass of the sheet?

Solution

Let $S$ be the surface that describes the sheet. Then, the mass of the sheet is given by $\displaystyle m = \iint_S x^2 yx \, dS.$ To compute this surface integral, we first need a parameterization of $S$. Since $S$ is given by the function $f(x,y) = 1 + x + 2y$, a parameterization of $S$ is $\mathbf r(x,y) = \langle x, \, y, \, 1 + x + 2y \rangle, \, 0 \leq x \leq 4, \, 0 \leq y \leq 2$.

The tangent vectors are $\mathbf t_x = \langle 1,0,1 \rangle$ and $\mathbf t_y = \langle 1,0,2 \rangle$. Therefore, $\mathbf t_x + \mathbf t_y = \langle -1,-2,1 \rangle$ and $||\mathbf t_x \times \mathbf t_y|| = \sqrt{6}$.

By the definition of the line integral (Section 16.2), $$\begin{align*} m &= \iint_S x^2 yz \, dS \\[4pt] &= \sqrt{6} \int_0^4 \int_0^2 x^2 y (1 + x + 2y) \, dy \,dx \\[4pt] &= \sqrt{6} \int_0^4 \dfrac{22x^2}{3} + 2x^3 \,dx \\[4pt] &= \dfrac{2560 \sqrt{6}}{9} \approx 696.74. \end{align*}$$

Exercise $\PageIndex{3}$

A piece of metal has a shape that is modeled by paraboloid $z = x^2 + y^2, \, 0 \leq z \leq 4,$ and the density of the metal is given by $\rho (x,y,z) = z + 1$. Find the mass of the piece of metal.

Answer

$38.401 \pi \approx 120.640$

Orientation of a Surface

Recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. The same was true for scalar surface integrals: we did not need to worry about an “orientation” of the surface of integration.

On the other hand, when we defined vector line integrals, the curve of integration needed an orientation. That is, we needed the notion of an oriented curve to define a vector line integral without ambiguity. Similarly, when we define a surface integral of a vector field, we need the notion of an oriented surface. An oriented surface is given an “upward” or “downward” orientation or, in the case of surfaces such as a sphere or cylinder, an “outward” or “inward” orientation.

Let S be a smooth surface. For any point $(x,y,z)$ on $S$, we can identify two unit normal vectors $\mathbf N$ and $-\mathbf N$. If it is possible to choose a unit normal vector $\mathbf N$ at every point $(x,y,z)$ on $S$ so that $\mathbf N$ varies continuously over $S$, then $S$ is “orientable.” Such a choice of unit normal vector at each point gives the orientation of a surface $S$. If you think of the normal field as describing water flow, then the side of the surface that water flows toward is the “negative” side and the side of the surface at which the water flows away is the “positive” side. Informally, a choice of orientation gives $S$ an “outer” side and an “inner” side (or an “upward” side and a “downward” side), just as a choice of orientation of a curve gives the curve “forward” and “backward” directions.

Closed surfaces such as spheres are orientable: if we choose the outward normal vector at each point on the surface of the sphere, then the unit normal vectors vary continuously. This is called the positive orientation of the closed surface (Figure $\PageIndex{18}$). We also could choose the inward normal vector at each point to give an “inward” orientation, which is the negative orientation of the surface.

Figure $\PageIndex{4}$: An oriented sphere with positive orientation. Two normal vectors $\mathbf N_1$ and $\mathbf N_2$ points outward.

A portion of the graph of any smooth function $z = f(x,y)$ is also orientable. If we choose the unit normal vector that points “above” the surface at each point, then the unit normal vectors vary continuously over the surface. We could also choose the unit normal vector that points “below” the surface at each point. To get such an orientation, we parameterize the graph of $f$ in the standard way: $\mathbf r(x,y) = \langle x,\, y, \, f(x,y)\rangle$, where $x$ and $y$ vary over the domain of $f$. Then, $\mathbf t_x = \langle 1,0,f_x \rangle$ and $\mathbf t_y = \langle 0,1,f_y \rangle$, and therefore the cross product $\mathbf t_x \times \mathbf t_y$ (which is normal to the surface at any point on the surface) is $\langle -f_x, \, -f_y, \, 1 \rangle$Since the $z$-component of this vector is one, the corresponding unit normal vector points “upward,” and the upward side of the surface is chosen to be the “positive” side.

Let $S$ be a smooth orientable surface with parameterization $\mathbf r(u,v)$. For each point $\mathbf r(a,b)$ on the surface, vectors $\mathbf t_u$ and $\mathbf t_v$ lie in the tangent plane at that point. Vector $\mathbf t_u \times \mathbf t_v$ is normal to the tangent plane at $\mathbf r(a,b)$ and is therefore normal to $S$ at that point. Therefore, the choice of unit normal vector

$$\mathbf N = \dfrac{\mathbf t_u \times \mathbf t_v}{||\mathbf t_u \times \mathbf t_v||}$$

gives an orientation of surface $S$.

Example $\PageIndex{5}$:Choosing an Orientation

Give an orientation of cylinder $x^2 + y^2 = r^2, \, 0 \leq z \leq h$.

Solution

This surface has parameterization $\mathbf r(u,v) = \langle r \, \cos u, \, r \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq h.$

The tangent vectors are $\mathbf t_u = \langle -r \, \sin u, \, r \, \cos u, \, 0 \rangle$ and $\mathbf t_v = \langle 0,0,1 \rangle$. To get an orientation of the surface, we compute the unit normal vector

$$\mathbf N = \dfrac{\mathbf t_u \times \mathbf t_v}{||\mathbf t_u \times \mathbf t_v||} \nonumber$$

In this case, $\mathbf t_u \times \mathbf t_v = \langle r \, \cos u, \, r \, \sin u, \, 0 \rangle$ and therefore

$$||\mathbf t_u \times \mathbf t_v|| = \sqrt{r^2 \cos^2 u + r^2 \sin^2 u} = r. \nonumber$$

An orientation of the cylinder is

$$\mathbf N(u,v) = \dfrac{\langle r \, \cos u, \, r \, \sin u, \, 0 \rangle }{r} = \langle \cos u, \, \sin u, \, 0 \rangle. \nonumber$$

Notice that all vectors are parallel to the $xy$-plane, which should be the case with vectors that are normal to the cylinder. Furthermore, all the vectors point outward, and therefore this is an outward orientation of the cylinder (Figure $\PageIndex{5}$).

Figure $\PageIndex{5}$: A normal vector pointing outward at the middle the cylinder.

Exercise $\PageIndex{4}$

Give the “upward” orientation of the graph of $f(x,y) = xy$.

Hint

Parameterize the surface and use the fact that the surface is the graph of a function.

Answer

$$\mathbf{N}(x,y) = \left\langle \dfrac{-y}{\sqrt{1+x^2+y^2}}, \, \dfrac{-x}{\sqrt{1+x^2+y^2}}, \, \dfrac{1}{\sqrt{1+x^2+y^2}} \right\rangle \nonumber$$

Since every curve has a “forward” and “backward” direction (or, in the case of a closed curve, a clockwise and counterclockwise direction), it is possible to give an orientation to any curve. Hence, it is possible to think of every curve as an oriented curve. This is not the case with surfaces, however. Some surfaces cannot be oriented; such surfaces are called nonorientable. Essentially, a surface can be oriented if the surface has an “inner” side and an “outer” side, or an “upward” side and a “downward” side. Some surfaces are twisted in such a fashion that there is no well-defined notion of an “inner” or “outer” side.

The classic example of a nonorientable surface is the Möbius strip. To create a Möbius strip, take a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together (Figure $\PageIndex{20}$). Because of the half-twist in the strip, the surface has no “outer” side or “inner” side. If you imagine placing a normal vector at a point on the strip and having the vector travel all the way around the band, then (because of the half-twist) the vector points in the opposite direction when it gets back to its original position. Therefore, the strip really only has one side.

Figure $\PageIndex{6}$: A Möbius strip by taking a rectangular strip of paper, give the piece of paper a half-twist, and the glue the ends together.

Since some surfaces are nonorientable, it is not possible to define a vector surface integral on all piecewise smooth surfaces. This is in contrast to vector line integrals, which can be defined on any piecewise smooth curve.

Surface Integral of a Vector Field

With the idea of orientable surfaces in place, we are now ready to define a surface integral of a vector field. The definition is analogous to the definition of the flux of a vector field along a plane curve. Recall that if $\mathbf{F}$ is a two-dimensional vector field and $C$ is a plane curve, then the definition of the flux of $\mathbf{F}$ along $C$ involved chopping $C$ into small pieces, choosing a point inside each piece, and calculating $\mathbf{F} \cdot \mathbf{N}$ at the point (where $\mathbf{N}$ is the unit normal vector at the point). The definition of a surface integral of a vector field proceeds in the same fashion, except now we chop surface $S$ into small pieces, choose a point in the small (two-dimensional) piece, and calculate $\mathbf{F} \cdot \mathbf{N}$ at the point.

To place this definition in a real-world setting, let $S$ be an oriented surface with unit normal vector $\mathbf{N}$. Let $\mathbf{v}$ be a velocity field of a fluid flowing through $S$, and suppose the fluid has density $\rho(x,y,z)$ Imagine the fluid flows through $S$, but $S$ is completely permeable so that it does not impede the fluid flow (Figure $\PageIndex{7}$). The mass flux of the fluid is the rate of mass flow per unit area. The mass flux is measured in mass per unit time per unit area. How could we calculate the mass flux of the fluid across $S$?

Figure $\PageIndex{7}$: Fluid flows across a completely permeable surface $S$.

The rate of flow, measured in mass per unit time per unit area, is $\rho \mathbf N$. To calculate the mass flux across $S$, chop $S$ into small pieces $S_{ij}$. If $S_{ij}$ is small enough, then it can be approximated by a tangent plane at some point $P$ in $S_{ij}$. Therefore, the unit normal vector at $P$ can be used to approximate $\mathbf N(x,y,z)$ across the entire piece $S_{ij}$ because the normal vector to a plane does not change as we move across the plane. The component of the vector $\rho v$ at P in the direction of $\mathbf{N}$ is $\rho \mathbf v \cdot \mathbf N$ at $P$. Since $S_{ij}$ is small, the dot product $\rho v \cdot N$ changes very little as we vary across $S_{ij}$ and therefore $\rho \mathbf v \cdot \mathbf N$ can be taken as approximately constant across $S_{ij}$. To approximate the mass of fluid per unit time flowing across $S_{ij}$ (and not just locally at point $P$), we need to multiply $(\rho \mathbf v \cdot \mathbf N) (P)$ by the area of $S_{ij}$. Therefore, the mass of fluid per unit time flowing across $S_{ij}$ in the direction of $\mathbf{N}$ can be approximated by $(\rho \mathbf v \cdot \mathbf N)\Delta S_{ij}$ where $\mathbf{N}$, $\rho$ and $\mathbf{v}$ are all evaluated at $P$ (Figure $\PageIndex{22}$). This is analogous to the flux of two-dimensional vector field $\mathbf{F}$ across plane curve $C$, in which we approximated flux across a small piece of $C$ with the expression $(\mathbf{F} \cdot \mathbf{N}) \,\Delta s$. To approximate the mass flux across $S$, form the sum

$$\sum_{i=1}m \sum_{j=1}^n (\rho \mathbf{v} \cdot \mathbf{N}) \Delta S_{ij}. \nonumber$$

As pieces $S_{ij}$ get smaller, the sum

$$\sum_{i=1}m \sum_{j=1}^n (\rho \mathbf{v} \cdot \mathbf{N}) \Delta S_{ij} \nonumber$$

gets arbitrarily close to the mass flux. Therefore, the mass flux is

$$\iint_s \rho \mathbf v \cdot \mathbf N \, dS = \lim_{m,n\rightarrow\infty} \sum_{i=1}^m \sum_{j=1}^n (\rho \mathbf{v} \cdot \mathbf{N}) \Delta S_{ij}.$$

This is a surface integral of a vector field. Letting the vector field $\rho \mathbf{v}$ be an arbitrary vector field $\mathbf{F}$ leads to the following definition.

Figure $\PageIndex{8}$: The mass of fluid per unit time flowing across $S_{ij}$ in the direction of $\mathbf N$ can be approximated by $(\rho \mathbf v \cdot \mathbf N) \Delta S_{ij}$.

Definition: Surface Integrals

Let $\mathbf{F}$ be a continuous vector field with a domain that contains oriented surface $S$ with unit normal vector $\mathbf{N}$. The surface integral of $\mathbf{F}$ over $S$ is

$$\iint_S \mathbf{F} \cdot \mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{N} \,dS. \label{surfaceI}$$

Notice the parallel between this definition and the definition of vector line integral $\displaystyle \int_C \mathbf F \cdot \mathbf N\, dS$. A surface integral of a vector field is defined in a similar way to a flux line integral across a curve, except the domain of integration is a surface (a two-dimensional object) rather than a curve (a one-dimensional object). Integral $\displaystyle \iint_S \mathbf F \cdot \mathbf N\, dS$ is called the flux of $\mathbf{F}$ across $S$, just as integral $\displaystyle \int_C \mathbf F \cdot \mathbf N\,dS$ is the flux of $\mathbf F$ across curve $C$. A surface integral over a vector field is also called a flux integral.

Just as with vector line integrals, surface integral $\displaystyle \iint_S \mathbf F \cdot \mathbf N\, dS$ is easier to compute after surface $S$ has been parameterized. Let $\mathbf r(u,v)$ be a parameterization of $S$ with parameter domain $D$. Then, the unit normal vector is given by $\mathbf N = \dfrac{\mathbf t_u \times \mathbf t_v}{||\mathbf t_u \times \mathbf t_v||}$ and, from Equation $\ref{surfaceI}$, we have

$$\begin{align*} \int_C \mathbf F \cdot \mathbf N\, dS &= \iint_S \mathbf F \cdot \dfrac{\mathbf t_u \times \mathbf t_v}{||\mathbf t_u \times \mathbf t_v||} \,dS \\[4pt] &= \iint_D \left(\mathbf F (\mathbf r (u,v)) \cdot \dfrac{\mathbf t_u \times \mathbf t_v}{||\mathbf t_u \times \mathbf t_v||} \right) || \mathbf t_u \times \mathbf t_v || \,dA \\[4pt] &= \iint_D (\mathbf F(\mathbf r(u,v)) \cdot (\mathbf t_u \times \mathbf t_v))\,dA. \end{align*}$$

Therefore, to compute a surface integral over a vector field we can use the equation

$$\iint_S \mathbf F \cdot \mathbf N\, dS = \iint_D (\mathbf F (\mathbf r (u,v)) \cdot (\mathbf t_u \times \mathbf t_v)) \,dA. \label{equation 5}$$

Example $\PageIndex{6}$: Calculating a Surface Integral

Calculate the surface integral

$$\iint_S \mathbf F \cdot \mathbf N\,dS, \nonumber$$

where $\mathbf{F} = \langle -y,x,0\rangle$ and $S$ is the surface with parameterization

$$\mathbf r(u,v) = \langle u,v^2 - u, \, u + v\rangle, \, 0 \leq u \leq 3, \, 0 \leq v \leq 4. \nonumber$$

Solution

The tangent vectors are $\mathbf t_u = \langle 1,-1,1\rangle$ and $\mathbf t_v = \langle 0,2v,1\rangle$. Therefore,

$$\mathbf t_u \times \mathbf t_v = \langle -1 -2v, -1, 2v\rangle. \nonumber$$

By Equation $\ref{equation 5}$,

$$\begin{align*} \iint_S \mathbf F \cdot dS &= \int_0^4 \int_0^3 F (\mathbf r(u,v)) \cdot (\mathbf t_u \times \mathbf t_v) \, du \,dv \\[4pt] &= \int_0^4 \int_0^3 \langle u - v^2, \, u, \, 0\rangle \cdot \langle -1 -2v, \, -1, \, 2v\rangle \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 [(u - v^2)(-1-2v) - u] \, du\,dv \\[4pt] &= \int_0^4 \int_0^3 (2v^3 + v^2 - 2uv - 2u) \, du\,dv \\[4pt] &= \int_0^4 \left.[2v^3u + v^2u - vu^2 - u^2]\right|_0^3 \, dv \\[4pt] &= \int_0^4 (6v^3 + 3v^2 - 9v - 9) \, dv \\[4pt] &= \left[ \dfrac{3v^4}{2} + v^3 - \dfrac{9v^2}{2} - 9v\right]_0^4\\[4pt] &= 340. \end{align*}$$

Therefore, the flux of $\mathbf{F}$ across $S$ is 340.

Exercise $\PageIndex{5}$

Calculate surface integral $$\iint_S \mathbf F \cdot \mathbf N \, dS, \nonumber$$ where $\mathbf F = \langle 0, -z, y \rangle$ and $S$ is the portion of the unit sphere in the first octant with outward orientation.

Answer

0

Example $\PageIndex{7}$:Calculating Mass Flow Rate

Let $\mathbf v(x,y,z) = \langle 2x, \, 2y, \, z\rangle$ represent a velocity field (with units of meters per second) of a fluid with constant density 80 kg/m³. Let $S$ be hemisphere $x^2 + y^2 + z^2 = 9$ with $z \leq 0$ such that $S$ is oriented outward. Find the mass flow rate of the fluid across $S$.

Solution

A parameterization of the surface is

$$\mathbf r(\phi, \theta) = \langle 3 \, \cos \theta \, \sin \phi, \, 3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle, \, 0 \leq \theta \leq 2\pi, \, 0 \leq \phi \leq \pi/2.$$

As in Example, the tangent vectors are $\mathbf t_{\theta} = \langle -3 \, \sin \theta \, \sin \phi, \, 3 \, \cos \theta \, \sin \phi, \, 0 \rangle$ and $\mathbf t_{\phi} = \langle 3 \, \cos \theta \, \cos \phi, \, 3 \, \sin \theta \, \cos \phi, \, -3 \, \sin \phi \rangle,$ and their cross product is

$$\mathbf t_{\phi} \times \mathbf t_{\theta} = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle.$$

Notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. Therefore we use the orientation

$\mathbf N = \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle$

for the sphere.

By \label{surfaceI},

$$\begin{align*} \iint_S \rho v \cdot \,dS &= 80 \int_0^{2\pi} \int_0^{\pi/2} v (r(\phi, \theta)) \cdot (t_{\phi} \times t_{\theta}) \, d\phi \, d\theta \\ &= 80 \int_0^{2\pi} \int_0^{\pi/2} \langle 6 \, \cos \theta \, \sin \phi, \, 6 \, \sin \theta \, \sin \phi, \, 3 \, \cos \phi \rangle \cdot \langle 9 \, \cos \theta \, \sin^2 \phi, \, 9 \, \sin \theta \, \sin^2 \phi, \, 9 \, \sin \phi \, \cos \phi \rangle \, d\phi \, d\theta \\ &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 \, \sin^3 \phi + 27 \, \cos^2 \phi \, \sin \phi \, d\phi \, d\theta \\ &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54 (1 - \cos^2\phi) \, \sin \phi + 27 \cos^2\phi \, \sin \phi \, d\phi \, d\theta \\ &= 80 \int_0^{2\pi} \int_0^{\pi/2} 54\, \sin \phi - 27 \, \cos^2 \phi \, \sin \phi \, d\phi \,d\theta \\ &= 80 \int_0^{2\pi} \Big[-54 \, \cos \phi + 9 \, \cos^3 \phi \Big]_{\phi=0}^{\phi=2\pi} \, d\theta \\ &=80 \int_0^{2\pi} 45 \, d\theta \\ &= 7200\pi.\end{align*}$$

Therefore, the mass flow rate is $7200\pi \, \text{kg/sec/m}^2$.

Exercise $\PageIndex{6}$

Let $\mathbf v(x,y,z) = \langle x^2 + y^2, \, z, \, 4y \rangle$ m/sec represent a velocity field of a fluid with constant density 100 kg/m³. Let $S$ be the half-cylinder $\mathbf r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u \leq \pi, \, 0 \leq v \leq 2$ oriented outward. Calculate the mass flux of the fluid across $S$.

Hint

Use \label{surfaceI}.

Answer

400 kg/sec/m

If we want to find the flow rate (measured in volume per time), we can use flux integral

$$\iint_S \mathbf v \cdot \mathbf N \, dS,$$

which leaves out the density. Since the flow rate of a fluid is measured in volume per unit time, flow rate does not take mass into account. Therefore, we have the following characterization of the flow rate of a fluid with velocity $\mathbf v$ across a surface $S$:

$$\text{Flow rate of fluid across S} = \iint_S \mathbf v \cdot dS. \nonumber$$

To compute the flow rate of the fluid in Example, we simply remove the density constant, which gives a flow rate of $90 \pi \, m^3/sec$.

Both mass flux and flow rate are important in physics and engineering. Mass flux measures how much mass is flowing across a surface; flow rate measures how much volume of fluid is flowing across a surface.

In addition to modeling fluid flow, surface integrals can be used to model heat flow. Suppose that the temperature at point $(x,y,z)$ in an object is $T(x,y,z)$. Then the heat flow is a vector field proportional to the negative temperature gradient in the object. To be precise, the heat flow is defined as vector field $F = - k \nabla T$, where the constant k is the thermal conductivity of the substance from which the object is made (this constant is determined experimentally). The rate of heat flow across surface S in the object is given by the flux integral

$$\iint_S \mathbf F \cdot dS = \iint_S -k \mathbf \nabla T \cdot dS.$$

Example $\PageIndex{8}$: Calculating Heat Flow

A cast-iron solid cylinder is given by inequalities $x^2 + y^2 \leq 1, \, 1 \leq z \leq 4$. The temperature at point $(x,y,z)$ in a region containing the cylinder is $T(x,y,z) = (x^2 + y^2)z$. Given that the thermal conductivity of cast iron is 55, find the heat flow across the boundary of the solid if this boundary is oriented outward.

Solution

Let $S$ denote the boundary of the object. To find the heat flow, we need to calculate flux integral $$\iint_S -k\mathbf \nabla T \cdot dS.$$ Notice that $S$ is not a smooth surface but is piecewise smooth, since $S$ is the union of three smooth surfaces (the circular top and bottom, and the cylindrical side). Therefore, we calculate three separate integrals, one for each smooth piece of $S$. Before calculating any integrals, note that the gradient of the temperature is $\mathbf \nabla T = \langle 2xz, \, 2yz, \, x^2 + y^2 \rangle$.

First we consider the circular bottom of the object, which we denote $S_1$. We can see that $S_1$ is a circle of radius 1 centered at point $(0,0,1)$ sitting in plane $z = 1$. This surface has parameterization $\mathbf r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 1 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.$

Therefore,

$\mathbf t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle$ and $\mathbf t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle$, and $\mathbf t_u \times \mathbf t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0, \, 0, -v \rangle$.

Since the surface is oriented outward and $S_1$ is the bottom of the object, it makes sense that this vector points downward. By Equation, the heat flow across $S_1$ is

$$\begin{align*}\iint_{S_1} -k \mathbf \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_0^1 \mathbf \nabla T(u,v) \cdot (\mathbf t_u \times \mathbf t_v) \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv \,du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos u, \, 2v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} -\dfrac{1}{4} du \\[4pt] &= \dfrac{55\pi}{2}.\end{align*}$$

Now let’s consider the circular top of the object, which we denote $S_2$. We see that $S_2$ is a circle of radius 1 centered at point $(0,0,4)$, sitting in plane $z = 4$. This surface has parameterization $\mathbf r(u,v) = \langle v \, \cos u, \, v \, \sin u, \, 4 \rangle, \, 0 \leq u < 2\pi, \, 0 \leq v \leq 1.$

Therefore, $\mathbf t_u = \langle -v \, \sin u, \, v \, \cos u, \, 0 \rangle$ and $\mathbf t_v = \langle \cos u, \, v \, \sin u, \, 0 \rangle$, and $\mathbf t_u \times \mathbf t_v = \langle 0, \, 0, -v \, \sin^2 u - v \, \cos^2 u \rangle = \langle 0,0,-v\rangle$.

Since the surface is oriented outward and $S_1$ is the top of the object, we instead take vector $\mathbf t_v \times \mathbf t_u = \langle 0,0,v\rangle$. By Equation, the heat flow across $S_1$ is

$$\begin{align*}\iint_{S_2} -k \mathbf \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_0^1 \mathbf \nabla T(u,v) \cdot\, (\mathbf t_u \times \mathbf t_v) \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2 \cos^2 u + v^2 \sin^2 u \rangle \cdot \langle 0,0, -v\rangle \, dv\,du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 8v \, \cos u, \, 8v \, \sin u, \, v^2\rangle \cdot \langle 0, 0, -v \rangle\, \, dv \,du\\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 -v^3 \, dv \,du = - 55 \int_0^{2\pi} -\dfrac{1}{4} \,du = - \dfrac{55\pi}{2}.\end{align*}$$

Last, let’s consider the cylindrical side of the object. This surface has parameterization $\mathbf r(u,v) = \langle \cos u, \, \sin u, \, v \rangle, \, 0 \leq u < 2\pi, \, 1 \leq v \leq 4$. By Example, we know that $\mathbf t_u \times \mathbf t_v = \langle \cos u, \, \sin u, \, 0 \rangle$. By Equation,

$$\begin{align*} \iint_{S_3} -k \mathbf \nabla T \cdot dS &= - 55 \int_0^{2\pi} \int_1^4 \mathbf \nabla T(u,v) \cdot (\mathbf t_u \times \mathbf t_v) \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_1^4 \langle 2v \, \cos u, \, 2v \, \sin u, \, \cos^2 u + \sin^2 u \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 \langle 2v \, \cos^2 u, \, 2v \, \sin u, \, 1 \rangle \cdot \langle \cos u, \, \sin u, \, 0 \rangle \, dv\,\, du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 (2v \, \cos^2 u + 2v \, \sin^2 u ) \, dv \,du \\[4pt] &= - 55 \int_0^{2\pi} \int_0^1 2v \, dv \,du \\[4pt] &= -55 \int_0^{2\pi} du \\[4pt] &= -110\pi. \end{align*}$$

Therefore, the rate of heat flow across $S$ is

$$\dfrac{55\pi}{2} - \dfrac{55\pi}{2} - 110\pi = -110\pi. \nonumber$$

Exercise $\PageIndex{7}$

A cast-iron solid ball is given by inequality $x^2 + y^2 + z^2 \leq 1$. The temperature at a point in a region containing the ball is $T(x,y,z) = \dfrac{1}{3}(x^2 + y^2 + z^2)$. Find the heat flow across the boundary of the solid if this boundary is oriented outward.

Answer

$-\dfrac{440\pi}{3}$

Key Concepts

• Surfaces can sometimes be oriented, just as curves can be oriented. Some surfaces, such as a Möbius strip, cannot be oriented.
• A surface integral is like a line integral in one higher dimension. The domain of integration of a surface integral is a surface in a plane or space, rather than a curve in a plane or space.
• The integrand of a surface integral can be a scalar function or a vector field. To calculate a surface integral with an integrand that is a function, use Equation. To calculate a surface integral with an integrand that is a vector field, use Equation.

Key Equations

• Scalar surface integral

$$\iint_S f(x,y,z) \,dS = \iint_D f (\mathbf r(u,v)) ||\mathbf t_u \times \mathbf t_v||\,dA \nonumber$$

• Flux integral

$$\iint_S \mathbf F \cdot \mathbf N \, dS = \iint_S \mathbf F \cdot dS = \iint_D \mathbf F (\mathbf r (u,v)) \cdot (\mathbf t_u \times \mathbf t_v) \, dA \nonumber$$

Glossary

Flux integral

another name for a surface integral of a vector field; the preferred term in physics and engineering

Grid curves

curves on a surface that are parallel to grid lines in a coordinate plane

Heat flow

a vector field proportional to the negative temperature gradient in an object

Mass flux

the rate of mass flow of a fluid per unit area, measured in mass per unit time per unit area

Orientation of a surface

if a surface has an “inner” side and an “outer” side, then an orientation is a choice of the inner or the outer side; the surface could also have “upward” and “downward” orientations

Surface integral

an integral of a function over a surface

Surface integral of a scalar-valued function

a surface integral in which the integrand is a scalar function

Surface integral of a vector field

a surface integral in which the integrand is a vector field

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.