Math 344: Calculus III

16.7 Surface Integrals

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16.7.1 Parametric Surfaces

As we did for path integrals along a path, we can think about using surface integrals to measure a function with respect to a surface. Just like when we integrated with respect to arc length for a path integral, we will now integrate with respect to surface area for surface integrals.

The surface integral is motivated by the same sort of Riemann sum we see for surface area. Consider a subrectangle on the parameter domain $D$. If we think about the part of the surface $S_{ij}$ this subrectangle corresponds to, then we can take a sample point $P_{ij}^*$ to plug into $f$.

Definition

If $f(x,y,z)$ is a scalar function and a surface $S$ has a parametrization $\mathbf{r}$ on a rectangular parameter domain $D$ given by

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pzi}[2][]{\dfrac{\partial #1}{\partial #2}} \mathbf{r}(u,v) = x(u,v)\,\ihat + y(u,v)\,\jhat + z(u,v)\,\khat\qquad\qquad (u,v)\in D $$
then the surface integral of $f$ over $S$ is given by

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

As in last section, we make the approximation

$$ \Delta S_{ij} \approx \left|\mathbf{r}_r\times\mathbf{r}_v\right|\Delta u\Delta v $$
where $\mathbf{r}_u = \pzi[x]{u}\ihat + \pzi[y]{u}\jhat + \pzi[z]{u}\khat$ and $\mathbf{r}_v = \pzi[x]{v}\ihat + \pzi[y]{v}\jhat + \pzi[z]{v}\khat$ are the tangent vectors as a corner of $S_{ij}$. This cross product yields a vector whose magnitude is approximately equal to the area of $\Delta S_{ij}$.

From here, we may write the surface integral in terms of the parametrization $\mathbf{r}(u,v)$, yielding the much more practical formula

$$ \iint_S f(x,y,z)\,dS = \iint_D f(\mathbf{r}(u,v))\left|\mathbf{r}_r\times\mathbf{r}_v\right|\,dA $$
We compare with the formula for arc length

$$ \int_C f(x,y,z)\,ds = \int_a^b f(\mathbf{r}(t))\left|\mathbf{r}'(t)\right|\,dt $$
In addition, this gives us another way to compute surface area

$$ A(S) = \iint_S 1\,dS = \iint_D \left|\mathbf{r}_r\times\mathbf{r}_v\right|\,dA $$

Example 1

Compute the surface integral $ \iint_S y^2\,dS $, where $S$ is the unit sphere $x^2 + y^2 + z^2 = 1$.

Follow Along

We being with the standard parametrization
$$ x = \sin\phi \cos\theta\qquad y = \sin\phi \sin\theta\qquad z = \cos\phi\qquad\qquad \phi\in[0,\pi],\ \theta\in[0,2\pi] $$
or equivalently,
$$ \mathbf{r}(\phi,\theta) = \sin\phi \cos\theta\,\ihat + \sin\phi \sin\theta\,\jhat + \cos\phi\,\khat $$
We begin by finding
$$ \begin{align*} \mathbf{r}_\phi\times\mathbf{r}_\theta &= \begin{vmatrix} \ihat & \jhat & \khat \\ \pzi[x]{\phi} & \pzi[y]{\phi} & \pzi[z]{\phi} \\ \pzi[x]{\theta} & \pzi[y]{\theta} & \pzi[z]{\theta} \end{vmatrix} \\ \\ &= \begin{vmatrix} \ihat & \jhat & \khat \\ \cos\phi \cos\theta & \cos\phi \sin\theta & -\sin\phi \\ -\sin\phi \sin\theta & \sin\phi \cos\theta & 0 \end{vmatrix} \\ \\ &= \sin^2\!\phi \cos\theta\,\ihat + \sin^2\!\phi\sin\theta\,\jhat + \sin\phi\cos\phi\,\khat \end{align*} $$
and then
$$ \begin{align*} \left|\mathbf{r}_\phi\times\mathbf{r}_\theta\right| &= \sqrt{\sin^4\!\phi \cos^2\!\theta + \sin^4\!\phi \sin^2\!\theta + \sin^2\!\phi \cos^2\!\phi} \\ \\ &= \sqrt{\sin^4\!\phi + \sin^2\!\phi \cos^2\!\phi} \\ \\ &= \sqrt{\sin^2\!\phi} \\ \\ &= \sin\phi \end{align*} $$
Finally, we form the surface integral
$$ \begin{align*} \iint_S y^2\,dS &= \iint_D \left(\sin\phi\sin\theta\right)^2 \left|\mathbf{r}_\phi\times\mathbf{r}_\theta\right|\,dA \\ \\ &= \int_0^{2\pi} \int_0^\pi \sin^2\!\phi \sin^2\!\theta \sin\phi\,d\phi d\theta \\ \\ &= \int_0^{2\pi} \sin^2\!\theta\,d\theta \int_0^\pi \sin^3\!\phi\,d\phi \\ \\ &= \int_0^{2\pi} \frac{1}{2}\left(1 - \cos 2\theta\right)d\theta \cdot \int_0^\pi \underbrace{\left(1-\cos^2\!\phi\right)}_{\sin^2\!\phi} \sin\phi\,d\phi \\ \\ &= \frac{1}{2}\left.\left[\theta - \frac{1}{2}\sin 2\theta\right]\right|_0^{2\pi} \cdot \left.\left[-\cos\phi + \frac{1}{3}\cos^3\!\phi\right]\right|_0^\pi \\ \\ &= \left[ 2\pi \right]\cdot\left[\frac{2}{3}\right] \\ \\ &= \frac{4\pi}{3} \end{align*} $$

16.7.2 Graphs of Functions

If a surface $S$ may be written as a function $z = z(x,y)$, then it may be thought of as a parametric surface with parametrization

$$ x = x\qquad y = y\qquad z = z(x,y) $$
where

$$ \mathbf{r}_x = \ihat + \left(\pzi[z]{x}\right)\khat \qquad\qquad \mathbf{r}_x = \jhat + \left(\pzi[z]{y}\right)\khat $$
Under this parametrization, we have

$$ \mathbf{r}_x \times \mathbf{r}_y = -\pzi[z]{x}\ihat - \pzi[z]{y}\jhat + \khat $$
and

$$ \left|\mathbf{r}_x \times \mathbf{r}_y\right| = \sqrt{\left(\pzi[z]{x}\right)^2 + \left(\pzi[z]{y}\right)^2 + 1 } $$

Definition

The surface integral of the graph of $f$ over $S$ is given by

$$ \iint_S f(x,y,z)\,dS = \iint_D f(x,y,z(x,y))\sqrt{\left(\pzi[z]{x}\right)^2 + \left(\pzi[z]{y}\right)^2 + 1 }\,dA $$

Example 2

Compute the integral

$$ \iint_S x\,dS $$
where $S$ is the triangular region $(1,0,0)$, $(0,4,0)$, and $(0,0,2)$.

Follow Along

The region in question looks like
and an equation of the plane may be found by taking the cross product of the vectors $\langle -1,4,0 \rangle$ and $\langle -1,0,2 \rangle$ and using this as the normal vector. This gives $z = 2 - 2x - \frac{1}{2}y$ as an equation for the plane. Hence,

$$ \pzi[z]{x} = -2 \qquad\qquad \pzi[z]{y} = -\frac{1}{2} $$
Setting $z = 0$, we can see that the bounds of the region are $y = 0$ and $y = 4 - 4x$ with $x\in[0,1]$, and so our integral is

$$ \begin{align*} \iint_S f(x,y,z)\,dS &= \iint_D f(x,y,z(x,y))\sqrt{\left(\pzi[z]{x}\right)^2 + \left(\pzi[z]{y}\right)^2 + 1 }\,dA \\ \\ &= \int_0^1 \int_0^{4-4x} x\sqrt{\left(-2\right)^2 + \left(-\frac{1}{2}\right)^2 + 1 }\,dydx \\ \\ &= \dfrac{\sqrt{21}}{2} \int_0^1 \int_0^{4-4x} x\,dydx \\ \\ &= \dfrac{\sqrt{21}}{2} \int_0^1 \left.\Big[ xy \Big]\right|_{\,y\,=\,0}^{\,y\,=\, 4 - 4x} dx \\ \\ &= \dfrac{\sqrt{21}}{2} \int_0^1 4x - 4x^2\, dx \\ \\ &= \dfrac{\sqrt{21}}{2} \left.\left[ 2x^2 - \frac{4}{3}x^3 \right]\right|_{\,0}^{\,1} \\ \\ &= \dfrac{\sqrt{21}}{2}\left[\dfrac{2}{3}\right] \\ \\ &= \dfrac{\sqrt{21}}{3} \end{align*} $$

16.7.3 Oriented Surfaces

So far, we have been working with surface integrals with respect to scalar fields. To be able to consider surface integrals in vector fields, we need to understand the orientation of the surfaces we work with. Orientable surfaces have two sides which we track with a normal vector $\mathbf{n_1}$ and its opposite $\mathbf{n}_2 = -\mathbf{n}_1$. We make choices so that one of these vectors can be thought of as point in the positive and negative direction with respect to the surface. The figure below shows the positive and negative normal vectors on a surface.

If the surface is defined as a graph with $z = f(x,y)$, then the normal vector is given by

$$ \mathbf{n} = \dfrac{-f_x\,\ihat - f_y\,\jhat + \khat}{\sqrt{(f_x)^2 + (f_y)^2 + 1}} $$
The surface here is oriented upward since the component in the $\khat$ direction is positive. This definition expands to any parametrized orientable surface.

Definition

Orientation of a Parametrized Surface
Is $S$ is a smooth orientable surface parametrized by the function $\mathbf{r}(u,v)$, then its orientation may be expressed with the unit normal vector

$$ \mathbf{n} = \dfrac{\mathbf{r}_u \times \mathbf{r}_v}{\left| \mathbf{r}_u \times \mathbf{r}_v \right|} $$
with the opposite orientation given by $-\mathbf{n}$.

Nonorientable Surfaces

Some surfaces are nonorientable , which means that they cannot represented in a way that is two-sided. Graphs of functions do not have this problem, but many objects like the famous Möbius strip do.

To create a Möbius strip, take a strip a paper longer than it is wide, make a "half twist" in it, then connect the two shorter sides.

The arrow indicators on the sides are meant to represent the twist. If you create the shape, you find that if you trace your finger along the surface that is only has "one side." It is possible to to have normal vectors transition from pointing "outward" to "inward" without ever crossing the edge of the paper strip (now thought of as the boundary of the surface).

This is undesirable, as we cannot come up with consistent definitions for calculus operations on such surfaces. Therefore, we will only be working with orientable surfaces and exclude these cases.

Closed Surfaces

Some surfaces are orientable but have no boundary curve. Such surfaces are called closed surfaces . Actually, closed surfaces can be thought of as the boundary of a solid which is important for several theorems.

A sphere is a classical closed surface, as is gives us a clear notion of what the positive and negative orientation of closed surface represents.

16.7.4 Surface Integrals Over Vector Fields

Next, we consider an oriented surface $S$ with unit normal vector $\mathbf{n}$ in the presence of a vector field. As we typically do, imagine that the vector field is a fluid. This fluid has density $\rho(x,y,z)$ and is flowing with velocity $\mathbf{v}(x,y,z)$ through $S$. The flow is unaffected by $S$, so $S$ may be thought of as a fishing net or similar object. Our goal is to determine how much of the fluid is flowing through the surface.

The mass per unit time per unit area flowing through the net is $\rho\:\!\mathbf{n}$. What we want to do is find the small bit of mass flowing through one "hole" $S_{ij}$ of this imagined net. If the mesh of the net is really fine, then we may approximate the small area with a parallelogram. The flow in the direction $\mathbf{n}$ per unit time is given by $(\rho\:\!\mathbf{v}\cdot\mathbf{n})A(S_{ij})$ where the quantities $\rho$, $\mathbf{v}$, and $\mathbf{n}$ are evaluated at a sample point on $S_{ij}$. Adding all $S_{ij}$ and taking the limit gives us the integral

$$ \iint_S \rho\:\!\mathbf{v}\cdot\mathbf{n}\,dS = \iint_S \rho(x,y,z)\mathbf{v}(x,y,z)\cdot\mathbf{n}(x,y,z)\,dS $$
the rate of flow through the surface $S$. Lastly, we may write a general vector field in $\mathbb{R}^3$ as $\mathbf{F} = \rho\,\mathbf{v}$. This gives us definition of the surface integral of $\mathbf{F}$ over $S$.

Definition

Surface Integral (or Flux) of $\,\mathbf{F}$ Over $S$
If $\,\mathbf{F}$ is a continuous vector field defined on an oriented surface $S$ with unit normal $\mathbf{n}$, then the surface integral of $\,\mathbf{F}$ over $S$ is given by

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_S \mathbf{F}\cdot\mathbf{n}\,dS $$
Alternatively, this is called the flux of $\,\mathbf{F}$ across $S$.

This formula is the one that tends to show up in textbooks and formula lists, but it is much more usably expressed in terms of a parametrization $\mathbf{r}(u,v)$ of the surface $S$ with normal vector $\mathbf{n}$. If the parameter domain for $\mathbf{r}$ is $D$, then

$$ \begin{align*} \iint_S \mathbf{F}\cdot d\mathbf{S} &= \iint_S \mathbf{F}\cdot \dfrac{\mathbf{r}_u \times \mathbf{r}_v}{\left| \mathbf{r}_u \times \mathbf{r}_v \right|}\,dS \\ \\ &= \iint_D \left( \mathbf{F}\left(\mathbf{r}(u,v)\right)\cdot \dfrac{\mathbf{r}_u \times \mathbf{r}_v}{\left| \mathbf{r}_u \times \mathbf{r}_v \right|} \right) \left| \mathbf{r}_u \times \mathbf{r}_v \right|\,dA \\ \\ &= \iint_D \mathbf{F}\cdot \left( \mathbf{r}_u \times \mathbf{r}_v \right)\,dA \end{align*} $$

Example 3

Find the flux of $\mathbf{F}$ across the surface $S$ when

$$ \mathbf{F} = x\,\ihat + y\,\jhat + z^2\,\khat $$
and $S$ is the sphere with radius 1 centered at the origin.

Follow Along

The sphere is parametrized using similar formulas to spherical coordinates

$$ \mathbf{r}(\phi,\theta) = \sin\phi\cos\theta\,\ihat + \sin\phi\sin\theta\,\jhat + \cos\phi\,\khat $$
which means that the vector field may be written as

$$ \mathbf{F}(\phi,\theta) = \sin\phi\cos\theta\,\ihat + \sin\phi\sin\theta\,\jhat + \cos^2\!\phi\,\khat $$
We need to find the normal vector $\mathbf{r}_\phi\times\mathbf{r}_\theta$, which is computed similarly to what is shown in Example 2 from section 16.6.

$$ \mathbf{r}_\phi\times\mathbf{r}_\theta = \sin^2\!\phi\cos\theta\,\ihat + \sin^2\!\phi\sin\theta\,\jhat + \sin\phi\cos\phi\,\khat $$
Next, we want to compute the integrand from our "usable" formula above for a parametrized surface

$$ \begin{align*} \mathbf{F}\left(\mathbf{r}(\phi,\theta)\right)\cdot \left( \mathbf{r}_\phi \times \mathbf{r}_\theta \right) &= \sin^3\!\phi\cos^2\!\theta + \sin^3\!\phi\sin^2\!\theta + \sin\phi\cos^3\!\phi \\ \\ &= \sin^3\!\phi + \sin\phi\cos^3\!\phi \end{align*} $$
Now we compute the integral

$$ \begin{align*} \iint_S \mathbf{F}\cdot d\mathbf{S} &= \iint_D \mathbf{F}\cdot \left( \mathbf{r}_u \times \mathbf{r}_v \right)\,dA \\ \\ &= \int_0^{2\pi}\int_0^\pi \left(\sin^3\!\phi + \sin\phi\cos^3\!\phi\right)d\phi d\theta \\ \\ &= \int_0^{2\pi} d\theta \int_0^\pi \left(\underbrace{1 - \cos^2\!\phi}_{\sin^2\!\phi} + \cos^3\!\phi\right)\sin\phi\,d\phi \\ \\ &= \left(2\pi\right)\left.\left[-\cos\phi + \frac{1}{3}\cos\phi - \frac{1}{4}\cos\phi\right]\right|_{\,0}^{\,\pi} \\ \\ &= 2\pi\left[\left(1 - \frac{1}{3} - \frac{1}{4}\right) - \left(-1 + \frac{1}{3} - \frac{1}{4}\right)\right] \\ \\ &= \frac{8\pi}{3} \end{align*} $$

Exercise 1

Find the flux of

$$ \mathbf{F}(x,y,z) = \sin(xyz)\,\ihat + x^2 y\,\jhat + z^2e^\frac{x}{5}\,\khat $$
across the part of the cylinder $4y^2 + z^2 = 4$ that lies above the $xy$-plane and between the planes $x = -2$ and $x = 2$ with upward orientation. Use a computer algebra system to plot the cylinder and the vector field on the same coordinate system.

Follow Along

Because the surface is above the $xy$-plane, we can use its graph as a parametrization. We solve the given equation $4y^2 + z^2 = 4$ for $z$, yielding

$$ \mathbf{r}(x,y) = x\,\ihat + y\,\jhat + 2\sqrt{1-y^2}\,\khat $$
We need the normal vector for our flux integral,
$$ \begin{align*} \mathbf{r}_x \times \mathbf{r}_y &= -\pzi{x}\left(2\sqrt{1-y^2}\right)\ihat - \pzi{y}\left(2\sqrt{1-y^2}\right)\jhat + \khat \\ \\ &= \dfrac{-2y}{\sqrt{1-y^2}}\,\jhat + \khat \end{align*} $$
Therefore, the flux integral is

$$ \begin{align*} \iint_S \mathbf{F}\cdot d\mathbf{S} &= \iint_D \mathbf{F}\cdot\left(\mathbf{r}_x\times\mathbf{r}_y\right)dA \\ \\ &= \int_{-2}^2 \int_{-1}^1 \left[-x^2y\left(\dfrac{-2y}{\sqrt{1-y^2}}\right) + \left(2\sqrt{1-y^2}\right)^2 e^\frac{x}{5}\right]dxdy \\ \\ &= \int_{-1}^1 \dfrac{2y^2}{\sqrt{1-y^2}}\,dy \int_{-2}^2 x^2\,dx + \int_{-1}^1 4\left(1-y^2\right)\,dy \int_{-2}^2 e^\frac{x}{5}\,dx \end{align*} $$
The step here splits the integral at the sum because both of the integrals you have after dividing it this way are separable.

$$ \begin{align*} &= \left.\left[\arcsin(y)-y\sqrt{1-y^2}\right]\right|_{\,-1}^{\,1}\cdot\left.\left[\dfrac{x^3}{3}\right]\right|_{\,-2}^{\,1} + \left.\left[4\left(y - \dfrac{y^3}{3}\right)\right]\right|_{\,-1}^{\,1} \cdot \left.\left[ 5\left(e^\frac{x}{5}\right)\right]\right|_{\,-2}^{\,2} \\ \\ &= \left[\left(\dfrac{\pi}{2} - 0\right) - \left(-\dfrac{\pi}{2} - 0\right)\right]\cdot\left[\dfrac{8}{3}+\dfrac{8}{3}\right] + 4\left[\left(1-\dfrac{1}{3}\right) - \left(-1+\dfrac{1}{3}\right)\right]\cdot 5\left[e^\frac{2}{5} - e^{-\frac{2}{5}}\right] \\ \\ &= \dfrac{16\pi}{3} + \dfrac{80}{3}\left(e^\frac{2}{5} - e^{-\frac{2}{5}}\right) \end{align*} $$

16.7.5 Surface Integrals with Graphs of Functions

When working with surfaces, it is often the case that the parametrization will be given as the graph of a function. So when you have

$$ x = x\qquad y = x\qquad z = z(x,y) $$
parameterizing $S$ and a vector field $\mathbf{F} = P\,\ihat + Q\,\jhat + R\,\khat$, you can rewrite the expression $ \mathbf{F}\cdot \left(\mathbf{r}_x \times \mathbf{r}_y\right) $ using our work from above:

$$ \mathbf{F}\cdot \left(\mathbf{r}_x \times \mathbf{r}_y\right) = \left(P\,\ihat + Q\,\jhat + R\,\khat\right)\cdot\left(-\pzi[z]{x}\ihat - \pzi[z]{y}\jhat + \khat\right) $$
Plugging this into our surface integral formula gives a very convenient expression

Surface Integral of $\,\mathbf{F}$ over the Graph of $\,\mathbf{S}$

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \left(-P\pzi[z]{x} - Q\pzi[z]{y} + R \right)\,dA $$

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