Surface Integral
How-To w/ Step-by-Step Examples!

What’s the difference between line integrals and surface integrals?

A line integral evaluates a function of two variables along a line, whereas a surface integral calculates a function of three variables over a surface.

And just as line integrals has two forms for either scalar functions or vector fields, surface integrals also have two forms:

• Surface integrals of scalar functions.
• Surface integrals of vector fields.

Let’s take a closer look at each form of the surface integral.

Scalar Field

The surface integral of a function \(f\left( {x,y,z} \right)\) over a surface \(S\) is written \(\iint_{S} f(x, y, z) d S\), where \(dS\) stands for the infinitesimal amount of surface area. Thus, the surface integral of a function can be written as:

\begin{equation}
\iint_{S} f(x, y, z) d S=\iint_{D} f(x, y, z) \sqrt{1+\left(\frac{\partial f}{\partial x}\right)^{2}+\left(\frac{\partial f}{\partial x}\right)^{2}} d A
\end{equation}

…where \(D\) is the projection onto the xy-plane.

But sometimes, even this formula is difficult to calculate.

Thankfully, we have another trick up our sleeves…

…Parameterization!

Parametric Surface

Let’s see how to compute the surface integral of a function over a parametric surface.

Let \(S\) be a smooth surface described parametrically by \(\vec r\left( {u,v} \right)\) where \(\left( {u,v} \right) \in D\), and suppose the domain of \(f\left( {x,y,z} \right)\) includes \(S\). Then the surface integral of \(f\) over \(S\) with respect to surface area is:

\begin{equation}
\iint_{S} f(x, y, z) d S=\iint_{D} f(\vec{r}(u, v))\left\|\vec{r}_{u} \times \vec{r}_{v}\right\| d A
\end{equation}

Example

Let’s look at an example.

Compute the surface integral of \(f\left( {x,y,z} \right) = yz\) over \(S\) with respect to surface area if \(S\) is described parametrically as \(\vec r\left( {u,v} \right) = \left\langle {{u^2},u\sin v,u\cos v} \right\rangle \) with \(u \in \left[ {0,1} \right],v \in \left[ {0,\frac{\pi }{2}} \right]\).

First, we need to find our first order partials.

\begin{equation}
\vec{r}_{u}=\langle 2 u, \sin v, \cos v\rangle \text { and } \vec{r}_{v}=\langle 0, u \cos v,-u \sin v\rangle
\end{equation}

Next, we will compute the cross product \({\vec r_u} \times {\vec r_v}\).

\begin{equation}
\vec{r}_{u} \times \vec{r}_{v}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
2 u & \sin v & \cos v \\
0 & u \cos v & -u \sin v
\end{array}\right|
\end{equation}

\begin{equation}
=\left(-\sin ^{2} v-u \cos ^{2} v\right) \vec{i}-\left(-2 u^{2} \sin v\right) \vec{j}+\left(2 u^{2} \cos v\right) \vec{k}
\end{equation}

\begin{equation}
\begin{aligned}
&=\left(-u\left(\sin ^{2} v+\cos ^{2} v\right)\right) \vec{i}-\left(-2 u^{2} \sin v\right) \vec{j}+\left(2 u^{2} \cos v\right) \vec{k} \\
&=(-u) \vec{i}-\left(-2 u^{2} \sin v\right) \vec{j}+\left(2 u^{2} \cos v\right) \vec{k} \\
&=\left\langle-u, 2 u^{2} \sin v, 2 u^{2} \cos v\right\rangle
\end{aligned}
\end{equation}

Now we will find the magnitude of the cross product \(\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\).

\begin{equation}
\left\|\vec{r}_{u} \times \vec{r}_{v}\right\|=\sqrt{(-u)^{2}+\left(2 u^{2} \sin v\right)^{2}+\left(2 u^{2} \cos v\right)^{2}}
\end{equation}

\begin{equation}
=\sqrt{u^{2}+4 u^{4} \sin ^{2} v+4 u^{4} \cos ^{2} v}
\end{equation}

\begin{equation}
\begin{aligned}
&=\sqrt{u^{2}+4 u^{4}\left(\sin ^{2} v+\cos ^{2} v\right)}=\sqrt{u^{2}+4 u^{4}} \\
&=\sqrt{u^{2}\left(1+4 u^{2}\right)} \\
&=u \sqrt{1+4 u^{2}}
\end{aligned}
\end{equation}

Now we will parameterize our integrand.

If\(f\left( {x,y,z} \right) = yz\) and \(\vec r\left( {u,v} \right) = \left\langle {\underbrace {{u^2}}_x,\underbrace {u\sin v}_y,\underbrace {u\cos v}_z} \right\rangle \) , then

\begin{equation}
f(x, y, z)=f(\vec{r}(u, v))=(u \sin v)(u \cos v)=u^{2} \sin v \cos v
\end{equation}

Lastly, we substitute everything into our surface integral formula and solve.

\begin{equation}
\iint_{S} f(x, y, z) d S=\iint_{D} f(\vec{r}(u, v))\left\|\vec{r}_{u} \times \vec{r}_{v}\right\| d A
\end{equation}

\begin{equation}
\text { with } 0 \leq u \leq 1,0 \leq v \leq \frac{\pi}{2}
\end{equation}

\begin{equation}
\iint_{S}(y z) d S=\int_{0}^{1} \int_{0}^{\frac{\pi}{2}}\left(u^{2} \sin v \cos v\right)\left(u \sqrt{1+4 u^{2}}\right) d v d u
\end{equation}

\begin{equation}
=\frac{1}{64}\left(\frac{50 \sqrt{5}-2}{5}+\frac{10 \sqrt{5}-2}{3}\right) \approx 0.449
\end{equation}

And that’s it.

All we have to do is follow the steps.

Surface Integral Orientation

Now it’s time to turn our attention to vector fields and the flow of vector fields through permeable surfaces placed in the field.

But before we do, we need to talk about orientation. In particular, finding the positive orientation for a closed-oriented surface.

Alright, so most surfaces have two sides (i.e., inside and outside), but some surfaces, such as a Mobius strip, are only one-sided and thus deemed nonorientable.

Thankfully, we will strictly be dealing with orientable or two-sided surfaces!

Why?

Because they come armed with a unit normal vector that provides two orientations for a surface quite nicely.

For a surface \(z = g\left( {x,y} \right)\) the orientation by the unit normal vector is:

\begin{equation}
\vec{n}=\frac{\left\langle-\frac{\partial g}{\partial x},-\frac{\partial g}{\partial x}, 1\right\rangle}{\sqrt{1+\left(\frac{\partial g}{\partial x}\right)^{2}+\left(\frac{\partial g}{\partial y}\right)^{2}}}
\end{equation}

And a parametric smooth surface \(S:\left\{ {\vec r\left( {u,v} \right):\left( {u,v} \right) \in D} \right\}\) the orientation of the unit normal vector is:

\begin{equation}
\vec{n}=\frac{\vec{r}_{u} \times \vec{r}_{v}}{\left\|\vec{r}_{u} \times \vec{r}\right\|}
\end{equation}

So, for a surface that is the boundary of a solid E (i.e., closed surface), the convention is that the positive orientation is the one for which the normal vectors point outward from \(E\). Inward-pointing normal vectors give the negative orientation.

Vector Fields

Okay, so now we are ready to talk about surface integrals of vector fields!

If \(\vec F\) is a continuous vector field defined on an oriented surface \(S\) with unit normal vector \(\vec n\), then the surface integral of \(\vec F\) over \(S\) is:

\begin{equation}
\iint_{S} \vec{F} \cdot d \vec{S}=\iint_{S} \vec{F} \cdot \vec{n} d S
\end{equation}

This integral is also called the flux of \(\vec F\) across \(S\).

Why?

Because the surface integral of a vector field represents the amount of fluid flowing through the surface per unit of time:

\begin{equation}
\text { Flux }=(\text { flow rate })(\text { segment length })
\end{equation}

So, if a surface integral measures the total rate of flow, then we can simply call it a flux integral, because that’s exactly what we are calculating…the flux!

But once again, our current formula is still a bit difficult to use. Thankfully, there is an alternate form for calculating the flux integral.

Flux Integral – Alternate Form

Suppose \(S\) is a smooth, two-sided surface given by \(z=f(x, y, z)\) where \(\left( {x,y} \right)\) is in \(D\) and let \(\vec n\) denote the upward unit normal vector on \(S\). And if \(f\) has continuous first partial derivatives and \(\vec F = \left\langle {P,Q,R} \right\rangle \) is a continuous vector field, then the flux of \(\vec F\) across \(S\) is given by:

\begin{equation}
\iint_{S} \vec{F} \cdot d \vec{S}=\iint_{D}\left(-P \frac{\partial f}{\partial x}-Q \frac{\partial f}{\partial y}+R\right) d A
\end{equation}

OR

\begin{equation}
\iint_{S} \vec{F} \cdot d \vec{S}=\iint_{D} \vec{F}(\vec{r}(u, v)) \cdot\left(\vec{r}_{u} \times \vec{r}_{v}\right) d A
\end{equation}

And together, we will learn how to use these formulas to evaluate the flux of a vector field across a surface in our video lesson.

So, together we will learn why we use surface integrals and which form or formula to utilize when finding the flow rate across a surface.

Let’s jump right in.

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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