Why do you add a function to the integral of surface integrals?

Well because surface integrals can be used for much more than just computing surface areas. It's like with triple integrals, how you use them for volume computations a lot, but in their full glory they can associate any function with a 3-d region, not just the function f(x,y,z)=1, which is how the volume computation ends up going. Double integrals also can compute volume, but if you let f(x,y)=1, then double integrals boil down to the capabilities of a plain single-variable definite integral (which can compute areas). Having an integrand allows for more possibilities with what the integral can do for you.

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Multivariable calculus

Course: Multivariable calculus > Unit 4

Lesson 12: Surface integrals (articles)

Surface integrals

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How do you add up infinitely many infinitely small quantities associated with points on a surface?

Background

Not strictly required, but useful for intuition and analogy:

Line integrals

What we're building to

In principle, the idea of a surface integral is the same as that of a double integral, except that instead of "adding up" points in a flat two-dimensional region, you are adding up points on a surface in space, which is potentially curved. The abstract notation for surface integrals looks very similar to that of a double integral:

$\begin{aligned} \underbrace{ \iint_S }_{\text{$S$ represents a surface}} \!\!\!\!\!\!\!\!\!\!\!\! f(x, y, z)\, \!\!\!\!\!\!\!\!\!\!\!\! \overbrace{ d\Sigma }^{\text{Tiny piece of area in $S$}} \end{aligned}$

Computing a surface integral is almost identical to computing surface area using a double integral, except that you stick a function inside the integral:

$\begin{aligned} \iint_T f(\vec{\textbf{v}}(\blueE{t}, \redE{s})) \underbrace{ \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right| \,\blueE{dt}\,\redE{ds} }_{\text{Tiny piece of area}} \end{aligned}$

Here, start bold text, v, end bold text, with, vector, on top, left parenthesis, start color #0c7f99, t, end color #0c7f99, comma, start color #bc2612, s, end color #bc2612, right parenthesis is a function parameterizing the surface S from the region T of the start color #0c7f99, t, end color #0c7f99, start color #bc2612, s, end color #bc2612-plane.

(This is analogous to how computing line integrals is basically the same as computing arc length integrals, except that you throw a function inside the integral itself.)

You can find an example of working through one of these integrals in the next article.

The idea of surface integrals

If you understand double integrals, and you understand how to compute the surface area of a parametric surface, you basically already understand surface integrals. It's just a matter of smooshing the two intuitions together. I'll go over the computation of a surface integral with an example in just a bit, but first, I think it's important for you to have a good grasp on what exactly a surface integral does.

Refresher of double integrals

Recall what a double integral does:

$\begin{aligned} \iint_R f(x, y)\,dA \end{aligned}$

Here, R represents some region of the x, y-plane, and f, left parenthesis, x, comma, y, right parenthesis is a way to associate each point of R with a number.

Maybe R represents a metal sheet, and f, left parenthesis, x, comma, y, right parenthesis represents the density at each point.
Or perhaps R represents a geographic region, and f, left parenthesis, x, comma, y, right parenthesis represents the temperature at each point.

The double integral provides a way to "add up" the values of f on this region. However, the idea of "adding up" points in a continuous region is vague, so I like to imagine the following process:

Chop up the region R into many tiny pieces.
Multiply the area of each piece, thought of as d, A, by the value of f at one of the points inside that piece.
Add up the resulting values.

For example,

If R represents a metal sheet, and f, left parenthesis, x, comma, y, right parenthesis is a density function, the double integral will give you the mass of the sheet. (Why?)
If R represents a geographic region, and f, left parenthesis, x, comma, y, right parenthesis give the temperature at each location, taking this double integral then dividing by the area of R will give you the average temperature in that region. (Why?)

Double integrals over curved regions

However, why stay so flat? This idea of adding up values over a continuous two-dimensional region can be useful for curved surfaces as well.

What if you are considering the surface of a curved airplane wing with variable density, and you want to find its total mass?
What if you have the temperature for every point on the curved surface of the earth, and you want to figure out the average temperature?

This time, the function f, which represents density, temperature, etc., must take in point of three dimensions since points on the surface live in three dimensions. The abstract notation for integrating a three-variable function f, left parenthesis, x, comma, y, comma, z, right parenthesis over a surface is pretty much the same as the abstract notation for double integrals:

$\begin{aligned} \underbrace{ \iint_S }_{\text{$S$ represents some surface}} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! f(x, y, z)\, \!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{d\Sigma}^{\text{Tiny piece of area on $S$}} \end{aligned}$

(Different authors might use different notation).

This is called a surface integral. The little S under the double integral sign represents the surface itself, and the term d, \Sigma represents a tiny bit of area piece of this surface. You can think about surface integrals the same way you think about double integrals:

Chop up the surface S into many small pieces.
Multiply the area of each tiny piece by the value of the function f on one of the points in that piece.
Add up those values.

Why write d, \Sigma instead of d, A? There's no real difference; each one represents a tiny bit of area of the thing you are integrating over. However, when it comes time to compute things, the way to handle a tiny bit of area on a curved surface is fundamentally different from doing it on a flat surface, so it's worth emphasizing this difference by using a different variable.

How to compute a surface integral

Abstract notation and visions of chopping up airplane wings are all well and good, but how do you actually compute one of these surface integrals? The trick is to sneakily turn it into an ordinary, flat, double integral.

Specifically, the way you tend to represent a surface mathematically is with a parametric function. You'll have some vector-valued function start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, which takes in points on the two-dimensional t, s-plane (lovely and flat), and outputs points in three-dimensional space. You also need to specify the region T of the t, s-plane which maps onto the surface S.

The trick for surface integrals, then, is to find a way of integrating over the flat region T that gives the same effect as integrating over the curved surface S. This requires describing "tiny piece of area" of S in terms of something inside the parameter.

Almost all of the work for this was done in the article on surface area. There, we saw how a tiny rectangle inside T with area start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612 gets transformed into a parallelogram on S with area

\begin{aligned} \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right| \,\blueE{dt}\,\redE{ds} \end{aligned}

For our surface integral desires, this means you expand d, \Sigma as follows:

d, \Sigma, equals, open vertical bar, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #0c7f99, t, end color #0c7f99, end fraction, times, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #bc2612, s, end color #bc2612, end fraction, close vertical bar, start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612

Specifically, here's how to write a surface integral with respect to the parameter space:

$\displaystyle \iint_S f(x, y, z)\, d\Sigma = \iint_T f(\vec{\textbf{v}}(t, s)) \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right| \,\blueE{dt}\,\redE{ds}$ \iint, start subscript, S, end subscript, f, left parenthesis, x, comma, y, comma, z, right parenthesis, d, \Sigma, equals, \iint, start subscript, T, end subscript, f, left parenthesis, start bold text, v, end bold text, with, vector, on top, left parenthesis, t, comma, s, right parenthesis, right parenthesis, open vertical bar, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #0c7f99, t, end color #0c7f99, end fraction, times, start fraction, \partial, start bold text, v, end bold text, with, vector, on top, divided by, \partial, start color #bc2612, s, end color #bc2612, end fraction, close vertical bar, start color #0c7f99, d, t, end color #0c7f99, start color #bc2612, d, s, end color #bc2612

Let's break that down a bit:

$\begin{aligned} \underbrace{ \iint_S }_{\text{Integral over surface}} \!\!\!\!\!\!\!\!\!\!\!\!\! f(x, y, z)\, \!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{ d\Sigma }^{\text{Area of a tiny piece of $S$}} \!\!\!\!\!\!\!\!\!\!\!\!\! = \underbrace{ \iint_T }_{\substack{ \text{Integral in}\\ \text{parameter space} }} \!\!\!\! \overbrace{ f(\vec{\textbf{v}}(t, s)) }^{\substack{ \text{See where each point}\\ \text{$(t, s)$ lands on $S$, then}\\ \text{evalute $f$} }} \!\!\!\!\!\!\!\! \underbrace{ \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right| }_{\substack{ \text{How much a tiny piece of}\\ \text{$T$ is scaled after it is} \\ \text{mapped onto $S$ by $\vec{\textbf{v}}$} }} \!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{ \blueE{dt}\,\redE{ds} }^{\substack{ \text{Area of a tiny}\\ \text{piece of $T$} }} \end{aligned}$

The main thing to focus on here, and what makes computations particularly labor intensive, is the way to express d, \Sigma.

In the next article, you can go through a full example of one of these surface integrals.

Summary

Surface integrals are used anytime you get the sensation of wanting to add a bunch of values associated with points on a surface. This is the two-dimensional analog of line integrals. Alternatively, you can view it as a way of generalizing double integrals to curved surfaces.

$\begin{aligned} \underbrace{ \iint_S }_{\text{$S$ represents a surface}} \!\!\!\!\!\!\!\!\!\!\!\! f(x, y, z)\, \!\!\!\!\!\!\!\!\!\!\!\! \overbrace{ d\Sigma }^{\text{Tiny piece of area in $S$}} \end{aligned}$

Computing a surface integral is almost identical to computing surface area using a double integral, except that you stick a function inside the integral:

$\begin{aligned} \iint_T f(\vec{\textbf{v}}(\blueE{t}, \redE{s})) \left| \dfrac{\partial \vec{\textbf{v}}}{\partial \blueE{t}} \times \dfrac{\partial \vec{\textbf{v}}}{\partial \redE{s}} \right| \,\blueE{dt}\,\redE{ds} \end{aligned}$

Like so many things in multivariable calculus, while the theory behind surface integrals is beautiful, actually computing one can be painfully labor intensive.

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Andréas Elrandsson
Posted 6 years ago. Direct link to Andréas Elrandsson's post “I almost went crazy over ...”
I almost went crazy over this but note that when you are looking for the SURFACE AREA (not surface integral) over some scalar field (z = f(x, y)), meaning that the vector V(x, y) of which you take the cross-product of becomes V(x, y) = (x, y, f(x, y)).

This also means that the cross-product is evaluated too:
sqrt( (partial-x-of-f(x,y))² + (partial-y-of-f(x,y))² + 1 )

Since we are looking for the area and not the integral, the equation then becomes:
double-integral-over-S (dS) =
double-integral-over-S (sqrt( (partial-x-of-f(x,y))² + (partial-y-of-f(x,y))² + 1 ) dxdy )

Better explanation: http://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceArea.aspx
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Aiman
Posted 3 years ago. Direct link to Aiman's post “Why do you add a function...”
Why do you add a function to the integral of surface integrals?
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- 𝜏 Is Better Than 𝝅
  Posted 3 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “Well because surface inte...”
  Well because surface integrals can be used for much more than just computing surface areas. It's like with triple integrals, how you use them for volume computations a lot, but in their full glory they can associate any function with a 3-d region, not just the function f(x,y,z)=1, which is how the volume computation ends up going. Double integrals also can compute volume, but if you let f(x,y)=1, then double integrals boil down to the capabilities of a plain single-variable definite integral (which can compute areas). Having an integrand allows for more possibilities with what the integral can do for you.
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Surya Raju
Posted 3 years ago. Direct link to Surya Raju's post “What about surface integr...”
What about surface integrals over a vector field?
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- 1564538
  Posted 2 months ago. Direct link to 1564538's post “That is known as flux”
  That is known as flux
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Qasim Khan
Posted 5 years ago. Direct link to Qasim Khan's post “Wow thanks guys! I unders...”
Wow thanks guys! I understood this even though I'm just a senior at high school and I haven't read the background material on double integrals or even Calc II
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- benvessely
  Posted 2 years ago. Direct link to benvessely's post “Wow what you're crazy sma...”
  Wow what you're crazy smart how do you get this without any of that background?
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