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Surface Integrals of Scalar Valued Functions

This section relies parameterized surfaces, which was first introduced in . While some of the motivation for this section came from our treatment of flux integrals, it is not necessary to cover flux integrals first.

Surface Integrals of Scalar Valued Functions

This section relies parameterized surfaces, which was first introduced in . While some of the motivation for this section came from our treatment of flux integrals, it is not necessary to cover flux integrals first.

Introduction

In we looked at flux integrals measure how much of a vector field flows through a given surface. In other words we calculated the accumulation of of the component of the vector field that is normal to the surface over all the points on the surface. In this section, we will generalize this ideas to calculate the accumluation of other scalar functions over a region given by a curved surface. The Preview Activity below will use our mining example that motivated several of our ideas related to line integrals to understand a few key ideas.

Exploration

In we looked at how to understand line integrals of scalar functions through the analogy of running a mining machine along a given path. In short, the amount of copium mined depended on the density of copium at points on the path and the length of the path driven by the mining rig.

The opening tasks of had you estimating the amount of copium mined by driving along the edge of a plot of land (drawn in yellow on ). This interpretation meant that the scalar function we were using measured the linear density of copium. Thus, the Riemann sum we computed was the product of linear density of copium with the distance traveled.

In this task, we will interpret as a contour graph of the density of copium per unit area. This will allow us to compute the total amount of copium in our mining area by setting up a double integral. To approximate the total amount of copium available in our mine, do the following:

Break the mining plot (the area inside the yellow segments of ) into three pieces. Estimate the area of the three pieces you are using. Write a few sentences explaining your methods of estimating the areas.

For each of the three pieces you used in part, estimate the average density of copium on the piece. Write a few sentences explaining your methods of estimating the average density on each piece.

Give an estimate for the total amount of copium on the mining plot and explain your computation.

What if instead of your mining plot being on a flat piece of land as represented in , your mining plot was on a hill as represented in . If we had the same copium density plot as a function of the \((x,y)\) coordinates, which of the following would you expect to be true?

  • the total amount of copium available on is greater than on
  • the total amount of copium available on is lesser than on
  • the total amount of copium available on is the same as on
Explain your reasoning for your choice.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Defining surface integrals of scalar functions

As you might expect, we can set a classic calculus approach to calculate the accumulation of our scalar valued function over the points on a curved surface. This process is given in more detail in the proof of but proceeds like in our work for calculating surface areas and flux integrals. Namely, we will split our curved surface into smaller piecese, then we will utilize a parameterization to measure the area of the piece and convert the \(xyz\)-coordinates of our point and the associated scalar function to a double integral in our parameters, usually \(s\) and \(t\).

In the following example, we both reason about the value of a scalar surface integral without working through computations and apply to compute values of scalar surface integrals.

Before moving on to an activity that gives you a chance to practice reasoning about scalar surface integrals, we consider an additional example that will highlight why our algebraic calcultions are vital in cases where there is not a simple geometric argument.

In the next activity, we consider some situations where we can reason about the sign of a surface integral of a scalar function.

Activity

In this activity, we will try to understand the scalar surface integral by looking at whether the value of the scalar surface integral will be positive, negative, or zero over common surfaces. In each part below, you are given a function and a surface. For each surface, first draw a plot of the surface and make sure you have labeled a proper scale for each coordinate direction. Then reason if the given surface integral is positive, negative, or zero. Be sure to justify your answers in terms of the function being integrated and the particulars of the surface of integration.

For \(S_1\) defined as the top half (\(z \geq 0\)) of the sphere of radius one centered at the origin, consider the surface integral \(\iint_{S_1} x \, dS\).

For \(S_2\) defined as the bottom half (\(z \leq 0\)) of the sphere of radius one centered at the origin, consider the surface integral \(\iint_{S_2} z \, dS\).

For \(S_3\) defined as the disc of radius one centered at \((1,0,0)\) on the plane \(x=1\), consider the surface integral \(\iint_{S_3} x+z \, dS\).

For \(S_1\) as defined above, consider the surface integral \(\iint_{S_1} x+z \, dS\).

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Properties of Scalar Surface Integrals

Before stating some useful properties of scalar line integrals, we will recall some convenient notation. If \(S_1\) and \(S_2\) are disjoint surfacesTechnically, the surfaces may intersect, but there are restrictions on the manner in which they can intersect, and we will not go into the details here., we denote by \(S_1+S_2\) the surface containing every point that is in \(S_1\) or \(S_2\). Also, if \(S_1\) is a surface, then \(-S_1\) denotes the same surface but parameterized in such a way that the normal vector points in the opposite direction. The list below summarizes some other properties of scalar surface integrals, each of which has a familiar analogue amongst the properties of other integrals we have studied.

For a constant scalar \(k\), scalar valued functions\(f\) and \(g\), and oriented surfaces \(S_1\)and \(S_2\), the following properties hold:

  1. \(\displaystyle \iint_{S_1} (k f) \, dS = k \iint_{S_1} f \, dS\)

  2. \(\displaystyle\iint_{S_1} (f+g) \, dS = \iint_{S_1} f dS + \iint_{S_1} g \, dS\)

  3. \(\displaystyle\iint_{-S_1} f \, dS = \iint_{S_1} f \, dS\)

  4. \(\displaystyle\iint_{S_1+S_2} f \, dS = \iint_{S_1} f \, dS + \iint_{S_2} f \, dS\)

We will conclude this section with one more example of computing a scalar surface integral and then an activity that asks you to compute one for yourself using a parameterization of a surface.

Condensed — the full section is in Boelkins et al., Active Calculus Multivariable.

Symbols used here

\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Surface Integrals of Scalar Valued Functions

  1. How can we measure the accumulation of a scalar-valued function along a surface in space?
  2. What does that accumulation measure?
  3. How can we efficiently calculate scalar surface integrals?

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Poskusi sam.

Parts of this page are adapted from Boelkins et al., Active Calculus Multivariable (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Več v Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems