Applications of unoriented line and surface integrals

Section 7.3 Applications of unoriented line and surface integrals

While most applications of integration over curves and surfaces involve the oriented line and surface integrals that we studied throughout this course, unoriented line and surface integrals can also be useful in applications, when we want to calculate a quantity associated to a curve or a surface that should not depend on the orientation. In this section we study a few applications of unoriented line and surface integrals, such as calculating the centre of mass of a wire and a thin sheet of material.

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Objectives

You should be able to:

Determine and evaluate unoriented line and surface integrals in the context of applications in science.

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Subsection 7.3.1 Centre of mass of a wire

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In the previous sections we studied unoriented line and surface integrals of functions. Those are useful to calculate quantities associated to curves and surfaces that should not depend on a choice of orientation. The prototypical examples were the arc length of a curve and surface area of a surface, but there are many other applications.

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Our first application concerns the calculation of the centre of mass of a wire in

R^{n} .

Let us first recall the physical concept of centre of mass.

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Suppose that there are

k

point particles of masses

m_{1}, \dots, m_{k}

at positions

X_{1}, \dots, X_{k} \in R

on a line. The centre of mass of the system of particles is located at

\bar{x} = \frac{\sum_{i = 1}^{k} m_{i} X_{i}}{m},

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where

m = \sum_{i = 1}^{k} m_{i}

is the total mass of the system. The numerator is sometimes called the “first moment of the system about the origin” and written as

M .

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If we are given instead a rod in

R

between

x = a

and

x = b

with mass density

ρ (x),

then to get its centre of mass we use the slicing principle. We slice the rod into small line segments. Let

d x

be of a typical line segment located at the point

x .

Its mass is

d m = ρ (x) d x,

and its first moment about the origin is

d M = x ρ (x) d x .

We then sum over slices and take the limit of an infinite number of infinitesimal slices. This turns the calculation of the total mass and first moment of the rod as a definite integral, and its centre of mass is:

\bar{x} = \frac{\int_{a}^{b} x ρ (x) d x}{\int_{a}^{b} ρ (x) d x} = \frac{1}{m} \int_{a}^{b} x ρ (x) d x .

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Now what if we want to calculate the centre of mass of a wire that is bent and twisted in

R^{n} ?

We apply the same idea, but thinking of the wire as a parametric curve

α : [a, b] \to R^{n},

with

α (t) = (x_{1} (t), \dots, x_{n} (t))

and image curve

C = α ([a, b]) .

Let

ρ : C \to R

be the mass density of the wire, which we assume to be continuous. We slice the wire (the image curve

C

) into small curve segments. Let

d s

(the line element from Subsection 7.1.1) be the length of a typical curve segment located at

α (t) .

Its mass is

d m = ρ (α (t)) d s .

By summing over all curve segments and taking the limit of an infinite number of segments of infinitesimal length, we calculate the total mass of the wire as an unoriented line integral:

m = \int_{C} ρ d s = \int_{a}^{b} ρ (α (t)) | T (t) | d t .

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To get the centre of mass, we also need to calculate the first moments of the wire. Here, we are working in

R^{n} .

There are

n

first moments, one in each coordinate

x_{1}, \dots, x_{n} .

The first moments of the curve segment are given by

x_{k} (t) ρ (α (t)) d s, k = 1, \dots, n .

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Summing over curve segments, and taking the limit as usual, we obtain that the position of the centre of mass of the wire is given by the point in

R^{n}

with coordinates

{\bar{x}}_{1}, \dots, {\bar{x}}_{n},

with:

{\bar{x}}_{k} = \frac{1}{m} \int_{C} x_{k} ρ d s = \frac{1}{m} \int_{a}^{b} x_{k} (t) ρ (α (t)) | T (t) | d t, k = 1, \dots, n .

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In other words, to find the centre of mass of the wire, we need to evaluate the unoriented line integrals corresponding to the total mass of the wire and its

n

first moments.

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Note that the centre of mass of the wire will not generally be located on the wire itself, since the wire is twisted and bent in the ambient space

R^{n};

this will be clear in the next example.

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Example 7.3.1. Finding the centre of mass of a wire in $R^{2}$ .

A wire is bent into the semi-circle $x^{2} + y^{2} = 4,$ $y \geq 0 .$ Its mass density is given by the function

ρ (x, y) = 4 - y

on the wire. Find the centre of mass of the wire.

Let us see what we expect first. The wire is heavier near its base ( $y = 0$ ) than at the top ( $y = 2$ ). The mass density is however symmetric about the $y$ -axis (it does not vary in $x$ ). We thus expect the centre of mass to be located at a point $(0, \bar{y}),$ with $\bar{y} < 2,$ since it should be below the top of the wire.

To calculate the centre of mass, we will need to calulate unoriented line integrals along the curve, so we first parametrize the curve as $α : [0, π] \to R^{2}$ with

α (θ) = (2 \cos (θ), 2 \sin (θ)) .

The tangent vector is

T_{θ} = (- 2 \sin (θ), 2 \cos (θ)),

whose norm is

| T_{θ} | = \sqrt{4 \sin^{2} (θ) + 4 \cos^{2} (θ)} = 2.

The line element is then

d s = | T_{θ} | d θ = 2 d θ .

We first calculate the total mass of the wire. It is given by the unoriented line integral:

\begin{aligned} m = & \int_{C} ρ d s \\ = & \int_{0}^{π} (4 - 2 \sin (θ)) 2 d θ \\ = & 4 (2 π - 2) \\ = & 8 (π - 1) . \end{aligned}

We then calculate the coordinates $\bar{x}$ and $\bar{y}$ of the centre of mass.

\begin{aligned} \bar{x} = & \frac{1}{m} \int_{C} x ρ d s \\ = & \frac{1}{8 (π - 1)} \int_{0}^{π} (2 \cos (θ)) (4 - 2 \sin (θ)) 2 d θ \\ = & 0, \end{aligned}

since the trigonometric integral vanishes. As for $\bar{y},$ we get:

\begin{aligned} \bar{y} = & \frac{1}{m} \int_{C} y ρ d s \\ = & \frac{1}{8 (π - 1)} \int_{0}^{π} (2 \sin (θ)) (4 - 2 \sin (θ)) 2 d θ \\ = & \frac{8}{8 (π - 1)} (4 - \frac{π}{2}) \\ = & \frac{8 - π}{2 (π - 1)} \end{aligned}

Therefore, the centre of mass of the wire is located at the point

(\bar{x}, \bar{y}) = (0, \frac{8 - π}{2 (π - 1)}) .

To make sure that this is consistent with our expectation, we can find a numerical value for the location of the centre of mass. We get:

(\bar{x}, \bar{y}) ≃ (0, 1.134) .

This is consistent with our expectation. Phew!

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Subsection 7.3.2 Centre of mass of a thin sheet

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We can do a very similar calculation to obtain the centre of mass of a thin sheet of material (such as aluminium foil, or paper) in

R^{3} .

Suppose that the sheet of material takes the shape of a parametric surface

α : D \to R^{3},

with

α (u, v) = (x (u, v), y (u, v), z (u, v))

and image surface

S = α (D) .

Suppose that

ρ : S \to R

is the mass density of the sheet (mass per unit area). Using the same slicing approach, but with small pieces of surface of area

d S,

we obtain that the total mass of the sheet can be written as the unoriented surface integral

m = \iint_{S} ρ d S = \iint_{D} ρ (α (u, v)) | T_{u} \times T_{v} | d A .

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We calculate the centre of mass as before, by calculating the first moments in the three coordinates

x, y, z .

Using the slicing process, those become unoriented surface integrals. The result is that position

(\bar{x}, \bar{y}, \bar{z})

of the centre of mass is given by

\begin{aligned} \bar{x} = & \frac{1}{m} \iint_{S} x ρ d S = \frac{1}{m} \iint_{D} x (u, v) ρ (α (u, v)) | T_{u} \times T_{v} | d A, \\ \bar{y} = & \frac{1}{m} \iint_{S} y ρ d S = \frac{1}{m} \iint_{D} y (u, v) ρ (α (u, v)) | T_{u} \times T_{v} | d A, \\ \bar{z} = & \frac{1}{m} \iint_{S} z ρ d S = \frac{1}{m} \iint_{D} z (u, v) ρ (α (u, v)) | T_{u} \times T_{v} | d A . \end{aligned}

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Note that, as was the case for the wire, the centre of mass of the sheet is not generally expected to lie on the sheet itself, since the sheet can be bent and twisted in the ambient space

R^{3} .

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Example 7.3.2. Finding the centre of mass of a sheet in $R^{3}$ .

Suppose the a sheet of paper is bent in the shape of the cylinder $x^{2} + y^{2} = 9,$ with $z \in [0, 1] .$ Suppose that its mass density is given by the function

ρ (x, y, z) = z + 1.

Find the centre of mass of the cylinder.

First, let us see what we expect. The cylinder is heavier at the top than at the bottom. However, its mass density has circular symmetry above the $z$ -axis, as it only depends on the height $z$ on the cylinder. Thus we expect the centre of mass to be in the middle of the cylinder, i.e. with coordinates $(0, 0, \bar{z}) .$ Furthermore, we expect its $z$ -coordinate to be a little bit higher than half-way up the cylinder, since the cylinder is heavier at the top than at the bottom. So we expect $0.5 < \bar{z} < 1 .$

To calculate the required unoriented surface integrals, we first parametrize the surface as $α : D \to R^{3}$ with

D = {(u, θ) \in R^{2} | u \in [0, 1], θ \in [0, 2 π]}

and

α (u, θ) = (3 \cos (θ), 3 \sin (θ), u) .

The tangent vectors are

T_{u} = (0, 0, 1), T_{θ} = (- 3 \sin (θ), 3 \cos (θ), 0),

and the cross product is

T_{u} \times T_{θ} = (- 3 \cos (θ), - 3 \sin (θ), 0) .

Its norm is

| T_{u} \times T_{θ} | = \sqrt{9 \cos^{2} (θ) + 9 \sin^{2} (θ)} = 3,

and thus the surface element is

d S = 3 d u d θ .

We calculate the total mass of the sheet of paper. We get:

\begin{aligned} m = & \iint_{S} ρ d S \\ = & \int_{0}^{1} \int_{0}^{2 π} (u + 1) 3 d θ d u \\ = & 6 π \int_{0}^{1} (u + 1) d u \\ = & 9 π . \end{aligned}

As for the coordinates of the centre of mass, we get:

\begin{aligned} \bar{x} = & \frac{1}{m} \iint_{S} x ρ d S \\ = & \frac{1}{9 π} \int_{0}^{1} \int_{0}^{2 π} (3 \cos (θ)) (u + 1) 3 d θ d u \\ = & 0, \end{aligned}

since the integral over $θ$ is zero. Similarly,

\begin{aligned} \bar{y} = & \frac{1}{m} \iint_{S} y ρ d S \\ = & \frac{1}{9 π} \int_{0}^{1} \int_{0}^{2 π} (3 \sin (θ)) (u + 1) 3 d θ d u \\ = & 0. \end{aligned}

As for $\bar{z},$ we get:

\begin{aligned} \bar{z} = & \frac{1}{m} \iint_{S} z ρ d S \\ = & \frac{1}{9 π} \int_{0}^{1} \int_{0}^{2 π} (u) (u + 1) 3 d θ d u \\ = & \frac{2}{3} \int_{0}^{1} (u^{2} + u) d u \\ = & \frac{5}{9} . \end{aligned}

As a result, the centre of mass of the sheet of paper is located at the point

(\bar{x}, \bar{y}, \bar{z}) = (0, 0, \frac{5}{9}) .

Since $\frac{5}{9} ≃ 0.556,,$ this is consistent with our expectation that the centre of mass should be on the $z$ -axis, a little bit higher than half-way up the cylinder, which is at $z = 0.5 .$

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Exercises 7.3.3 Exercises

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1.

Find the centre of mass of a wire that is bent in the shape of a circle of radius $R$ centered at the origin in $R^{2},$ with mass density $ρ (x, y) = 2 R - y .$ Is your result consistent with your expectations?

Solution.

Let us first see what we expect. As the wire is bent in a circle centered at the origin, if the mass density was constant (or symmetric about the $x$ - and $y$ -axes), the centre of mass would be at the origin. However, the mass density is not constant here: it decreases linearly with $y .$ It is constant in the $x$ -direction, so we expect that centre of mass to lie on the $y$ -axis. As the mass decreases as $y$ increases, we expect the centre of mass to be at a position $(0, \bar{y}),$ with $- R < \bar{y} < 0 .$

We parametrize the circle of radius $R$ as $α : [0, 2 π] \to R^{2}$ with $α (θ) = (R \cos (θ), R \sin (θ)) .$ The tangent vector is

T (θ) = (- R \sin (θ), R \cos (θ)) .

Its norm is

| T (θ) | = \sqrt{R^{2} \sin^{2} (θ) + R^{2} \cos^{2} (θ)} = R .

The total mass of the wire is

\begin{aligned} m = \int_{C} ρ d s = & \int_{0}^{2 π} (2 R - R \sin (θ)) R d θ \\ = & 4 π R^{2} . \end{aligned}

The $\bar{x}$ -coordinate of the centre of mass is

\begin{aligned} \bar{x} = & \frac{1}{m} \int_{C} x ρ d s \\ = & \frac{1}{4 π R^{2}} \int_{0}^{2 π} R \cos (θ) (2 R - R \sin (θ)) R d θ \\ = & 0, \end{aligned}

as expected. As for the $\bar{y}$ -coordinate, we get:

\begin{aligned} \bar{y} = & \frac{1}{m} \int_{C} y ρ d s \\ = & \frac{1}{4 π R^{2}} \int_{0}^{2 π} R \sin (θ) (2 R - R \sin (θ)) R d θ \\ = & \frac{1}{4 π R^{2}} (- R^{3} π) \\ = & - \frac{R}{4} . \end{aligned}

Therefore, the position of the centre of mass is

(\bar{x}, \bar{y}) = (0, - \frac{R}{4}) .

As expected, it is on the $y$ -axis, and the $y$ -coordinate is between $- R$ and $0 .$

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2.

Given a wire with density $ρ (x, y)$ that lies on a curve $C \subset R^{2},$ its moments of inertia about the $x$ - and the $y$ -axes are defined by

I_{x} = \int_{C} y^{2} ρ d s, I_{y} = \int_{C} x^{2} ρ d s .

Find the moments of inertia of a wire that lies along the line $2 x + y = 5$ between $x = 0$ and $x = 1,$ with density $ρ (x, y) = x .$

Solution.

We parametrize the line as $α : [0, 1] \to R^{2}$ with

α (t) = (t, 5 - 2 t) .

The tangent vector and its norm are:

T (t) = (1, - 2), | T (t) | = \sqrt{1 + 2^{2}} = \sqrt{5} .

We can then evaluate the moments of inertia. First,

\begin{aligned} I_{x} = & \int_{C} y^{2} ρ d s \\ = & \sqrt{5} \int_{0}^{1} (5 - 2 t)^{2} t d t \\ = & \sqrt{5} \int_{0}^{1} (25 t - 20 t^{2} + 4 t^{3}) d t \\ = & \sqrt{5} (\frac{25}{2} - \frac{20}{3} + 1) \\ = & \frac{41 \sqrt{5}}{6} . \end{aligned}

Second,

\begin{aligned} I_{y} = & \int_{C} x^{2} ρ d s \\ = & \sqrt{5} \int_{0}^{1} t^{3} d t \\ = & \frac{\sqrt{5}}{4} . \end{aligned}

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3.

Consider a sheet of aluminium foil that is bent in the shape of a paraboloid $z = x^{2} + y^{2}$ with $z \in [0, 1] .$ Suppose that it mass density is $ρ = k$ where $k$ is a constant. Find the total mass of the sheet and its centre of mass. Does it agree with your expectation?

Solution.

As the mass density is constant, and the paraboloid has circular symmetry about the $z$ -axis, we expect the centre of mass to lie on the $z$ -axis, somewhere between $z = 0$ and $z = 1 .$

We parametrize the paraboloid as $α : D \to R^{3}$ with

D = {(r, θ) \in R^{2} | r \in [0, 1], θ \in [0, 2 π]}

and

α (r, θ) = (r \cos (θ), r \sin (θ), r^{2}) .

The tangent vectors are

T_{r} = (\cos (θ), \sin (θ), 2 r), T_{θ} = (- r \sin (θ), r \cos (θ), 0) .

The cross product is

T_{r} \times T_{θ} = (- 2 r^{2} \cos (θ), - 2 r^{2} \sin (θ), r) .

Its norm is

| T_{r} \times T_{θ} | = \sqrt{4 r^{4} \cos^{2} (θ) + 4 r^{4} \sin^{2} (θ) + r^{2}} = r \sqrt{4 r^{2} + 1} .

We first calculate the total mass. We get:

\begin{aligned} m = & \iint_{S} ρ d S \\ = & k \int_{0}^{1} \int_{0}^{2 π} r \sqrt{4 r^{2} + 1} d θ r \\ = & 2 π k \int_{0}^{1} r \sqrt{4 r^{2} + 1} d r \\ = & \frac{π k}{4} \int_{1}^{5} \sqrt{u} d u \\ = & \frac{π k}{6} (5^{3 / 2} - 1) . \end{aligned}

In the process of evaluating the integral we did the substitution $u = 4 r^{2} + 1 .$

Next we calculate the coordinates of the centre of mass. First,

\begin{aligned} \bar{x} = & \frac{1}{m} \iint_{S} x ρ d S \\ = & \frac{k}{m} \int_{0}^{1} \int_{0}^{2 π} r^{2} \cos (θ) \sqrt{4 r^{2} + 1} d θ r \\ = & 0, \end{aligned}

as expected. Next,

\begin{aligned} \bar{y} = & \frac{1}{m} \iint_{S} y ρ d S \\ = & \frac{k}{m} \int_{0}^{1} \int_{0}^{2 π} r^{2} \sin (θ) \sqrt{4 r^{2} + 1} d θ r \\ = & 0, \end{aligned}

as expected. Finally,

\begin{aligned} \bar{z} = & \frac{1}{m} \iint_{S} z ρ d S \\ = & \frac{6 k}{π k (5^{3 / 2} - 1)} \int_{0}^{1} \int_{0}^{2 π} r^{3} \sqrt{4 r^{2} + 1} d θ r \\ = & \frac{12}{(5^{3 / 2} - 1)} \int_{0}^{1} r^{3} \sqrt{4 r^{2} + 1} d r \\ = & \frac{3}{8 (5^{3 / 2} - 1)} \int_{1}^{5} (u - 1) \sqrt{u} d u \\ = & \frac{3}{8 (5^{3 / 2} - 1)} (\frac{2}{5} 5^{5 / 2} - \frac{2}{5} - \frac{2}{3} 5^{3 / 2} + \frac{2}{3}) \\ = & \frac{1}{10} \frac{5^{5 / 2} + 1}{5^{3 / 2} - 1} . \end{aligned}

As before, to evaluate the integral we did the substitution $u = 4 r^{2} + 1 .$

Therefore, the centre of mass is located at

(\bar{x}, \bar{y}, \bar{z}) = (0, 0, \frac{1}{10} \frac{5^{5 / 2} + 1}{5^{3 / 2} - 1}) ≃ (0, 0, 0.559) .

As expected, it is located on the $z$ -axis, somewhere between $z = 0$ and $z = 1 .$

Section 7.3 Applications of unoriented line and surface integrals

Objectives

Subsection 7.3.1 Centre of mass of a wire

Example 7.3.1. Finding the centre of mass of a wire in R2.

Subsection 7.3.2 Centre of mass of a thin sheet

Example 7.3.2. Finding the centre of mass of a sheet in R3.

Exercises 7.3.3 Exercises

1.

2.

3.

Example 7.3.1. Finding the centre of mass of a wire in $R^{2}$ .

Example 7.3.2. Finding the centre of mass of a sheet in $R^{3}$ .