Integrals of a Scalar Function

Section 7.3 Integrals of a Scalar Function

We may consider a curve as a deformed interval. We know how to integrate a function over an interval. In this section we want to learn how to integrate functions defined on a deformed interval, that is, on a curve. Denote the curve by

C,

and assume that a function,

f,

with values in

R

is defined on

C .

P

is a point on the curve, denote by

s = s (P)

the distance measured along the curve from one end of the curve to

P .

Suppose that the total length of the curve is

L .

The function

f

is now only a function of

s .

Hence we may look at

f

as a function on the interval

[0, L]

and define

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\int_{C} f d s := \int_{0}^{L} f (s) d s .

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Now suppose that

C

is a smooth curve, and that

γ (t),

t \in [a, b],

is a regular parametrisation of

C .

γ

is a regular parametrisation, every point,

P \in C,

has a representation

γ (t)

for a unique

t \in [a, b]

(recall that a parametrisation is one-to-one by definition). According to Proposition 7.8 the length of the curve between

γ (a)

and

γ (t) = P

is given by

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s (t) = \int_{a}^{t} ‖ γ^{'} (τ) ‖ d τ .

By the fundamental theorem of calculus it follows that

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s^{'} (t) = ‖ γ^{'} (t) ‖,

and thus by the substitution formula (5.5) we see that

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\begin{aligned} \int_{0}^{L} f (s) d s & = \int_{a}^{b} f (s (t)) s^{'} (t) d t \\ = \int_{a}^{b} f (γ (t)) ‖ γ^{'} (t) ‖ d t . \end{aligned}

Hence we make the following definition.

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Definition 7.10. Line integral of a scalar function.

Suppose that

C

is a smooth positively oriented curve, and that

γ (t),

t \in [a, b],

a regular parametrisation of

C .

Then for every continuous scalar function on

C

we define

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\int_{C} f d s := \int_{a}^{b} f (γ (t)) ‖ γ^{'} (t) ‖ d t .

The integral over a piecewise smooth curve is defined to be the sum over the integrals over the smooth parts of the curve.

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One can show that the above definition of line integral is independent of the particular parametrisation

γ

choosen.

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Example 7.11.

Compute

\int_{C} (x^{2} - y) d s

C

is given by

γ_{1} (t) = t

and

γ_{2} (t) = \cosh t

with

t \in [0, 1] .

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Solution.

We first compute

‖ γ (t) ‖ = \sqrt{(γ_{1}^{'} (t))^{2} + (γ_{2}^{'} (t))^{2}} .

γ_{1}^{'} (t) = 1

and

γ_{2}^{'} (t) = \sinh t

we get

‖ γ^{'} (t) ‖ = \sqrt{1 + \sinh^{2} t} = \sqrt{\cosh^{2} t} = \cosh t .

Hence,

\begin{aligned} \int_{C} (x^{2} - y) d s & = \int_{0}^{1} ((γ_{1} (t))^{2} - γ_{2} (t)) \sqrt{(γ_{1}^{'} (t))^{2} + (γ_{2}^{'} (t))^{2})} d t \\ = \int_{0}^{1} (t^{2} - \cosh t) \cosh t d t \\ = \int_{0}^{1} t^{2} \cosh t d t - \int_{0}^{1} \cosh^{2} t d t \end{aligned}

We now compute the last two integrals separately. For the first we integrate by parts twice:

\begin{aligned} \int_{0}^{1} t^{2} \cosh t d t = [t^{2} \sinh t]_{0}^{1} - 2 \int_{0}^{1} t \sinh t d t & = \sinh 1 - 2 ([t \cosh t]_{0}^{1} - \int_{0}^{1} \cosh t d t) \\ = \sinh 1 - 2 \cosh 1 + 2 [\sinh t]_{0}^{1} \\ = \sinh 1 - 2 \cosh 1 + 2 \sinh 1 = 3 \sinh 1 - 2 \cosh 1 . \end{aligned}

Using a well known identity and integration of parts we have

\begin{aligned} \int \cosh^{2} t d t & = \sinh t \cosh t - \int \sinh^{2} t d t \\ = \sinh t \cosh t - \int (\cosh^{2} t - 1) d t \\ = \sinh t \cosh t + t - \int \cosh^{2} t d t . \end{aligned}

Hence,

\int_{0}^{1} \cosh^{2} t d t = \frac{1}{2} [t + \sinh t \cosh t]_{0}^{1} = \frac{1 + \sinh 1 \cosh 1}{2},

and therefore

\int_{C} (x^{2} - y) d s = 3 \sinh 1 - 2 \cosh 1 - \frac{1 + \sinh 1 \cosh 1}{2} .

This is the required result.

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Insight 7.12. Physical Interpretation of Line Integrals.

We finish this section by giving some physical interpretations of the line integral. First consider the case

N = 2 .

We think of the curve,

C,

in the plane to be the boundary of a block of land. We want to put up a fence along that boundary. Its height at a point

(x, y) \in R^{2}

is supposed to be

h (x, y),

and may vary along different parts. (The fence to the bad neighbours is a bit higher than the fence to the good neighbours.) Before we go to the hardware store we want to know how much fencing we need to buy, so we want to compute the surface area of the fence. This is exactly what the line integral gives us, so the area is

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\int_{C} h d s .

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Another possible physical interpretation is the following. Suppose the curve

C

has the shape of a bent wire. The wire is made of different materials, or its thickness slightly varies. Fix a point,

P,

C

and let

S

be a piece of length

Δ s

C

containing

P .

Δ m

is the mass of

S

then

Δ m / Δ s

is the average mass density of

S .

Shrinking

Δ s

to zero we get the mass density of the wire at

P,

that is, the mass per length. Denote that mass density at

(x, y, z) \in C

ϱ (x, y, z) .

Then the total mass of the wire is given by

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\int_{C} ϱ d s .

In the next section we motivate and define line integrals of vector fields. The definition will be reduced to the one given here.

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