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 \(C\text{,}\) and assume that a function, \(f\text{,}\) with values in \(\mathbb R\) is defined on \(C\text{.}\) If \(P\) is a point on the curve, denote by \(s=s(P)\) the distance measured along the curve from one end of the curve to \(P\text{.}\) Suppose that the total length of the curve is \(L\text{.}\) The function \(f\) is now only a function of \(s\text{.}\) Hence we may look at \(f\) as a function on the interval \([0,L]\) and define
\begin{equation*}
\int_Cf\,ds:=\int_0^Lf(s)\,ds\text{.}
\end{equation*}
Now suppose that
\(C\) is a smooth curve, and that
\(\vect\gamma(t)\text{,}\) \(t\in[a,b]\text{,}\) is a regular parametrisation of
\(C\text{.}\) As
\(\vect\gamma\) is a regular parametrisation, every point,
\(P\in C\text{,}\) has a representation
\(\gamma(t)\) 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
\(\vect\gamma(a)\) and
\(\vect\gamma(t)=P\) is given by
\begin{equation*}
s(t)=\int_a^t\|\vect\gamma'(\tau)\|\,d\tau\text{.}
\end{equation*}
By the fundamental theorem of calculus it follows that
\begin{equation*}
s'(t)=\|\vect\gamma'(t)\|\text{,}
\end{equation*}
and thus by the substitution formula
(5.5) we see that
\begin{align*}
\int_0^Lf(s)\,ds
\amp=\int_a^bf(s(t))s'(t)\,dt\\
\amp=\int_a^bf(\vect\gamma(t))\|\vect\gamma'(t)\|\,dt\text{.}
\end{align*}
Hence we make the following definition.
Definition 7.10. Line integral of a scalar function.
Suppose that \(C\) is a smooth positively oriented curve, and that \(\vect\gamma(t)\text{,}\) \(t\in[a,b]\text{,}\) a regular parametrisation of \(C\text{.}\) Then for every continuous scalar function on \(C\) we define
\begin{equation*}
\int_Cf\,ds:=\int_a^bf(\vect\gamma(t))\|\vect\gamma'(t)\|\,dt\text{.}
\end{equation*}
The integral over a piecewise smooth curve is defined to be the sum over the integrals over the smooth parts of the curve.
One can show that the above definition of line integral is independent of the particular parametrisation \(\vect\gamma\) choosen.
Example 7.11.
Compute \(\int_C(x^2-y)\,ds\) if \(C\) is given by \(\gamma_1(t)=t\) and \(\gamma_2(t)=\cosh t\) with \(t\in[0,1]\text{.}\)
Solution.We first compute
\(\|\vect\gamma(t)\|=\sqrt{\bigl(\gamma_1'(t)\bigr)^2+\bigl(\gamma_2'(t)\bigr)^2}\text{.}\) As
\(\gamma_1'(t)=1\) and
\(\gamma_2'(t)=\sinh t\) we get
\begin{equation*}
\|\vect\gamma'(t)\|
=\sqrt{1+\sinh^2 t}=\sqrt{\cosh^2 t}=\cosh t\text{.}
\end{equation*}
Hence,
\begin{align*}
\int_C(x^2-y)\,ds
\amp=\int_0^1((\gamma_1(t))^2-\gamma_2(t))
\sqrt{(\gamma_1'(t))^2+(\gamma_2'(t))^2)}\,dt\\
\amp=\int_0^1(t^2-\cosh t)\cosh t\,dt\\
\amp=\int_0^1t^2\cosh t\,dt-\int_0^1\cosh^2 t\,dt
\end{align*}
We now compute the last two integrals separately. For the first we integrate by parts twice:
\begin{align*}
\int_0^1t^2\cosh t\,dt
=\bigl[t^2\sinh t\bigl]_0^1-2\int_0^1t\sinh t\,dt
\amp=\sinh 1-2\Bigl(\bigl[t\cosh t\bigl]_0^1-\int_0^1\cosh t\,dt\Bigr)\\
\amp=\sinh 1-2\cosh 1+2\bigl[\sinh t\bigl]_0^1\\
\amp=\sinh 1-2\cosh 1+2\sinh 1=3\sinh 1-2\cosh 1\text{.}
\end{align*}
Using a well known identity and integration of parts we have
\begin{align*}
\int\cosh^2 t\,dt
\amp=\sinh t\cosh t-\int\sinh^2 t\,dt\\
\amp=\sinh t\cosh t-\int(\cosh^2t-1)\,dt\\
=\sinh t\cosh t+t-\int\cosh^2 t\,dt\text{.}
\end{align*}
Hence,
\begin{equation*}
\int_0^1\cosh^2 t\,dt
=\frac{1}{2}\Bigl[t+\sinh t\cosh t\Bigr]_0^1
=\frac{1+\sinh 1\cosh 1}{2},
\end{equation*}
and therefore
\begin{equation*}
\int_C(x^2-y)\,ds
=3\sinh 1-2\cosh 1-\frac{1+\sinh 1\cosh 1}{2}\text{.}
\end{equation*}
This is the required result.
