Math2111: Chapter 3: Line integrals. Section 3: Fundamental theorem of line integrals

In this blog entry you can find lecture notes for Math2111, several variable calculus. See also the table of contents for this course. This blog entry printed to pdf is available here.

Fundamental theorem of line integrals

The fundamental theorem of calculus gives a relation between the integral of the derivative of a function and the value of the function at the boundary, that is,

\displaystyle \int_a^b g^\prime(x)\,\mathrm{d} t= g(b)-g(a). \qquad\qquad\qquad\qquad\qquad (1)

The aim is now to find an analogous formula for line integrals.

We have seen previously in the section on vector line integrals that the line integral of a vector field \boldsymbol{F}: D \subseteq \mathbb{R}^3\to\mathbb{R}^3 over a curve \boldsymbol{r}:[a,b]\to\mathbb{R} is given by

\displaystyle \int_{\mathcal{C}} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{s} = \int_a^b \boldsymbol{F}(\boldsymbol{r}(t)) \cdot \boldsymbol{r}^\prime(t) \,\mathrm{d} t.

The last integral is used for evaluating line integrals and is of the form (1). Thus, if there is a function g:[a,b]\to\mathbb{R} such that

\displaystyle g^\prime(t)=\boldsymbol{F}(\boldsymbol{r}(t)) \cdot \boldsymbol{r}^\prime(t), \qquad\qquad\qquad\qquad (2)

then we can use (1) directly to obtain a fundamental theorem of line integrals. Hence the question arises for which vector fields \boldsymbol{F} and curves \boldsymbol{r} this property holds.

Notice that Equation (1) holds for all, say, continuously differentiable functions g and all intervals [a,b] (where {}g is continuously differentiable). Similarly, we want property (2) to hold for all curves \boldsymbol{r} and only put a restriction on the vector field \boldsymbol{F} (to be more precise: all curves whose range \{\boldsymbol{r}(t): \in\mathbb{R}^3: a\le t \le b\} is in a region where the property of \boldsymbol{F} holds).

Fundamental Theorem of Line Integrals
Let f:\mathbb{R}^3\to\mathbb{R} be a continuously differentiable function and let \boldsymbol{r}:[a,b]\to\mathbb{R}^3 be a continuously differentiable parameterisation of a curve \mathcal{C}. Then

\int_{\mathcal{C}} \nabla f \cdot \mathrm{d} \boldsymbol{s} = f(\boldsymbol{r}(b)) - f(\boldsymbol{r}(a)).

Note that the same result holds for f:\mathbb{R}^2\to\mathbb{R} and curves in \mathbb{R}^2 which satisfy the assumptions.

Let f:\mathbb{R}^3\to\mathbb{R} be a continuously differentiable function.

  1. Let \mathcal{C} be a closed curve. Then \int_{\mathcal{C}} \nabla f\cdot \mathrm{d} \boldsymbol{r} = 0.
  2. Let \mathcal{C}_1 and \mathcal{C}_2 be two curves with the same starting and end points. Then \int_{\mathcal{C}_1} \nabla f\cdot \mathrm{d} \boldsymbol{s} = \int_{\mathcal{C}_2} \nabla f \cdot \mathrm{d}\boldsymbol{s}.

Let f(x,y,z)= xyz. Calculate \int_{\mathcal{C}} \nabla f \cdot \mathrm{d} \boldsymbol{r} along a curve with starting point (0,0,0) and end point (1,-1,1).

Notice that the fundamental theorem of calculus only holds for vector fields \boldsymbol{F} which are gradient fields, i.e. for which there exists a function f:\mathbb{R}^3\to\mathbb{R} such that \boldsymbol{F}=\nabla f.

The following example shows that the fundamental theorem of calculus does not hold in \mathbb{R}^2 if the vector field is not defined everywhere.
Let f:\mathbb{R}^2\setminus\{(0,0)\}\to\mathbb{R} be given by f(x,y)=\arctan(y/x). Let the curve \mathcal{C} be the unit circle around the origin, which is parameterised by \boldsymbol{r}:[0,2\pi]\to\mathbb{R}^2 given by \boldsymbol{r}(t)=(\cos t, \sin t). Calculate \int_{\mathcal{C}} \nabla f \cdot \mathrm{d}\boldsymbol{r}.

Conservative vector fields

A vector field \boldsymbol{F}:\mathbb{R}^3\to\mathbb{R}^3 for which there exists a function f:\mathbb{R}^3\to\mathbb{R} such that \boldsymbol{F}=\nabla f is called conservative. The function {}f is called potential function.

Examples of conservative vector fields are gravitational forces and electric fields.

The gravitational force \boldsymbol{F} acting on a mass {}m due to a mass {}M is given by

\displaystyle \boldsymbol{F}= -G m M \frac{\widehat{\boldsymbol{d}}}{r^2},

where {}G is the gravitational constant, \widehat{\boldsymbol{d}} is a unit vector pointing from {}M to {}m and {}r is the distance between the two mass points. \Box

Two questions now arise:

  1. Given a vector field \boldsymbol{F}. How can we find out whether \boldsymbol{F} is conservative?
  2. Assuming we know that \boldsymbol{F} is conservative, how can we find a function {}f such that \boldsymbol{F} =\nabla f?

To answer the first question, recall that in Section 2 on divergence and curl that \mathrm{curl} \, \mathrm{grad} f = \boldsymbol{0}. Let \boldsymbol{F} be twice continuously differentiable. Then, if there is a function {}f such that \boldsymbol{F}=\nabla f, then

\displaystyle \mathrm{curl}\,\boldsymbol{F} = \mathrm{curl} \nabla f = \boldsymbol{0}.

Thus, we can test whether a given vector field is conservative by calculating its curl. If \mathrm{curl}\,\boldsymbol{F}\neq \boldsymbol{0}, then we know that \boldsymbol{F} is not conservative.

Show that the vector field \boldsymbol{F}(x,y)= y^2 \widehat{\boldsymbol{i}} + x^2 \widehat{\boldsymbol{j}} is not conservative. \Box

The converse also holds under some assumptions. This is a special case of Poincare’s lemma. For instance, Poincare’s lemma implies that if \boldsymbol{F} is twice continuously differentiable and \mathrm{curl}\, \boldsymbol{F} = \boldsymbol{0} everywhere, then \boldsymbol{F} is conservative.

Let \boldsymbol{F}(x,y)=(y \cos xy, \mathrm{e}^y + x \cos xy). Find a function {}f such that \boldsymbol{F}=\nabla f.

Let \boldsymbol{F}(x,y,z)=(-y z \sin xz +\mathrm{e}^x y z + \cos x \cos z, \cos xz + \mathrm{e}^x z, -xy \sin xz +\mathrm{e}^x y-\sin x\sin z). Find a function {}f such that \boldsymbol{F}=\nabla f.


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