Line Integrals (Real Vector Fields)#

TODO: Make this more rigorous

Definition: Line Integral of a Real Vector Field

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}^n\) be a real vector field and let \(\gamma: I \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve which is differentiable on the interior of the interval \(I\).

The line integral of \(f\) along \(\gamma\) is the (potentially improper) Riemann integral

\[\int_I f(\gamma(t)) \cdot \gamma'(t) \, \mathrm{d}t,\]

of the dot product between \(f \circ \gamma\) and \(\gamma'\), provided that it exists.

Notation

We denote the line integral of \(f\) along \(\gamma\) as follows:

\[\int_{\gamma} f \cdot \mathrm{d}\boldsymbol{r}\]

If \(\gamma\) is closed, we write:

\[\oint_{\gamma} f \cdot \mathrm{d}\boldsymbol{r}\]

Example

Consider the real vector field \(f: \mathbb{R}^3 \to \mathbb{R}^3\) defined as

\[f(x,y,z) = \begin{bmatrix}xy \\ x - z \\ xz\end{bmatrix}\]

and the parametric curve \(\gamma: [0,2] \to \mathbb{R}^3\) defined as follows:

\[\gamma(t) = \begin{bmatrix}t \\ t^3 \\ 3\end{bmatrix}\]

We have:

\[f(\gamma(t)) = \begin{bmatrix}t^4 \\ t - 3 \\ 3t\end{bmatrix} \qquad \gamma'(t) = \begin{bmatrix}1 \\ 3t^2 \\ 0 \end{bmatrix}\]

Both \(f\circ \gamma\) and \(\gamma'\) are continuous. Therefore, \((f \circ \gamma) \cdot \gamma'\) is also continuous and we have

\[\begin{aligned}\int_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} & = \int_0^2 f(\gamma(t)) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_0^2 (t^4 \cdot 1 + (t-3)\cdot 3t^2 + 3t\cdot 0) \,\mathrm{d}t \\ & = \int_0^2 (t^4 + 3t^3 - 9t^2) \,\mathrm{d}t \\ & = \left.\left( \frac{t^5}{5} + \frac{3t^4}{4} - 3t^3 \right)\right\vert_0^2 \\ & = \left( \frac{32}{5} + 12 - 24 \right) - 0 \\ & = \frac{32}{5} - 12 \\ & = -\frac{28}{5}\end{aligned}\]

for the line integral of \(f\) along \(\gamma\).

Example

Consider the real vector field \(f: \mathbb{R}^2 \to \mathbb{R}^2\) defined as

\[f(x,y) = \begin{bmatrix}-y \\ x\end{bmatrix}\]

and the parametric curve \(\gamma: [0,2\uppi] \to \mathbb{R}^2\) defined as follows:

\[\gamma(t) = \begin{bmatrix}\cos t \\ \sin t\end{bmatrix}\]

Since \(\gamma(0) = \gamma(2\uppi)\), we know that \(\gamma\) is closed.

We have:

\[f(\gamma(t)) = \begin{bmatrix}-\sin t \\ \cos t\end{bmatrix} \qquad \gamma'(t) = \begin{bmatrix}-\sin t \\ \cos t\end{bmatrix}\]

Both \(f\circ \gamma\) and \(\gamma'\) are continuous. Therefore, \((f \circ \gamma) \cdot \gamma'\) is also continuous and we have

\[\begin{aligned}\oint_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} & = \int_0^{2\uppi} f(\gamma(t)) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_0^{2\uppi} ((-\sin t)(-\sin t) + (\cos t)(\cos t)) \,\mathrm{d}t \\ & = \int_0^{2\uppi} (\sin^2 t + \cos^2 t) \,\mathrm{d}t \\ & = \int_0^{2\uppi} 1 \,\mathrm{d}t \\ & = \left. t \right\vert_0^{2\uppi} \\ & = 2\uppi - 0 \\ & = 2\uppi\end{aligned}\]

for the line integral of \(f\) along \(\gamma\).

Example

Consider the real vector field \(f: \mathbb{R}^2 \to \mathbb{R}^2\) defined as

\[f(x,y) = \begin{bmatrix}3x \\ 0\end{bmatrix}\]

and the parametric curve \(\gamma: [0,2\uppi] \to \mathbb{R}^2\) defined as follows:

\[\gamma(t) = \begin{bmatrix}\cos t \\ \sin t\end{bmatrix}\]

Since \(\gamma(0) = \gamma(2\uppi)\), we know that \(\gamma\) is closed.

We have:

\[f(\gamma(t)) = \begin{bmatrix}3\cos t \\ 0\end{bmatrix} \qquad \gamma'(t) = \begin{bmatrix}-\sin t \\ \cos t\end{bmatrix}\]

Both \(f\circ \gamma\) and \(\gamma'\) are continuous. Therefore, \((f \circ \gamma) \cdot \gamma'\) is also continuous and we have

\[\begin{aligned}\int_{\gamma} f \cdot \mathrm{d}\boldsymbol{r} & = \int_0^{2\uppi} f(\gamma(t)) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_0^{2\uppi} ((3\cos t)(-\sin t) + (0)(\cos t)) \,\mathrm{d}t \\ & = \int_0^{2\uppi} -3\sin t \cos t \,\mathrm{d}t \\ & = \left. -\frac{3}{2}\sin^2 t \right\vert_0^{2\uppi} \\ & = 0 - 0 \\ & = 0\end{aligned}\]

for the line integral of \(f\) along \(\gamma\).

Theorem: Linearity of Line Integrals

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}^n\) and \(g: \mathcal{D}_g \subseteq \mathbb{R}^n \to \mathbb{R}^n\) be real vector fields and let \(\gamma: I \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve which is differentiable on the interior of the interval \(I\) with \(\gamma(I) \subseteq \mathcal{D}_f \cap \mathcal{D}_g\).

If the line integrals of \(f\) and \(g\) along \(\gamma\) exist, then so does the line integral of \(\alpha f + \beta g\) for all \(\alpha, \beta \in \mathbb{R}\):

\[\int_{\gamma} (\alpha f + \beta g) \cdot \mathrm{d}\mathbf{r} = \alpha \int_{\gamma} f \cdot \mathrm{d}\mathbf{r} + \beta \int_{\gamma} g \cdot \mathrm{d}\mathbf{r}\]

Proof

\[\begin{aligned}\int_{\gamma} (\alpha f + \beta g) \cdot \mathrm{d}\mathbf{r} & = \int_I((\alpha f + \beta g) \circ \gamma)(t) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_I(\alpha f (\gamma(t)) + \beta g(\gamma(t))) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_I \left( \alpha (f (\gamma(t)) \cdot \gamma'(t)) + \beta (g(\gamma(t)) \cdot \gamma'(t)) \right) \,\mathrm{d}t \\ & = \int_I\alpha (f (\gamma(t)) \cdot \gamma'(t)) \,\mathrm{d}t + \int_I\beta (g(\gamma(t)) \cdot \gamma'(t)) \,\mathrm{d}t \end{aligned}\]

Since the line integrals of \(f\) and \(g\) along \(\gamma\) exist, we know that

\[\int_I f (\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t= \int_{\gamma} f \cdot \mathrm{d}\mathbf{r}\]

and

\[\int_I g (\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t = \int_{\gamma} g \cdot \mathrm{d}\mathbf{r}\]

exist. Therefore:

\[\begin{aligned}\int_{\gamma} (\alpha f + \beta g) \cdot \mathrm{d}\mathbf{r} & = \alpha \int_If (\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t + \beta \int_I g (\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t \\ & = \alpha \int_{\gamma} f \cdot \mathrm{d}\mathbf{r} + \beta \int_{\gamma} g \cdot \mathrm{d}\mathbf{r}\end{aligned}\]

Theorem: Vector Line Integral \(=\) Scalar Line Integral

Let \(f: \mathcal{D}_f \subseteq \mathbb{R}^n \to \mathbb{R}^n\) be a real vector field and let \(\gamma: I \subseteq \mathbb{R} \to \mathbb{R}^n\) be a real parametric curve which is differentiable on the interior of the interval \(I\).

The line integral of \(f\) over \(\gamma\) (provided it exists) is equal to the line integral of the dot product between \(f\) and \(\gamma\)'s unit tangent vector over \(\gamma\) (provided it exists):

\[\int_{\gamma} f \cdot \,\mathrm{d}\boldsymbol{r} = \int_{\gamma} f \cdot \boldsymbol{T} \, \mathrm{d}s\]

Proof

TODO

Green's Theorem

TODO: Make more rigorous

Let \(\mathcal{D} \subset \mathbb{R}^2\) be Lebesgue-measurable and such that its boundary can be parameterized by a piecewise regular, simple closed parametric curve \(\gamma: [a,b] \subseteq \mathbb{R} \to \mathbb{R}^2\) with positive orientation. Let \(f: \mathcal{D} \to \mathbb{R}^2\) be a real vector field with component functions \(f_1\) and \(f_2\).

If \(f\) is continuously differentiable on \(\mathcal{D}\), then its line integral over \(\gamma\) is given by the following integral of the partial derivatives of \(f_1\) and \(f_2\):

\[\int_{\gamma} f \cdot \,\mathrm{d}\boldsymbol{\gamma} = \iint_{\mathcal{D}} \partial_1 f_2 - \partial_2 f_1 \, \mathrm{d}\mathcal{D}\]

Example

Consider the square \(\mathcal{D} = [0,1]^2 \subset \mathbb{R}^2\). The boundary of \(\mathcal{D}\) can be parameterized by the following parametric curve:

\[\gamma:[0,4] \to \mathbb{R}^2 \qquad \gamma(t) = \begin{cases} \begin{bmatrix} t & 0 \end{bmatrix}^{\mathsf{T}}, & t \in [0,1] \\ \begin{bmatrix} 1 & t-1 \end{bmatrix}^{\mathsf{T}}, & t \in [1,2] \\ \begin{bmatrix} 3 - t & 1 \end{bmatrix}^{\mathsf{T}}, & t \in [2,3] \\ \begin{bmatrix} 0 & 4 - t \end{bmatrix}^{\mathsf{T}}, & t \in [3,4]\end{cases}\]

We see that \(\gamma\) is piecewise regular, simple, closed and with positive orientation.

Let \(f: \mathcal{D} \to \mathbb{R}^2\) be a real vector field defined as

\[f(x, y) = \begin{bmatrix}f_1(x, y) \\ 0\end{bmatrix}\]

for some continuously differentiable real scalar field \(f_1\).

The conditions for the theorem are satisfied and for the line Integral of \(f\) over \(\gamma\), we have:

\[\begin{aligned} \int_{\gamma} f \cdot \,\mathrm{d}\boldsymbol{\gamma} & = \iint_{\mathcal{D}} \partial_1 f_2 - \partial_2 f_1 \, \mathrm{d}\mathcal{D} \\ & = \iint_{[0,1]^2} - \partial_2 f_1(x,y) \, \mathrm{d}\mathcal{D} \end{aligned}\]

Since \(\partial_2 f_1(x,y)\) is continuous, we get the following iterated integral:

\[\begin{aligned}\iint_{[0,1]^2} - \partial_2 f_1(x,y) \, \mathrm{d}\mathcal{D} & = \int_0^1 \left( \int_0^1 - \partial_2 f_1(x,y) \, \mathrm{d}y \right) \, \mathrm{d}x \\ & = \int_0^1 -(f_1(x, 1)- f_1(x,0) )\,\mathrm{d}x \\ & = \int_0^1 f_1(x, 0) -f_1(x,1)\,\mathrm{d}x \end{aligned}\]

We can also verify this via the definition of the line integral:

\[\begin{aligned} \int_{\gamma} f \cdot \mathrm{d}\boldsymbol{\gamma} &= \int_{0}^{4} f(\gamma(t)) \cdot \gamma'(t) \,\mathrm{d}t \\ &= \int_{0}^{1} f(t, 0) \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} \,\mathrm{d}t + \int_{1}^{2} f(1, t-1) \cdot \begin{bmatrix} 0 \\ 1 \end{bmatrix} \,\mathrm{d}t + \int_{2}^{3} f(3-t, 1) \cdot \begin{bmatrix} -1 \\ 0 \end{bmatrix} \,\mathrm{d}t + \int_{3}^{4} f(0, 4-t) \cdot \begin{bmatrix} 0 \\ -1 \end{bmatrix} \,\mathrm{d}t \\ &= \int_{0}^{1} f_1(t, 0) \,\mathrm{d}t - \int_{2}^{3} f_1(3-t, 1) \,\mathrm{d}t \\ &= \int_{0}^{1} f_1(t, 0) \,\mathrm{d}t - \int_{0}^{1} f_1(s, 1) \,\mathrm{d}s \\ & = \int_0^1 f_1(x, 0) -f_1(x,1)\,\mathrm{d}x \end{aligned}\]

Example: Area via Line Integral

The theorem can also be used to find the area of a bounded Lebesgue-measurable subset \(\mathcal{D} \subseteq \mathbb{R}^2\).

Specifically, if \(f: \mathcal{D} \to \mathbb{R}^2\) is a continuously differentiable real vector field such that the partial derivatives of its component functions \(f_1\), \(f_2\) obey \(\partial_1 f_2 - \partial_2 f_1 = 1\), then the area of \(\mathcal{D}\) is given by the line integral of \(f\) over any parametric curve which parameterizes the boundary of \(\mathcal{D}\).

For example, we can pick \(f: \mathcal{D} \to \mathbb{R}^2\) defined as follows:

\[f(x,y) = \frac{1}{2}\begin{bmatrix}-y \\ x\end{bmatrix}\]

We have:

\[\partial_x f_2(x,y) - \partial_y f_1(x,y) = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1\]

Now consider the following ellipse:

\[\mathcal{D} = \left\{\begin{bmatrix} x \\ y \end{bmatrix} \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1\right\}\]

Its boundary can be parameterized by the following parametric curve:

\[\gamma: [0, 2\uppi] \to \mathbb{R}^2 \qquad \gamma(t) = \begin{bmatrix}a \cos t \\ b \sin t\end{bmatrix}\]

We see that the conditions for the theorem are fulfilled. Therefore:

\[\begin{aligned}A(\mathcal{D}) & = \iint_{\mathcal{D}} 1 \, \mathrm{d}\mathcal{D} \\ & = \iint_{\mathcal{D}} \partial_x f_2(x,y) - \partial_y f_1(x,y) \, \mathrm{d}\mathcal{D} \\ & = \int_{\gamma}f \cdot \,\mathrm{d}\boldsymbol{\gamma} \\ & = \int_{0}^{2\uppi} f(\gamma(t)) \cdot \gamma'(t)\,\mathrm{d}t \\ & = \int_{0}^{2\uppi} \frac{1}{2} \begin{bmatrix}-b \sin t \\ a \cos t\end{bmatrix} \cdot \begin{bmatrix} - a \sin t \\ b \cos t \end{bmatrix} \, \mathrm{d}t \\ & = \frac{1}{2}\int_0^{2\uppi} ab (\sin^2 t + \cos^2 t) \, \mathrm{d}t \\ & = \frac{1}{2}abt\vert_{0}^{2\uppi} \\ & = \frac{1}{2}\cdot 2 \uppi ab \\ & = \uppi ab\end{aligned}\]

Proof

TODO