2.5. Vector Fields and Line integral
Vector Fields
A vector field assigns a vector to every point in space. Examples include hemodynamics, where vector fields are used to understand blood flow patterns and detect abnormalities such as stenosis or aneurysms, and respiratory dynamics, where vector fields model airflow in the lungs to study ventilation efficiency.
Let \(\mathbf{r} = (x, y, z)\). A vector field is defined as: $$F(\mathbf{r}) = F_x(\mathbf{r}) \hat{i} + F_y(\mathbf{r}) \hat{j} + F_z(\mathbf{r}) \hat{k},$$ where:
- \( F_x(\mathbf{r}) \): The \(x\)-component of the vector field.
- \( F_y(\mathbf{r}) \): The \(y\)-component of the vector field.
- \( F_z(\mathbf{r}) \): The \(z\)-component of the vector field.
- \( \hat{i}, \hat{j}, \hat{k} \): Unit vectors in the \(x\)-, \(y\)-, and \(z\)-directions, respectively.
Line Integral Along a Vector Field
The line integral of a vector field \(\mathbf{F}(\mathbf{r})\) along a curve \(C\) measures the work done by the field along the path. It is given by: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (F_x \, dx + F_y \, dy + F_z \, dz), \] where:
- \(C\) is the curve along which the integral is evaluated.
- \(\mathbf{F}(\mathbf{r}) = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}\) is the vector field.
- \(d\mathbf{r} = dx \, \hat{i} + dy \, \hat{j} + dz \, \hat{k}\) is the differential element of the curve. \end{itemize}
- To reduce this line integral to a 1D integral, parameterize the curve \(C\) by a parameter \(t\), such that: \[ \mathbf{r}(t) = (x(t), y(t), z(t)), \quad t \in [a, b]. \] The differentials \(dx\), \(dy\), and \(dz\) are expressed as: \[ dx = \frac{dx}{dt} \, dt, \quad dy = \frac{dy}{dt} \, dt, \quad dz = \frac{dz}{dt} \, dt. \] Substituting these into the line integral, we have: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_a^b \left[ F_x(x(t), y(t), z(t)) \frac{dx}{dt} + F_y(x(t), y(t), z(t)) \frac{dy}{dt} + F_z(x(t), y(t), z(t)) \frac{dz}{dt} \right] dt. \]
- Example: Work Done Along a Helix. Let \(\mathbf{F}(\mathbf{r}) = z \, \hat{i} + x \, \hat{j} + y \, \hat{k}\), and consider the helix parameterized by: \[ \mathbf{r}(t) = (\cos t, \sin t, t), \quad t \in [0, 2\pi]. \] Then: \[ \frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t, \quad \frac{dz}{dt} = 1. \] Substituting these into the line integral: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left[ z(-\sin t) + x(\cos t) + y(1) \right] dt, \] where \(x = \cos t\), \(y = \sin t\), and \(z = t\). Simplifying: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left[ t(-\sin t) + \cos t (\cos t) + \sin t (1) \right] dt. \] Further simplifications yield the final 1D integral: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \left( -t\sin t + \cos^2 t + \sin t \right) dt. \]
- Example: Blood Flow Work Along an Artery. The line integral can quantify the work done by blood pressure as blood moves along a curved artery. Suppose the pressure field \(\mathbf{F}(x, y) = -y \hat{i} + x \hat{j}\) acts along a path \(C\), parameterized as \(\mathbf{r}(t) = t \hat{i} + t^2 \hat{j}\), \(0 \leq t \leq 1\). The line integral is: \[\int_C \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \mathbf{F}(\mathbf{r}(t)) \cdot \frac{d\mathbf{r}}{dt} \, dt.\] Substitute:\[\mathbf{F}(\mathbf{r}(t)) = -t^2 \hat{i} + t \hat{j}, \quad \frac{d\mathbf{r}}{dt} = \hat{i} + 2t \hat{j}.\] The dot product is:\[\mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = (-t^2)(1) + (t)(2t) = -t^2 + 2t^2 = t^2.\]The integral becomes:\[\int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}.\]The work done by the pressure field along the artery is \( \frac{1}{3} \, \text{units} \).
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