Question

A vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{C} \vec{F} \cdot \vec{n} d s,\) the flux of \(\vec{F}\) over \(C\). \(\vec{F}=\langle x+y, x-y\rangle ; C\) is the curve with initial and terminal points (3,-2) and (3,2) , respectively, parametrized by \(\vec{r}(t)=\langle 3, t\rangle\) on \(-2 \leq t \leq 2\).

Short answer

The flux of \(\vec{F}\) over \(C\) is 12.

Step-by-step solution

Understanding the Curve Parameterization

The curve \(C\) is parametrized by \(\vec{r}(t) = \langle 3, t \rangle\) where \(-2 \leq t \leq 2\). This indicates it's a vertical line segment from \((3, -2)\) to \((3, 2)\).

Derive the Tangent Vector

The tangent vector to the curve \(C\) is found by differentiating \(\vec{r}(t)\) with respect to \(t\). This gives \(\frac{d\vec{r}}{dt} = \langle 0, 1 \rangle\).

Find the Normal Vector

For a vertical line segment, the normal vector \(\vec{n}\) can be \(\pm\langle 1, 0 \rangle\). We'll choose \(\vec{n} = \langle 1, 0 \rangle\) to point in the positive x-direction.

Evaluate \(\vec{F} \cdot \vec{n}\)

Substitute \(x = 3\) and \(y = t\) into the vector field \(\vec{F}\). Hence, \(\vec{F}(3, t) = \langle 3+t, 3-t \rangle\). We compute \(\vec{F} \cdot \vec{n} = \langle 3+t, 3-t \rangle \cdot \langle 1, 0 \rangle = 3 + t \).

Set Up the Line Integral

The line integral for the flux is \(\int_{C} \vec{F} \cdot \vec{n} \, ds = \int_{-2}^{2} (3 + t) \, dt \).

Evaluate the Integral

Compute the integral: \(\int_{-2}^{2} (3 + t) \, dt = \left[3t + \frac{t^2}{2} \right]_{-2}^{2}\). Evaluate at the bounds: \([3(2) + \frac{2^2}{2}] - [3(-2) + \frac{(-2)^2}{2}] = (6 + 2) - (-6 + 2) = 8 - (-4) = 8 + 4 = 12\).

Key concepts

Line Integrals

Line integrals allow us to accumulate values along a curve, like summing up a function's value or components of a vector field. When dealing with vector fields, line integrals come in handy to evaluate the flow through a path or the work done by a force over a distance.
In this problem, we specifically focus on how the vector field \(\vec{F} = \langle x+y, x-y \rangle\) interacts with the curve \(C\), and we compute the flux across that path. The key task is to find how the vector field's components project onto the normal to the curve.
A typical line integral for vector fields requires the dot product \(\vec{F} \cdot \vec{n}\), followed by integration along \(ds\), the differential path of \(C\), from the starting to ending points. This integral essentially measures the cumulative effect of \(\vec{F}\) across the curve.

Parameterization of Curves

Parameterization simplifies dealing with curves by associating them with a continuous range of values. For instance, here we parametrize the curve \(C\) with \(\vec{r}(t) = \langle 3, t \rangle\).
This parameterization indicates a vertical line segment from \((3, -2)\) to \((3, 2)\), governed by the parameter \(t\) ranging from -2 to 2. It assigns a position vector \(\vec{r}(t)\) to every value of \(t\), enabling easy tracking along the curve.
By parametrizing, we can effectively dissect the curve into simpler computations, making both analysis and integration much easier. The straightforward equations such as \(x = 3\) and \(y = t\) within the parameterization are particularly useful for vector field evaluations, as they help express the vector field components in terms of the parameter \(t\).

Tangent and Normal Vectors

Tangent and normal vectors provide directions along and perpendicular to the path. Finding these vectors aids in understanding the curve's geometry better.
The tangent vector is determined by differentiating the parametric representation of \(C\) with respect to \(t\). Here, differentiating \(\vec{r}(t) = \langle 3, t \rangle\) yields \(\frac{d\vec{r}}{dt} = \langle 0, 1 \rangle\), pointing along the path direction.
Normals are perpendicular to tangents and are crucial for flux calculations, as they represent the direction of the projection of the vector field. For our vertical segment, the normal \(\vec{n} = \langle 1, 0 \rangle\) is chosen to point in the positive \(x\)-direction, simplifying the integral setup and evaluation process.
Understanding these vectors is essential for setting up correct integrations, as they determine how the vector field components contribute along or across the path.