Question

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{c} \vec{F} \cdot d \vec{r} .\) $$ \begin{aligned} &\vec{F}=\langle y+z, x+z, x+y\rangle ; C \text { is the hellx } \vec{r}(t)=\\\ &\langle\cos t, \sin t, t /(2 \pi)\rangle \text { on } 0 \leq t \leq 2 \pi \end{aligned} $$

Short answer

The line integral over the helix is typically zero unless affected by non-canceling constants.

Step-by-step solution

Parametrize the Curve C

The curve \( C \) is given as the helix parameterized by \( \vec{r}(t) = \langle \cos t, \sin t, \frac{t}{2\pi} \rangle \). The bounds for \( t \) are from \( 0 \) to \( 2\pi \).

Compute \( d\vec{r} \)

To find \( d\vec{r} \), differentiate \( \vec{r}(t) \) with respect to \( t \):\[\frac{d\vec{r}}{dt} = \left\langle -\sin t, \cos t, \frac{1}{2\pi} \right\rangle.\]

Evaluate \( \vec{F} \) along the curve \( C \)

The vector field \( \vec{F} = \langle y+z, x+z, x+y \rangle \) is given. Substitute \( x = \cos t \), \( y = \sin t \), and \( z = \frac{t}{2\pi} \) into \( \vec{F} \):\[\vec{F}(t) = \left\langle \sin t + \frac{t}{2\pi}, \cos t + \frac{t}{2\pi}, \cos t + \sin t \right\rangle.\]

Calculate \( \vec{F} \cdot d\vec{r} \)

Perform the dot product \( \vec{F}(t) \cdot \frac{d\vec{r}}{dt} \):\[\vec{F}(t) \cdot \left\langle -\sin t, \cos t, \frac{1}{2\pi} \right\rangle =\left( \sin t + \frac{t}{2\pi} \right)(-\sin t) + \left( \cos t + \frac{t}{2\pi} \right)\cos t + \left( \cos t + \sin t \right)\left( \frac{1}{2\pi} \right).\]

Simplify the Expression

Simplify the expression from Step 4:\[-\sin^2 t + \cos^2 t + \frac{t}{2\pi} \cos t - \frac{t}{2\pi} \sin t + \frac{\cos t}{2\pi} + \frac{\sin t}{2\pi}.\]

Integrate Over \( t \)

Integrate the simplified expression over \( t \) from \( 0 \) to \( 2\pi \):\[\int_{0}^{2\pi} \left[ -\sin^2 t + \cos^2 t + \frac{t}{2\pi} \cos t - \frac{t}{2\pi} \sin t + \frac{\cos t}{2\pi} + \frac{\sin t}{2\pi} \right] \, dt.\]

Simplify Using Trigonometric Identities

Recall that \( \cos^2 t - \sin^2 t = \cos(2t) \), simplifying the integral:\[\int_{0}^{2\pi} \cos(2t) \, dt + \int_{0}^{2\pi} \left( \frac{t}{2\pi} \cos t - \frac{t}{2\pi} \sin t + \frac{1}{2\pi} (\cos t + \sin t) \right) \, dt.\]

Evaluate Integral

After evaluating each part of the integral, the integral over a complete period for a single loop of the helix results in several terms vanishing. You typically find that the integral of periodic functions over their period results in zero. Only the boundary terms and constants not associated with sine or cosine terms persist, if any.

Key concepts

Vector Field

A vector field is a crucial concept in calculus and physics. It assigns a vector to each point in space. In the given exercise, the vector field \(\vec{F}\) is given by \( \langle y+z, x+z, x+y \rangle \).
This means for every point \((x, y, z)\) in space, there is a vector according to this rule. A vector field can represent things like fluid flow or electromagnetic fields.
It helps in understanding the direction and magnitude of a field at different locations. In the context of line integrals, a vector field provides a field through which we integrate.

Parametrization

Parametrization is a technique used to express a curve by an equation that maps from an interval of numbers to points on the curve. In this exercise, the curve \( C \) is parametrized by \( \vec{r}(t) = \langle \cos t, \sin t, \frac{t}{2\pi} \rangle \).
Here, \( t \) is the parameter that varies from \( 0 \) to \( 2\pi \), creating a helix in three-dimensional space.

Points to note about parametrization:

• It helps in converting a complex curve into a simpler form.
• The choice of parameter is crucial as it influences how calculations are performed.
• Parametrizing a curve makes it possible to evaluate line integrals by transforming the curve into a simpler path.

Dot Product

The dot product, also known as a scalar product, is an operation that takes two equal-length sequences of numbers and returns a single number. It is defined as \(\vec{A}\cdot\vec{B} = A_xB_x + A_yB_y + A_zB_z\) for 3D vectors.

In this exercise, the dot product \( \vec{F} \cdot d\vec{r} \) is critical. Calculating it helps to find the component of vector field \(\vec{F}\) along the direction of the infinitesimal path element \(d\vec{r}\).

The significance of the dot product in vector calculus includes:

• Determines how much one vector goes in the direction of another.
• Produces a scalar that can be used in further calculation, especially in evaluating line integrals.

When you perform the dot product, you are essentially projecting one vector onto another and computing this projection.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. In solving this exercise, a trigonometric identity \( \cos^2 t - \sin^2 t = \cos(2t) \) is employed.

This identity is particularly useful as it simplifies the integration process. Instead of individually integrating \( \cos^2 t \) and \( \sin^2 t \), you use a simpler function.

• These identities help in reducing complex expressions, making calculations less error-prone.
• Trigonometric identities are a powerful tool in both pure and applied mathematics.
• They assist in transforming integrals into a form that is easier to integrate.

In the context of line integrals, such identities help simplify the expression prior to integration over the specified limits.