91Ó°ÊÓ

In the context of line integrals, a vector field is a function that assigns a vector to every point in space. For example, in our exercise, we have a vector field defined as \( \vec{F} = x \vec{i} + 6 \vec{j} - \vec{k} \). This means that at every point \((x, y, z)\), the vector assigned is \((x, 6, -1)\).
A vector field can be thought of as describing forces, velocities, or other vector quantities acting throughout space. In practical terms, this could represent, for example, the flow of a fluid at different points or the force exerted by a magnetic field.
When dealing with a line integral, you essentially want to measure how much of the vector field is influencing an object as it traverses along a specific path.

Parametrization involves expressing the coordinates of points along a path using a parameter, usually denoted as \( t \). This method is crucial when calculating line integrals because it allows us to describe the path as a set of equations.
In our exercise, we need to parametrize the line segment from \((0, -2, 0)\) to \((0, -10, 0)\). A suitable parametrization for this straight line path is \( \vec{r}(t) = (0, -2 - 8t, 0) \) where \( t \) varies from 0 to 1.

• At \( t = 0 \): \( \vec{r}(0) = (0, -2, 0) \)
• At \( t = 1 \): \( \vec{r}(1) = (0, -10, 0) \)

Parametrization simplifies the process of integrating along a path, as it allows you to consider one variable (\( t \)) instead of multiple spatial components.

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation offers a way to measure how much one vector acts in the direction of another.
In the context of a line integral, the dot product helps to find the component of the vector field that affects the motion along the path. For our exercise, the expression \( \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt} \) computes this influence.

• \( \vec{F}(\vec{r}(t)) = (0, 6, -1) \)
• Derivative of the parametrization: \( \frac{d\vec{r}}{dt} = (0, -8, 0) \)
• Dot Product: \( (0, 6, -1) \cdot (0, -8, 0) = -48 \)

The result of \(-48\) tells us the cumulative effect of the vector field along our path in the direction specified by our parametrization.

The final step in computing the line integral involves integrating over the parameter \( t \), which represents our path. This process sums up the effects measured by the dot product over the entire path.
For our exercise, the line integral becomes \( \int_{0}^{1} -48 \, dt \). Integration over the range \( t = 0 \) to \( t = 1 \) calculates the total effect along the path:

• Integral: \( \int_{0}^{1} -48 \, dt = [-48t]_{0}^{1} \)
• Evaluation: \( [-48t]_{0}^{1} = -48 \cdot 1 - (-48 \cdot 0) \)
• Result: \(-48\)

This negative value indicates the direction and magnitude of work done by the vector field along the path, portraying how the vector field interacts with the path's orientation.