91Ó°ÊÓ

Understanding vector fields is crucial when working with line integrals. A vector field is a function that assigns a vector to each point in space. Picture it as an invisible field where each point has a direction and a magnitude. For example, in physics, vector fields can describe force fields, like gravity or magnetism.
In the given exercise, the vector field is described by \[\vec{F}(x, y) = \frac{y}{x+1} \vec{i} + \ln(x+1) \vec{j}\]This equation gives a vector at each point \((x, y)\), which will interact with any paths, or curves, through the field.
To find a line integral, we need to consider how this vector field acts along a given curve. Essentially, we're summing up all the influences of the field along that path.

Curve parameterization is a key step in solving line integrals. It involves expressing the path or curve with respect to a single variable, often denoted as \(t\) or \(\theta\). This simplifies the problem by reducing multiple variables to just one.
In the first part of the exercise, the curve is on the x-axis, thus it is parameterized as:

• \(x = t\)
• \(y = 0\)

where \(t\) varies from 0 to 3. This makes the entire process easier as it eliminates \(y\) from consideration in this segment.
For the circular path, which is part of a circle with radius 3, we use the parametric equations based on trigonometric functions:

• \(x = 3 \cos \theta\)
• \(y = 3 \sin \theta\)

where \(\theta\) ranges from 0 to \(\pi/4\). This transformation allows the curve to be expressed in terms of \(\theta\), simplifying further calculations.

Integration, particularly line integration, involves calculating the integral of a vector field along a curve. The key is to set up the integral properly using the parameterized path.
For the line segment on the x-axis, the line integral simplifies to zero due to the form of our vector field components, which lack an \(\vec{i}\) term while \(y = 0\). Therefore, the line integral turns out to be \[\int_0^3 0 \, dt = 0\].
In the case of the circular arc, you plug the parameterized equations into the vector field components and their derivatives. This involves processing the integral:\[\int_0^{\pi/4} \left( \frac{3 \sin \theta}{3 \cos \theta + 1} (-3 \sin \theta) + \ln(3 \cos \theta +1) 3 \cos \theta \right) d\theta\]This often requires substitution or using integration by parts, depending on the specific form.

Trigonometric identities are pivotal when parameterizing curves, especially if the curve is circular or involves sinusoidal functions. These identities help simplify expressions and can make integration possible.
In the exercise, the circular path utilizes the identities:

• \(\cos^2 \theta + \sin^2 \theta = 1\)
• The periodic and symmetric nature of sine and cosine functions.

These facilitate the parameterization of the curve, and consequently, the evaluation of the line integral. They make it possible to express locations on the unit circle in such a way that the calculus is manageable, transforming what might be complex integrations into more approachable forms.