Problem 3 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. The line integral of Exercise \(1 ; C\) : the \(x\) axis from the origin to \((2,0)\) and then the line \(x=2\) from \((2,0)\) to \((2,2)\).

Step-by-Step Solution

Answer

ewline \( \int_0^2 f(t, 0) dt + \int_0^2 f(2, t) dt \)

Step 1: Understand the Curve

The curve consists of two parts: 1. The segment along the x-axis from the origin (0,0) to (2,0).2. The vertical line from (2,0) to (2,2).

Step 2: Parametrize the Segments

For the segment along the x-axis: ewline Use the parameterization: ewline \( x = t, \ y = 0 \) where \(0 \leq t \leq 2 \) ewline For the vertical line: ewline Use the parameterization: ewline \( x = 2, \ y = t \) where \(0 \leq t \leq 2 \)

Step 3: Setup the Line Integrals

The line integral will be the sum of the integrals over each segment. ewline For the x-axis segment: ewline \( \int_C f(x, y) \cdot ds = \int_0^2 f(t, 0) \cdot \|r'(t)\| dt \) ewline For the vertical line segment: ewline \( \int_C f(x, y) \cdot ds = \int_0^2 f(2, t) \cdot \|r'(t)\| dt \)

Step 4: Compute the Magnitude of r'(t)

For the x-axis segment: ewline \( r(t) = (t, 0) \Rightarrow r'(t) = (1, 0) \Rightarrow \|r'(t)\| = 1 \) ewline For the vertical line segment: ewline \( r(t) = (2, t) \Rightarrow r'(t) = (0, 1) \Rightarrow \|r'(t)\| = 1 \)

Step 5: Integrate Over Each Segment

For the x-axis segment: ewline \( \int_0^2 f(t, 0) \cdot 1 \cdot dt = \int_0^2 f(t, 0) dt \) ewline For the vertical line segment: ewline \( \int_0^2 f(2, t) \cdot 1 \cdot dt = \int_0^2 f(2, t) dt \)

Step 6: Combine the Results

Add the results from both integral computations: ewline \( \int_C f(x, y) \cdot ds = \int_0^2 f(t, 0) dt + \int_0^2 f(2, t) dt \)

Key Concepts

Parametrization

When dealing with line integrals, parametrization is a crucial step. Parametrization involves expressing the coordinates of the points on the curve as functions of a parameter. In this exercise, we have two segments of the curve to parametrize:

• The x-axis segment from the origin \( (0,0) \) to \( (2,0) \) is parameterized by letting \( x = t \) and \( y = 0 \), where \( 0 \leq t \leq 2 \).


• The vertical line from \( (2,0) \) to \( (2,2) \) is parameterized by letting \( x = 2 \) and \( y = t \), where \( 0 \leq t \leq 2 \).

These parametrizations make setting up integrals simpler. We turn otherwise complex functions into manageable expressions.

Curve

In this context, a curve represents the path along which the integration takes place. Our curve consists of two segments:

• First segment: This is the horizontal line from point \( (0,0) \) to point \( (2,0) \). Because it’s along the x-axis, \( y \) remains zero throughout this segment.


• Second segment: This is the vertical line from point \( (2,0) \) to point \( (2,2) \). Here, \( x = 2 \) always, while \( y \) varies from \( 0 \) to \( 2 \).

Recognizing the curve and its segments helps us to correctly parametrize and integrate functions over them.

Integration

Integration in the context of line integrals requires summing up the values of a function along a curve. After parametrizing the curve, we set up and evaluate the integrals for each segment:

• For the x-axis segment, we get the integral:
  \[ \int_0^2 f(t, 0) \, dt \]


• For the vertical line segment, we get the integral:
  \[ \int_0^2 f(2, t) \, dt \]

Notice how we compute the magnitude of the derivative of the parametrized functions (which is 1 in both cases), simplifying the computations.
Finally, we combine these integrals to obtain the result:
\[\textstyle \int_C f(x, y) \, ds = \int_0^2 f(t, 0) \, dt + \int_0^2 f(2, t) \, dt \]
This completes our line integral evaluation. Understanding integration in this way showcases the power of parametrization and breaking down curves into integrable segments.