Problem 13
Question
Evaluate \(\int_{C} y d x+x d y\) on the given curve \(C\) between \((0,0)\) and \((1,1)\). \(C\) consists of the line segments from \((0,0)\) to \((0,1)\) and from \((0,1)\) to \((1,1)\).
Step-by-Step Solution
Verified Answer
The integral evaluates to 1.
1Step 1: Set Up Parameterization for line segment (0,0) to (0,1)
The first line segment is vertical. For this segment, parameterize using the parameter \( t \) such that \( x = 0 \) and \( y = t \), where \( 0 \leq t \leq 1 \). Then, \( dx = 0 \) and \( dy = dt \).
2Step 2: Evaluate Integral on First Segment
Substitute the parameterization into the integral. The integral becomes:\[\int_{0}^{1} (t \cdot 0) \, dt + 0 \cdot dt = \int_{0}^{1} 0 \, dt = 0.\]
3Step 3: Set Up Parameterization for line segment (0,1) to (1,1)
The second line segment is horizontal. For this segment, parameterize using \( t \) such that \( x = t \) and \( y = 1 \), where \( 0 \leq t \leq 1 \). Then, \( dx = dt \) and \( dy = 0 \).
4Step 4: Evaluate Integral on Second Segment
Substitute the parameterization into the integral for this segment. It becomes:\[\int_{0}^{1} 1 \cdot dt + t \cdot 0 = \int_{0}^{1} 1 \, dt = \left[ t \right]_{0}^{1} = 1.\]
5Step 5: Combine Integrals for Full Path
Add the integrals from the two segments to find the integral over the entire path \( C \):\[ 0 + 1 = 1.\]
Key Concepts
ParameterizationIntegral EvaluationVector Calculus
Parameterization
To solve a line integral, you must first understand the concept of parameterization. Parameterization is the process of expressing a curve as a function of a single variable, typically denoted as \( t \). This approach allows us to express curves on the plane in a simpler, more manageable form, making it easier to compute integrals along these paths.
In the exercise given, we have two line segments. Both segments require parameterization:
In the exercise given, we have two line segments. Both segments require parameterization:
- For the vertical segment from \((0,0)\) to \((0,1)\), we parameterize the line by setting \( x = 0 \), while \( y = t \), where \( 0 \leq t \leq 1 \).
- For the horizontal segment from \((0,1)\) to \((1,1)\), the parameterization is \( x = t \) and \( y = 1 \), again for \( 0 \leq t \leq 1 \).
Integral Evaluation
With the curve fully parameterized, the next step in evaluating a line integral is substituting these expressions into the integral. Each segment is handled separately at first and then combined:
- **First Segment (Vertical):** By substituting the parameterization for the vertical segment, the integral simplifies given that \( dx = 0 \) and \( dy = dt \).The calculation results in:\[ \int_{0}^{1} (t \cdot 0) \, dt + 0 \, dt = \int_{0}^{1} 0 \, dt = 0. \] This tells us that the contribution from the vertical segment of the path is zero.
- **Second Segment (Horizontal):** For the horizontal segment, we substitute similarly and find:\[ \int_{0}^{1} 1 \, dt + t \cdot 0 = \int_{0}^{1} 1 \, dt = \left[ t \right]_{0}^{1} = 1. \]This shows the only non-zero contribution comes from this segment.
Vector Calculus
Line integrals are a fundamental concept in vector calculus, a branch of mathematics that deals with vector fields and operations on them. Line integrals, specifically, calculate the accumulation of a vector field along a curve. In simpler terms, they help compute how a vector field acts on a path.
Key elements to remember about vector calculus in line integrals include:
Key elements to remember about vector calculus in line integrals include:
- **Vector Fields:** These are associations of vectors to each point in a plane. In this problem, vector fields don't change direction, but understanding them is crucial for more complex problems.
- **Curve Paths:** The path \( C \) is decomposed into segments, each requiring parameterization, to perform a line integral.
- **Fundamental Theorems:** Line integrals are connected deeply to concepts like work and circulation, often simplified using theorems such as Green's Theorem, which relate a double integral over a plane region to a line integral around its boundary.
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