Question

Consider the line integral
\[ \int_C (2x\, dx + 2y\, dy + 2z\, dz) \]
where \( C \) is a semi-circle in the \( z = 0 \) plane with start point at \( (0, 0, 0) \) and end point at \( (1, 0, 0) \).

The value of the line integral is ______ (in integer).

Hint:

Check if the vector field is conservative by finding a potential function \( \phi \) with \( \nabla \phi = \mathbf{F} \); then the integral is just \( \phi(\text{end}) - \phi(\text{start}) \).

Answer 1

Correct Answer: 1

Step 1: Understanding the Concept:
We must evaluate a line integral along a semicircular path that starts at the origin and ends at \( (1, 0, 0) \), lying entirely in the \( z = 0 \) plane.
Since \( z = 0 \) all along the path, the \( 2z\, dz \) term contributes nothing, so only the \( x \) and \( y \) terms matter.
We will parameterize the semicircle directly and integrate over the parameter.

Step 2: Key Formula or Approach:
The two endpoints \( (0,0,0) \) and \( (1,0,0) \) are the ends of a diameter of length 1, so the semicircle has centre \( (0.5, 0, 0) \) and radius \( 0.5 \).
A convenient parameterization is:
\[ x = 0.5 + 0.5\cos\theta, \quad y = 0.5\sin\theta, \quad z = 0, \quad \theta: \pi \to 0 \]

Step 3: Detailed Explanation:
Check the endpoints: at \( \theta = \pi \), \( x = 0.5 - 0.5 = 0 \), \( y = 0 \), matching the start point; at \( \theta = 0 \), \( x = 0.5 + 0.5 = 1 \), \( y = 0 \), matching the end point.
Differentiating: \( dx = -0.5\sin\theta\, d\theta \) and \( dy = 0.5\cos\theta\, d\theta \).
Substituting into the integrand:
\[ 2x\,dx + 2y\,dy = -0.5\sin\theta\,d\theta - 0.5\sin\theta\cos\theta\,d\theta + 0.5\sin\theta\cos\theta\,d\theta = -0.5\sin\theta\,d\theta \]
The two cross terms cancel exactly, leaving a simple integral:
\[ \int_\pi^0 -0.5\sin\theta\,d\theta = -0.5\big[-\cos\theta\big]_\pi^0 = -0.5\big[(-1) - (1)\big] = -0.5 \times (-2) = 1 \]

Final Answer:
The value of the line integral works out to 1, regardless of the exact curved path taken. \[ \boxed{1} \]