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:
Instead of tracing the exact shape of the semicircle, it helps to first check whether the vector field behind this integral is conservative.
If it is, the integral only depends on the start and end points, not on the actual path taken.
Here the field is $\mathbf{F} = 2x\mathbf{i} + 2y\mathbf{j} + 2z\mathbf{k}$.

Step 2: Key Formula or Approach:
A field is conservative if it can be written as the gradient of some scalar function $\phi$, that is $\mathbf{F} = \nabla \phi$.
For such a field, the Fundamental Theorem of Line Integrals gives:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \phi(\text{end point}) - \phi(\text{start point}) \]

Step 3: Detailed Explanation:
Try $\phi(x,y,z) = x^2 + y^2 + z^2$. Its partial derivatives are $\frac{\partial \phi}{\partial x} = 2x$, $\frac{\partial \phi}{\partial y} = 2y$ and $\frac{\partial \phi}{\partial z} = 2z$, which exactly match the field, so $\phi$ is a valid potential function and the field is conservative.
Since the path shape no longer matters, just evaluate $\phi$ at the two given endpoints, start $(0,0,0)$ and end $(1,0,0)$:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \phi(1,0,0) - \phi(0,0,0) = (1^2 + 0^2 + 0^2) - (0^2 + 0^2 + 0^2) \]
\[ = 1 - 0 = 1 \]

Final Answer:
Because the field is conservative, the integral equals the difference in potential between the endpoints. \[ \boxed{1} \]