Question

Let \( r : [0,1] \to \mathbb{R}^2 \) be a continuously differentiable path from \( (0,2) \) to \( (3,0) \) and let \[ \mathbf{F} : \mathbb{R}^2 \to \mathbb{R}^2 be  \mathbf{F}(x, y) = (1 - 2y, 1 - 2x). \] The line integral of \( \mathbf{F} \) along \( r \) is equal to \(\_\_\_\_\_\_\) (round off to TWO decimal places).

Hint:

For line integrals of vector fields, parameterize the path and compute \( \int \mathbf{F}(r(t)) \cdot r'(t) dt \).

Answer 1

1. Line Integral Formula:
The line integral of \( \mathbf{F} \) along \( r \) is given by: \[ \int \mathbf{F} \cdot d\mathbf{r} = \int_0^1 \left[ \mathbf{F}(r(t)) \cdot r'(t) \right] dt, \] where \( r(t) = (x(t), y(t)) \) represents the path and \( r'(t) = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) \).
2. Path Definition: Since \( r(t) \) is a straight-line path from \( (0,2) \) to \( (3,0) \), it can be parameterized as: \[ r(t) = (3t, 2 - 2t), \quad r'(t) = (3, -2). \]
3. Substitute \( r(t) \) into \( \mathbf{F} \): Substituting \( x = 3t \) and \( y = 2 - 2t \) into \( \mathbf{F}(x, y) = (1 - 2y, 1 - 2x) \), we get: \[ \mathbf{F}(r(t)) = \left( 1 - 2(2 - 2t), 1 - 2(3t) \right) = (-3 + 4t, 1 - 6t). \]
4. Dot Product \( \mathbf{F}(r(t)) \cdot r'(t) \): Compute the dot product: \[ \mathbf{F}(r(t)) \cdot r'(t) = (-3 + 4t)(3) + (1 - 6t)(-2) = -9 + 12t - 2 + 12t = -11 + 24t. \]
5. Evaluate the Integral: The integral becomes: \[ \int_0^1 (-11 + 24t) dt = \left[ -11t + 12t^2 \right]_0^1 = -11(1) + 12(1)^2 - (-11(0) + 12(0)^2) = -11 + 12 = 1. \] Final Answer: 1.0