Let's clarify and **solve the line integral along the curve C**, which is the **upper half of a circle of radius 2**. The parameterization is given as: - \( x(t) = 2\cos(t) \) - \( y(t) = 2\sin(t) \) Since it's the upper half of a circle, **\( t \) goes from \( 0 \) to \( \pi \)**. --- ### Step 1: Parameterize the Curve Given: - \( x(t) = 2\cos(t) \) - \( y(t) = 2\sin(t) \) - For the **upper half**, \( t \in [0, \pi] \) --- ### Step 2: Write the Differential Elements - \( dx = \frac{dx}{dt} dt = -2\sin(t) dt \) - \( dy = \frac{dy}{dt} dt = 2\cos(t) dt \) --- ### Step 3: Write the Integral Since the integrand is not given, let's assume the problem is to **evaluate the length of the upper half of the circle** (arc length), as that's a common question for a parameterized curve. **Arc length \( L \) is:** \[ L = \int_{C} ds = \int_{t=0}^{t=\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt \] Substitute: - \( \frac{dx}{dt} = -2\sin(t) \) - \( \frac{dy}{dt} = 2\cos(t) \) So, \[ \sqrt{(-2\sin(t))^2 + (2\cos(t))^2} = \sqrt{4\sin^2(t) + 4\cos^2(t)} = \sqrt{4(\sin^2(t) + \cos^2(t))} = \sqrt{4(1)} = 2 \] --- ### Step 4: Integrate \[ L = \int_{0}^{\pi} 2 \, dt = 2t \Big|_{0}^{\pi} = 2\pi - 0 = 2\pi \] --- ### **Final Answer** \[ \boxed{2\pi} \] This is the length of the upper half of a circle of radius 2. --- **If you are integrating a different function over C, please provide the integrand.** Otherwise, this is the standard calculation for the arc length of the upper half of a circle with the given parameterization.