Question

The value of the line integral \[ \oint (x^2 \, dx + 2x \, dy) \] along the ellipse \( 4x^2 + y^2 = 4 \), oriented in the counterclockwise sense, is:

• A. \( \pi \)
• B. \( 2\pi \)
• C. \( 4\pi \)
• D. \( 8\pi \)

Hint:

When evaluating line integrals over closed curves, consider using Green's Theorem if the curve encloses a well-defined region. This often simplifies calculations significantly.

Answer 1

The Correct Option is C

Step 1: Recall Green's Theorem. 
Green's Theorem relates a line integral over a closed curve \( C \) to a double integral over the region \( R \) enclosed by \( C \): \[ \oint_C \left( P \, dx + Q \, dy \right) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA, \] where \( P(x, y) = x^2 \) and \( Q(x, y) = 2x \) in this problem. 

Step 2: Compute the partial derivatives. 
The partial derivatives are: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2x) = 2, \quad \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2) = 0. \] Thus: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - 0 = 2. \] Step 3: Parametrize the ellipse and compute the area. 
The equation of the ellipse is \( \frac{x^2}{1} + \frac{y^2}{4} = 1 \), so the semi-major axis is \( a = 2 \), and the semi-minor axis is \( b = 1 \). The area of the ellipse is: \[ \text{Area} = \pi \cdot a \cdot b = \pi \cdot 2 \cdot 1 = 2\pi. \] Step 4: Evaluate the double integral. 
Using Green's Theorem, the line integral becomes: \[ \oint_C (x^2 \, dx + 2x \, dy) = \iint_R 2 \, dA = 2 \cdot (\text{Area of ellipse}) = 2 \cdot 2\pi = 4\pi. \] Conclusion: The value of the line integral is \( 4\pi \).