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 $4\pi$.