Question

Let \( C \) be the positively oriented boundary of the domain bounded by the curves \( y = 2x^2 \) and \( y^2 = 4x \). Then the value of the line integral \[ \oint_C \left( 2y^2 + 2xy + 4y \right) dx + \left( x^2 + 4xy + 8x \right) dy \] is equal to:

• A. \( \frac{8}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{4}{3} \)
• D. \( \frac{1}{3} \)

Hint:

Green's Theorem can simplify line integrals around closed curves by converting them into double integrals over the region enclosed by the curve. Remember to compute the partial derivatives correctly.

Answer 1

The Correct Option is A

This problem involves a line integral around a closed curve. We can solve this using Green's Theorem, which converts a line integral over a closed curve into a double integral over the region \( D \) enclosed by the curve. 
Green's Theorem states: \[ \oint_C P(x, y) dx + Q(x, y) dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \] Here, \( P(x, y) = 2y^2 + 2xy + 4y \) and \( Q(x, y) = x^2 + 4xy + 8x \). We need to compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = 2x + 4y + 8, \frac{\partial P}{\partial y} = 4y + 2x + 4 \] Thus, the integrand becomes: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (2x + 4y + 8) - (4y + 2x + 4) = 4 \] Now, we need to find the area of the region \( D \), which is bounded by the curves \( y = 2x^2 \) and \( y^2 = 4x \). The area can be computed as: \[ {Area} = \int_{x=0}^{1} \left( \sqrt{4x} - 2x^2 \right) dx \] Evaluating the integral gives the area as \( \frac{2}{3} \). Thus, the line integral is: \[ \oint_C \left( 2y^2 + 2xy + 4y \right) dx + \left( x^2 + 4xy + 8x \right) dy = 4 \times \frac{2}{3} = \frac{8}{3} \]