Question

Let \( C \) be the boundary of the region \( R : 0 \leq x \leq \pi, 0 \leq y \leq \sin x \) in the xy-plane and \( \alpha \) be the area of the region \( R \). If \( C \) traverses once in the counterclockwise direction, then the value of the line integral \[ \int_C (2y \, dx + 5x \, dy) \] is equal to

• A. \( \alpha \)
• B. \( 2\alpha \)
• C. \( 3\alpha \)
• D. \( 4\alpha \)

Hint:

Use Green's Theorem to convert a line integral into a double integral, which can simplify the calculation of the area or other properties of the region.

Answer 1

The Correct Option is C

We are given a line integral over the boundary \( C \) of a region \( R \) in the xy-plane. The formula for the line integral is: \[ \int_C (2y \, dx + 5x \, dy). \] To solve this, we can apply Green's Theorem, which relates a line integral over a closed curve to a double integral over the region it encloses: \[ \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA, \] where \( P = 2y \) and \( Q = 5x \). First, compute the partial derivatives: \[ \frac{\partial Q}{\partial x} = 5, \frac{\partial P}{\partial y} = 2. \] Thus, the integral becomes: \[ \iint_R (5 - 2) \, dA = \iint_R 3 \, dA = 3 \times \text{Area of } R. \] The area of region \( R \) is \( \alpha \). Therefore, the value of the line integral is: \[ 3\alpha. \] Final Answer: \text{(C) \( 3\alpha \)}