Question

In Green's theorem, \( \oint_C (x^2y dx + x^2 dy) = \iint_R f(x,y) dx dy \), where C is the boundary described counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1) and R is the region bounded by a simple closed curve C in the x-y plane, then \( f(x,y) \) is equal to:

• A. \( x - x^2 \)
• B. \( 2x - x^2 \)
• C. \( y - x^2 \)
• D. \( 2y - x^2 \)

Hint:

This is a straightforward application of the Green's theorem formula. Be careful to correctly identify P (the coefficient of dx) and Q (the coefficient of dy) and to perform the subtraction in the correct order: \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The question asks to identify the integrand \( f(x,y) \) that results from applying Green's theorem to a given line integral. Green's theorem relates a line integral around a simple closed curve to a double integral over the region it encloses.

Step 2: Key Formula or Approach:
Green's Theorem is given by the formula: \[ \oint_C P(x,y) dx + Q(x,y) dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA \] The question states this is equal to \( \iint_R f(x,y) dx dy \). By comparing the two forms, we can see that: \[ f(x,y) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \]
Step 3: Detailed Explanation:
From the given line integral \( \oint_C (x^2y dx + x^2 dy) \), we identify the functions \( P \) and \( Q \): \[ P(x,y) = x^2y \] \[ Q(x,y) = x^2 \] Now, we compute the required partial derivatives: \[ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x \] \[ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2y) = x^2 \] Substitute these into the formula for \( f(x,y) \): \[ f(x,y) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2x - x^2 \]
Step 4: Final Answer:
The function \( f(x,y) \) is equal to \( 2x - x^2 \).