Question

If \( E \) is the region bounded by a closed curve \( C \) in the \( xy \)-plane, then \( \iint_E dx\,dy = \)

• A. \( \displaystyle \oint_C \left( -\frac{y}{2} \, i + \frac{x}{2} \, j \right) \cdot d\vec{R} \)
• B. \( \displaystyle \oint_C \left( \frac{x}{2} \, i - \frac{y}{2} \, j \right) \cdot d\vec{R} \)
• C. \( \displaystyle \oint_C (xi + yj) \cdot d\vec{R} \)
• D. \( \displaystyle \oint_C (-yi + xj) \cdot d\vec{R} \)

Hint:

Green's Theorem lets you transform area integrals into line integrals around the boundary. For computing area, use the field \( \left( -\frac{y{2, \frac{x{2 \right) \).

Answer 1

The Correct Option is A

Apply Green’s Theorem
Green’s Theorem converts a double integral into a line integral: \[ \iint_E \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx\,dy = \oint_C (M\,dx + N\,dy) \] To compute \( \iint_E dx\,dy \), choose a vector field such that: \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 \] Select: \[ M = -\frac{y}{2}, \quad N = \frac{x}{2} \quad \Rightarrow \quad \frac{\partial N}{\partial x} = \frac{1}{2}, \quad \frac{\partial M}{\partial y} = -\frac{1}{2} \quad \Rightarrow \quad 1 \] Therefore, \[ \iint_E dx\,dy = \oint_C \left( -\frac{y}{2} dx + \frac{x}{2} dy \right) = \oint_C \left( -\frac{y}{2} i + \frac{x}{2} j \right) \cdot d\vec{R} \]