Question

The value of $\oint_C x \, dy - y \, dx$, where $C$ is the boundary of the rectangular region formed by $x = 5$, $x = 12$, $y = 0$, and $y = 8$ is:

• A. $56$
• B. $28$
• C. $112$
• D. $72$

Hint:

The line integral $\frac{1}{2} \oint_C (x \, dy - y \, dx)$ is a standard formula for computing the area of a closed region. Therefore, $\oint_C (x \, dy - y \, dx)$ is always equal to $2 \times \text{Area}$.

Answer 1

The Correct Option is C

Concept: By Green's Theorem in a plane, a line integral around a positively oriented, piecewise smooth, simple closed curve $C$ can be converted into a double integral over the region $R$ bounded by $C$: \[ \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx \, dy \] In this problem, the given integral is $\oint_C (-y \, dx + x \, dy)$. Comparing this with the standard form: \[ P = -y \implies \frac{\partial P}{\partial y} = -1 \] \[ Q = x \implies \frac{\partial Q}{\partial x} = 1 \]

Step 1: Apply Green's Theorem to set up the double integral.
Substituting the partial derivatives into Green's Theorem formula: \[ \oint_C x \, dy - y \, dx = \iint_R \left( 1 - (-1) \right) dx \, dy = \iint_R 2 \, dx \, dy = 2 \iint_R dx \, dy \] The double integral $\iint_R dx \, dy$ represents the geometric area of the region $R$.

Step 2: Calculate the area of the rectangular region $R$.
The region $R$ is a rectangle defined by the boundaries: \[ x \text{ goes from } 5 \text{ to } 12 \implies \text{Length} = 12 - 5 = 7 \] \[ y \text{ goes from } 0 \text{ to } 8 \implies \text{Width} = 8 - 0 = 8 \] Thus, the area of the rectangle is: \[ \text{Area}(R) = \text{Length} \times \text{Width} = 7 \times 8 = 56 \]

Step 3: Compute the final integral value.
Using our relation from
Step 1: \[ \oint_C x \, dy - y \, dx = 2 \times \text{Area}(R) = 2 \times 56 = 112 \]