Question

Consider a unit circle C in the xy plane, centered at the origin. The value of the integral \(\oint\)[(sin x - y)dx - (sin y - x)dy] over the circle C, traversed anticlockwise, is

• A. 0
• B. 2π
• C. 3π
• D. 4π

Answer 1

The Correct Option is B

To solve the integral \(\oint\) \([( \sin x - y ) \, dx - ( \sin y - x ) \, dy]\) over the unit circle \( C \) centered at the origin and traversed anticlockwise, we can employ Green's Theorem. This theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \).

Green's Theorem is expressed as:

\(\oint_{C} ( P \, dx + Q \, dy ) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\)

where \( P(x, y) = \sin x - y \) and \( Q(x, y) = -(\sin y - x) = -\sin y + x \). We can now calculate the partial derivatives:

• \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x - \sin y) = 1\)
• \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (\sin x - y) = -1\)

Substituting these into Green's Theorem, we get:

\(\oint_{C} ((\sin x - y) \, dx - (\sin y - x) \, dy) = \iint_{D} \left( 1 - (-1) \right) \, dA = \iint_{D} 2 \, dA\)

The region \( D \) is the unit circle, which has an area of \(\pi \cdot r^2\), where \( r = 1 \). Therefore, the area of the unit circle is \(\pi\).

Thus, the double integral over \( D \) becomes:

\(\iint_{D} 2 \, dA = 2 \times \pi = 2\pi\)

Therefore, the value of the given integral is 2π.