Green's Theorem — Question 4

Question 4 – Solution

Strategy. On a region with a hole, positive orientation is counterclockwise outside and clockwise around the hole.

Step 1: Orientation Traverse the radius-$2$ circle counterclockwise and the radius-$1$ circle clockwise. In both cases the annular region remains on the left.

Step 2: Apply Green’s Theorem On $D$, the field is continuously differentiable and $$Q_x-P_y=0.$$ Therefore $$\oint_{\partial D}\mathbf F\cdot d\mathbf r=\iint_D0\,dA=\boxed{0}.$$

Step 3: Check each component On any counterclockwise circle centered at the origin, the integral is $2\pi$, independent of radius. Hence $$\int_{C_{\mathrm{outer}}}\mathbf F\cdot d\mathbf r=2\pi, \qquad \int_{C_{\mathrm{inner}}}\mathbf F\cdot d\mathbf r=-2\pi$$ because the inner circle is clockwise. Their sum is zero.

Verification If both circles were traversed counterclockwise, the total would be $4\pi$; that is not the positively oriented boundary of the annulus.