Line Integrals of Vector Fields — Question 4

Question 4 – Solution

Strategy. On the unit circle this field is the counterclockwise unit tangent, making the dot product especially simple.

Step 1: Counterclockwise traversal Take $$\mathbf r(t)=\langle\cos t,\sin t\rangle,\quad 0\le t\le 2\pi.$$ Then $$\mathbf F(\mathbf r(t))=\langle-\sin t,\cos t\rangle=\mathbf r'(t).$$

Consequently, $$\oint_C\mathbf F\cdot d\mathbf r =\int_0^{2\pi}\mathbf r'(t)\cdot\mathbf r'(t)\,dt =\int_0^{2\pi}1\,dt=\boxed{2\pi}.$$

Step 2: Reverse the orientation Reversing a vector-field line integral changes its sign, so the clockwise value is $$\boxed{-2\pi}.$$

Verification The field assists counterclockwise motion at every point and opposes clockwise motion at every point, so the respective signs must be positive and negative.