Line Integrals - Part I — Question 6

Question 6 – Solution

Strategy. Reversing direction changes the parameter derivative’s sign but not its magnitude, so $ds$ is unchanged.

Step 1: Counterclockwise Let $\mathbf r_+(t)=\langle\cos t,\sin t\rangle$, $0\le t\le 2\pi$. Since $|\mathbf r_+'(t)|=1$, $$I_+=\int_0^{2\pi}(\cos t+2)\,dt=0+4\pi=4\pi.$$

Step 2: Clockwise Let $\mathbf r_-(t)=\langle\cos t,-\sin t\rangle$, $0\le t\le 2\pi$. Again the speed is $1$, and $x=\cos t$, so $$I_-=\int_0^{2\pi}(\cos t+2)\,dt=\boxed{4\pi}.$$

Step 3: General reason Under a reversed parametrization, $\widetilde{\mathbf r}(t)=\mathbf r(a+b-t)$, the derivative is $-\mathbf r'(a+b-t)$, but $$|\widetilde{\mathbf r}'(t)|=|\mathbf r'(a+b-t)|.$$ After substitution, the scalar values and arc-length factors are identical. Therefore $\int_C f\,ds$ does not depend on orientation.

Verification Here the $x$-term cancels by symmetry, while the constant $2$ contributes $2$ times the circumference $2\pi$, confirming $4\pi$.