Line Integrals - Part II — Question 4

Question 4 – Solution

Strategy. Compute once in the counterclockwise direction, then use the sign change caused by reversing $dx$.

Step 1: Forward orientation For $C_1$, use $$x=\cos\theta,\quad y=\sin\theta,\quad 0\le\theta\le\pi,\quad dx=-\sin\theta\,d\theta.$$

Therefore $$\int_{C_1}y\,dx =-\int_0^\pi\sin^2\theta\,d\theta =\boxed{-\frac{\pi}{2}}.$$

Step 2: Reverse orientation Reversing the same curve negates the coordinate differential, so $$\boxed{\int_{C_2}y\,dx=\frac{\pi}{2}}.$$

Step 3: Why the rules differ Under reversal, $dx=x'(t)dt$ changes sign because the derivative reverses. By contrast, $ds=|\mathbf r'(t)|dt$ uses a magnitude and remains nonnegative.

Verification On $C_1$, $y\ge 0$ while $x$ decreases, so $dx\le 0$ and the integral must be nonpositive. Its reverse must be nonnegative, matching the two answers.