Line Integrals - Part I — Question 3

Question 3 – Solution

Strategy. On a circle, the scalar integrand $x^2+y^2$ is constant, so the integral is that constant times arc length.

Step 1: Parametrization $$\mathbf r(\theta)=\langle 2\cos\theta,2\sin\theta\rangle, \qquad 0\le\theta\le\pi.$$ This begins at $(2,0)$ and ends at $(-2,0)$, as required.

Step 2: Evaluate Along $C$, $$x^2+y^2=4,\qquad |\mathbf r'(\theta)| =|\,\langle-2\sin\theta,2\cos\theta\rangle\,|=2.$$ Thus $$\int_C(x^2+y^2)\,ds =\int_0^\pi(4)(2)\,d\theta =\boxed{8\pi}.$$

Verification The upper semicircle has length $\frac 12(2\pi\cdot 2)=2\pi$. Since the integrand is constantly $4$, the integral must be $4(2\pi)=8\pi$.