Line Integrals - Part I — Question 8

Question 8 – Solution

Strategy. Parametrize the circle and divide the scalar line integral by total arc length.

Step 1: Integral and length Use $$\mathbf r(\theta)=\langle a\cos\theta,a\sin\theta\rangle, \qquad 0\le\theta\le 2\pi.$$ Then $ds=a\,d\theta$, so $$\int_Cx^2\,ds =a^3\int_0^{2\pi}\cos^2\theta\,d\theta =\pi a^3, \qquad L=2\pi a.$$

Step 2: Average value $$f_{\mathrm{avg}} =\frac 1L\int_Cf\,ds =\frac{\pi a^3}{2\pi a} =\boxed{\frac{a^2}{2}}.$$

Step 3: Points attaining the average We need $x^2=a^2/2$, so $x=\pm a/\sqrt 2$. The circle equation then gives $y=\pm a/\sqrt 2$, independently. Thus the four points are $$\boxed{\left(\frac{\pm a}{\sqrt 2},\frac{\pm a}{\sqrt 2}\right)},$$ with all four sign combinations.

Verification Symmetry gives equal averages for $x^2$ and $y^2$. Their sum is constantly $a^2$ on $C$, so each average must be $a^2/2$.