Line Integrals - Part I — Question 10

Question 10 – Solution

Strategy. Parametrize by polar angle; symmetry immediately predicts that the center of mass lies on the $y$-axis.

Step 1: Mass Let $x=R\cos\theta$, $y=R\sin\theta$, $0\le\theta\le\pi$, so $ds=R\,d\theta$ and $\delta=1+\sin\theta$. Then $$M=R\int_0^\pi(1+\sin\theta)\,d\theta =\boxed{R(\pi+2)}.$$

Step 2: Moments The moment about the $x$-axis is $$\begin{align*} M_x&=\int_Cy\delta\,ds =R^2\int_0^\pi\sin\theta(1+\sin\theta)\,d\theta\\ &=R^2\left(2+\frac{\pi}{2}\right). \end{align*}$$ For the $y$-axis, $$M_y=R^2\int_0^\pi\cos\theta(1+\sin\theta)\,d\theta=0.$$

Step 3: Center of mass $$\bar x=\frac{M_y}{M}=0,\qquad \bar y=\frac{M_x}{M} =\boxed{\frac{R(\pi+4)}{2(\pi+2)}}.$$ Thus $$\boxed{(\bar x,\bar y)= \left(0,\frac{R(\pi+4)}{2(\pi+2)}\right)}.$$

Verification Left-right symmetry forces $\bar x=0$. Also $0<(\pi+4)/[2(\pi+2)]<1$, so $0<\bar y<R$, placing the center of mass inside the semicircle’s vertical range.