= Solution
{c}
Write the <centroid> as $\overline{\mathbf r}=\overline x\,\mathbf i+\overline y\,\mathbf j$. The region has <reflection symmetry> in the $y$-axis: replacing $\theta$ by $\pi-\theta$ changes $x$ to $-x$ and preserves $y$ and $dA$. Thus <symmetry in integration> gives
$$
\iint_Rx\,dA=0,\qquad\overline x=0.
$$
For the other <first moment of area>, use $y=r\sin\theta$ in <polar coordinates>:
$$
\iint_Ry\,dA
=\int_0^\pi\int_0^{\sin^2\theta}r^2\sin\theta\,dr\,d\theta
=\frac13\int_0^\pi\sin^7\theta\,d\theta
=\frac{32}{105}.
$$
Here the last <integral> has the value allowed in the paper. Dividing the <first moment of area> by the <area> found in <i> gives
$$
\boxed{(\overline x,\overline y)=\left(0,\frac{512}{315\pi}\right)},
\qquad
\boxed{\overline{\mathbf r}=\frac{512}{315\pi}\,\mathbf j}.
$$
The positive vertical coordinate agrees with the fact that the entire region lies above the $x$-axis.
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