Find the centroid of the lamina bounded by the graph of \(y=f(x)=x^{2}+1\), the \(x\)-axis, and the lines \(x=-2\) and \(x=2\).

Solution Figure 73(a) shows the graph of the lamina.

We notice two properties of \(f\).

The graph of \(f\) is symmetric about the \(y\)-axis, so by the symmetry principle, \(\bar{x}=0.\)

\(f\) is an even function, so \(\displaystyle \int_{-2}^{2}f(x) ~{\it dx}=2\displaystyle \int_{0}^{2}f(x) ~{\it dx}\).

The area \(A\) of the region is \begin{equation*} A=2\displaystyle \int_{0}^{2}( x^{2}+1) ~{\it dx}=2\left[ \dfrac{x^{3}}{3} +x \right] _{0}^{2}=2\left( \dfrac{8}{3}+2\right) =\dfrac{28}{3}\ \end{equation*} Using (2), we get \begin{eqnarray*} \bar{y} &=&\dfrac{1}{2A}\displaystyle \int_{a}^{b}[ f(x) ] ^{2}\,{\it dx}=\dfrac{1}{2A}\displaystyle \int_{-2}^{2}[f(x)] ^{2}\,{\it dx}=\dfrac{1}{2A} \cdot 2\displaystyle \int_{0}^{2}[f(x)] ^{2}\,{\it dx}\\[5pt] &=&\dfrac{3}{28}\displaystyle \int_{0}^{2}( x^{2}+1) ^{2}\,{\it dx}=\dfrac{3}{28}\displaystyle \int_{0}^{2}( x^{4}+2x^{2}+1) ~{\it dx} \\[5pt] &=&\dfrac{3}{28}\left[ \dfrac{x^{5}}{5}+\dfrac{2x^{3}}{3}+x\right] _{0}^{2}= \dfrac{3}{28}\left( \dfrac{32}{5}+\dfrac{16}{3}+2\right) =\dfrac{103}{70} \approx 1.471 \end{eqnarray*}

The centroid of the lamina, as shown in Figure 73(b), is \(( \bar{x},\bar{y}) =\left( 0, \dfrac{103}{70}\right) .\)