Find $\iint\limits_{\kern-3ptR} xy\,dA$ if $R$ is the region enclosed by $y=x^{2}$ and $y=\sqrt{x}.$

Figure 13 The region $R$ is $x$-simple.

Solution We begin by graphing the region $R.$ See Figure 13. Observe that $R$ is a closed, bounded region. Also notice that the bottom and top boundaries of $R$ are smooth curves expressed as functions of $x$: $g_{1}(x) =x^{2}$, $g_{2}(x)=\sqrt{x}$, $0\leq x\leq 1$. That is, $R$ is $x$-simple, so we can use Fubini’s Theorem. $$\begin{eqnarray*} \displaystyle\iint\limits_{\kern-3ptR} xy\,{\it dA} &=&\int_{0}^{1}\left[ \int_{x^{2}}^{\sqrt{x}}xy\,{\it dy} \right] \! {\it dx}=\int_{0}^{1}x \left[ \dfrac{y^{2}}{2}\right] _{x^{2}}^{ \sqrt{x}}{\it dx}\\[4pt] &=&\int_{0}^{1}x\!\left(\dfrac{x-x^{4}}{2}\right)\!{\it dx}=\dfrac{1}{2} \int_{0}^{1}(x^{2}-x^{5})\,{\it dx}\\[4pt] &=&\dfrac{1}{2}\left[ \dfrac{x^{3}}{3}-\dfrac{x^{6}}{6}\right] _{0}^{1}=\dfrac{1 }{12} \end{eqnarray*}$$