## Calculus (3rd Edition)

$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \frac{{14}}{3}$
The domain ${\cal D}$ in Figure 29 is not a simple region. However, we can consider two horizontally simple regions ${{\cal D}_1}$ and ${{\cal D}_2}$ such that ${\cal D} = {{\cal D}_2} - {{\cal D}_1}$ (please see the figure attached). Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$ for ${\cal D}$ is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x{\rm{d}}A - \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x{\rm{d}}A$ 1. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x{\rm{d}}A$ for ${{\cal D}_1}$ The lower and upper boundaries are $y=-1$ and $y=1$, respectively. Whereas, the left boundary is $x=0$. To find the right boundary, we use the equation of the disk of radius $1$: ${x^2} + {y^2} = 1$ $x = \pm \sqrt {1 - {y^2}}$ Since ${{\cal D}_1}$ is located on the right-hand side of the $y$-axis, we choose $x = \sqrt {1 - {y^2}}$. Thus, the domain description of ${{\cal D}_1}$: ${{\cal D}_1} = \left\{ {\left( {x,y} \right)| - 1 \le y \le 1,0 \le x \le \sqrt {1 - {y^2}} } \right\}$ Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x{\rm{d}}A$ for ${{\cal D}_1}$ as an iterated integral: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x{\rm{d}}A = \mathop \smallint \limits_{y = - 1}^1 \left( {\mathop \smallint \limits_{x = 0}^{\sqrt {1 - {y^2}} } x{\rm{d}}x} \right){\rm{d}}y$ $= \frac{1}{2}\mathop \smallint \limits_{y = - 1}^1 \left( {{x^2}|_0^{\sqrt {1 - {y^2}} }} \right){\rm{d}}y$ $= \frac{1}{2}\mathop \smallint \limits_{y = - 1}^1 \left( {1 - {y^2}} \right){\rm{d}}y$ $= \frac{1}{2}\left( {\left( {y - \frac{1}{3}{y^3}} \right)|_{ - 1}^1} \right)$ $= \frac{1}{2}\left( {1 - \frac{1}{3} + 1 - \frac{1}{3}} \right)$ $= \frac{2}{3}$ 2. Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x{\rm{d}}A$ for ${{\cal D}_2}$ The lower and upper boundaries are $y=-2$ and $y=2$, respectively. Whereas, the left boundary is $x=0$. To find the right boundary, we use the equation of the disk of radius $2$: ${x^2} + {y^2} = 4$ $x = \pm \sqrt {4 - {y^2}}$ Since ${{\cal D}_2}$ is located on the right-hand side of the $y$-axis, we choose $x = \sqrt {4 - {y^2}}$. Thus, the domain description of ${{\cal D}_2}$: ${{\cal D}_2} = \left\{ {\left( {x,y} \right)| - 2 \le y \le 2,0 \le x \le \sqrt {4 - {y^2}} } \right\}$ Evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x{\rm{d}}A$ for ${{\cal D}_2}$ as an iterated integral: $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x{\rm{d}}A = \mathop \smallint \limits_{y = - 2}^2 \left( {\mathop \smallint \limits_{x = 0}^{\sqrt {4 - {y^2}} } x{\rm{d}}x} \right){\rm{d}}y$ $= \frac{1}{2}\mathop \smallint \limits_{y = - 2}^2 \left( {{x^2}|_0^{\sqrt {4 - {y^2}} }} \right){\rm{d}}y$ $= \frac{1}{2}\mathop \smallint \limits_{y = - 2}^2 \left( {4 - {y^2}} \right){\rm{d}}y$ $= \frac{1}{2}\left( {\left( {4y - \frac{1}{3}{y^3}} \right)|_{ - 2}^2} \right)$ $= \frac{1}{2}\left( {8 - \frac{8}{3} + 8 - \frac{8}{3}} \right)$ $= \frac{{16}}{3}$ Thus, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_2}}^{} x{\rm{d}}A - \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_1}}^{} x{\rm{d}}A$ $= \frac{{16}}{3} - \frac{2}{3}$ $= \frac{{14}}{3}$