Answer
We prove that the centroid of $\lambda {\cal D}$ is $\left( {\lambda \bar x,\lambda \bar y} \right)$.
Work Step by Step
Let $\left( {\bar x,\bar y} \right)$ be the centroid of ${\cal D}$. By definition (Section 16.5):
$\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$, ${\ \ \ \ \ }$ $\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A$,
where $A$ is the area of ${\cal D}$, that is $A = {\rm{Area}}\left( {\cal D} \right)$.
We define the dilation mapping $G$ using $p = \lambda x$ and $q = \lambda y$ such that $G\left( {x,y} \right) = \left( {\lambda x,\lambda y} \right)$.
Let ${{\cal D}_d}$ denote the dilation of ${\cal D}$, defined by
${{\cal D}_d} = \lambda {\cal D} = \left\{ {\left( {\lambda x,\lambda y} \right)|\left( {x,y} \right) \in {\cal D}} \right\}$
So, ${{\cal D}_d} = G\left( {\cal D} \right)$.
Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left( {\begin{array}{*{20}{c}}
{\frac{{\partial p}}{{\partial x}}}&{\frac{{\partial p}}{{\partial y}}}\\
{\frac{{\partial q}}{{\partial x}}}&{\frac{{\partial q}}{{\partial y}}}
\end{array}} \right) = \left| {\begin{array}{*{20}{c}}
\lambda &0\\
0&\lambda
\end{array}} \right| = {\lambda ^2}$
Let ${A_d}$ denote the area of ${{\cal D}_d}$. Using the Change of Variables Formula:
${A_d} = {\rm{Area}}\left( {{{\cal D}_d}} \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left| {{\rm{Jac}}\left( G \right)} \right|{\rm{d}}A$
${A_d} = {\lambda ^2}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A$
So, the area of ${{\cal D}_d}$ is ${A_d} = {\lambda ^2}A$.
Let $\left( {\overline {{x_d}} ,\overline {{y_d}} } \right)$ denote the centroid of ${{\cal D}_d}$.
By definition:
$\overline {{x_d}} = \frac{1}{{{A_d}}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_d}}^{} p{\rm{d}}A$, ${\ \ \ \ \ }$ $\overline {{y_d}} = \frac{1}{{{A_d}}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_d}}^{} q{\rm{d}}A$
Using the Change of Variables Formula, we evaluate these double integrals:
$\overline {{x_d}} = \frac{1}{{{A_d}}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_d}}^{} p{\rm{d}}A = \frac{1}{{{\lambda ^2}A}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \lambda x\left| {Jac\left( G \right)} \right|{\rm{d}}A$
$\overline {{x_d}} = \frac{\lambda }{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \lambda \bar x$
Similarly for $\overline {{y_d}} $:
$\overline {{y_d}} = \frac{1}{{{A_d}}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_d}}^{} q{\rm{d}}A = \frac{1}{{{\lambda ^2}A}}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \lambda y\left| {Jac\left( G \right)} \right|{\rm{d}}A$
$\overline {{y_d}} = \frac{\lambda }{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \lambda \bar y$
Hence, the centroid of $\lambda {\cal D}$ is $\left( {\lambda \bar x,\lambda \bar y} \right)$.
