Answer
(a) the statement is true.
(b) the statement is false.
(c) the statement is false.
(d) the statement is false.
Work Step by Step
We have ${\cal D}$ in ${\mathbb{R}^2}$ with a uniform mass density $\delta$ and a region symmetric with respect to the $y$-axis.
(a) ${x_{CM}} = 0$
We have
${x_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A = \frac{\delta }{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A$
Since ${\cal D}$ is symmetric with respect to the $y$-axis and the integrand is an odd function, ${x_{CM}} = 0$. So, the statement is true.
(b) ${y_{CM}} = 0$
We have
${y_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \frac{\delta }{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A$
Since ${\cal D}$ is not symmetric with respect to the $x$-axis, the integral is not zero. So, the statement is false.
(c) ${I_x} = 0$
We have
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}\delta \left( {x,y} \right){\rm{d}}A = \delta \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}{\rm{d}}A$
The integrand is an even function, so ${I_x}$ is not zero. So, the statement is false.
(d) ${I_y} = 0$
We have
${I_y} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}\delta \left( {x,y} \right){\rm{d}}A = \delta \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {x^2}{\rm{d}}A$
The integrand is an even function, so ${I_y}$ is not zero. So, the statement is false.
