The probability density for finding a molecule at distance $r$ from the origin at time $t$ in the spherical coordinates is denoted by $p(r,t)$

$p(r,t)=\frac{1}{8(Dt{)}^{\frac{3}{2}}}{e}^{-\frac{r^2}{4}Dt}$

show that:

$($a$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}pdV=1$

$($b$){\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}{r}^{2}pdV=6Dt$