Answer
a)
$M_{x}$ = $\frac{729}{10}$
$M_{y}$ = $\frac{243}{4}$
b) $(\frac{9}{4}, \frac{27}{10})$
Work Step by Step
a)
$M_{x}$ = $3\int_0^9{y(3-y)dy}$ = $(\frac{9y^{2}}{2}-\frac{6}{5}y^{\frac{5}{2}})|_0^9$ = $\frac{729}{10}$
$M_{y}$ = $3\int_0^3{x(x^{2})}dx$ = $\frac{3}{4}x^{4}|_0^3$ = $\frac{243}{4}$
b)
$A$ = $\int_0^3{x^{2}}dx$ = $\frac{x^{3}}{3}|_0^3$ = $9$ $cm^{2}$
With a constant density = 3 $g/cm^{2}$, the mass of the lamina is M = 27 grams, and the coordinates of the center of mass are
$(\frac{M_{y}}{M},\frac{M_{x}}{M})$ = $(\frac{\frac{243}{4}}{27}, \frac{\frac{729}{10}}{27})$ = $(\frac{9}{4}, \frac{27}{10})$
