Answer
$$(\bar{x}, \bar{y})= (9/8,18/5 ) $$
Work Step by Step
Since
\begin{aligned}
M_{y} &=\delta \int_{a}^{b} x f(x) d x \\
&=\int_{0}^{3} x\left(9-x^{2}\right) d x \\
&=\int_{0}^{3}\left(9 x-x^{3}\right) d x \\
&=\left.\left(\frac{9}{2} x^{2}-\frac{x^{4}}{4}\right)\right|_{0} ^{3} \\
&=\frac{81}{4}
\end{aligned}
and
\begin{aligned}
M_{x} &=\frac{1}{2} \delta \int_{a}^{b} f(x)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3}\left(9-x^{2}\right)^{2} d x \\
&=\frac{1}{2} \int_{0}^{3}\left(9-x^{2}\right)^{2} d x \\
&=\frac{1}{2}\left(81-18 x^{2}+x^{4}\right) d x \\
&=\left.\frac{1}{2}\left(81 x-6 x^{3}+\frac{x^{5}}{5}\right)\right|_{0} ^{3} \\
&=\frac{1}{2}\left(\frac{648}{5}\right) \\
&=\frac{324}{5}
\end{aligned}
and
$$M=\delta A=\int_{0}^{3}\left(9-x^{2}\right) d x=\left.\left(9 x-\frac{x^{3}}{3}\right)\right|_{0} ^{3}=18$$
Then
$$\bar{x}=\frac{M_{y}}{M}=\frac{81}{4 \times 18}=\frac{9}{8} \quad \text { and } \quad \bar{y}=\frac{M_{x}}{M}=\frac{324}{5 \times 18}=\frac{18}{5}$$
