## Thomas' Calculus 13th Edition

$\bar{x}=1$ and $\bar{y}=1$
Since \begin{align*} M&=\int_{0}^{3} \int_{0}^{3-x} d y d x\\ &=\int_{0}^{3}(3-x) d x\\ &=\frac{9}{2} \\ M_{y}&=\int_{0}^{3} \int_{0}^{3-x} x d y d x\\ &=\int_{0}^{3}[x y]_{0}^{3-x} d x\\ &=\int_{0}^{3}\left(3 x-x^{2}\right) d x\\ &=\frac{9}{2} \end{align*} Then by symmetry $\bar{x}=1$ and $\bar{y}=1$