Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.3 Center of Mass - Exercises - Page 483: 20

$(\frac{6}{15}, \frac{1}{2})$

$M_{x}$ = $\frac{1}{2}\int_0^1{(x-x^{2})}dx$ = $\frac{1}{12}$ $M_{y}$ = $\int_0^1{x(\sqrt x-x)}dx$ = $\frac{1}{15}$ $A$ = $\int_0^1{(\sqrt x-x)}dx$ = $\frac{1}{6}$ so the coordinates of the centroid are $(\frac{6}{15}, \frac{1}{2})$