## Calculus (3rd Edition)

$0.36 g.$
The total mass of the deposit is given by $$\int_{0}^{6} s(x) d x=\int_{0}^{6} 0.01 x(6-x) d x=\left.\left(0.03 x^{2}-\frac{0.01}{3} x^{3}\right)\right|_{0} ^{6}=0.36 g.$$