Answer
$\int^{4}_{0}$ $\int^{\sqrt y}_{0}$ $xy^{2} dx dy = 32$
Work Step by Step
$\int^{4}_{0}$ $\int^{\sqrt y}_{0}$ $xy^{2} dx dy = $$\int^{4}_{0}[\frac{1}{2}x^{2}y^{2}]^{x=\sqrt y}_{x=0}$$ dy = $ $\int^{4}_{0}$$ \frac{1}{2} y^{2} [(\sqrt y)^2 - 0^2] dy = $
$\frac{1}{2}$$\int^{4}_{0} y^{3} dy = \frac{1}{2}[\frac{1}{4}y^{4}]^{4}_{0} = \frac{1}{2}(64-0) = 32$