Answer
$$\dfrac{8}{3}$$
Work Step by Step
Let us consider $\iint_{R} f(s,t) dA= \int_{0}^{1} \int_{0}^{\sqrt{1-s^2}} 8t \space dt\space ds\\= \int_{0}^{1} [4t^2]_{0}^{\sqrt{1-s^2}} \space ds \\= \int_{0}^{1} 4(1-s^2) \space ds \\= [4s -\dfrac{4s^3}{3}]_0^1\\=4(1-0) -\dfrac{4}{3}(1-0) \\=\dfrac{8}{3}$
