Answer
$\dfrac{8}{3}$
Work Step by Step
Consider $\iint_{R} f(s,t) dA= \int_{0}^{1} \int_{0}^{\sqrt{1-s^2}} 8t dt ds= \int_{0}^{1} [4t^2]_{0}^{\sqrt{1-s^2}} ds$
or, $= \int_{0}^{1} 4(1-s^2) ds$
or, $= [4s -\dfrac{4s^3}{3}]_0^1$
Thus, we have $I=4(1-0) -\dfrac{4}{3}=\dfrac{8}{3}$
