Answer
$$2\sqrt 2 \sin \left( 2 \right) + \sqrt 2 \cos \left( 2 \right) - \sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_{\left( {1/2} \right){x^2}}^2 {\sqrt y \cos y} dydx} \cr
& y = \frac{1}{2}{x^2} \to x = \sqrt {2y} \cr
& {\text{Switching the order of integration, the new region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant x \leqslant \sqrt {2y} ,{\text{ }}0 \leqslant y \leqslant 2} \right\} \cr
& {\text{Therefore,}} \cr
& \int_0^2 {\int_{\left( {1/2} \right){x^2}}^2 {\sqrt y \cos y} dydx} = \int_0^2 {\int_0^{\sqrt {2y} } {\sqrt y \cos y} dxdy} \cr
& = \int_0^2 {\left[ {\int_0^{\sqrt {2y} } {\sqrt y \cos y} dx} \right]dy} \cr
& {\text{Integrate with respect to }}x \cr
& = \int_0^2 {\left[ {x\sqrt y \cos y} \right]_0^{\sqrt {2y} }dy} \cr
& = \int_0^2 {\left[ {\sqrt {2y} \sqrt y \cos y - 0\sqrt y \cos y} \right]dy} \cr
& = \sqrt 2 \int_0^2 {y\cos ydy} \cr
& {\text{Integrating by parts, we obtain}} \cr
& = \sqrt 2 \int_0^2 {y\cos ydy} \cr
& = \sqrt 2 \left[ {y\sin y + \cos y} \right]_0^2 \cr
& = \sqrt 2 \left[ {2\sin \left( 2 \right) + \cos \left( 2 \right)} \right] - \sqrt 2 \left[ {0\sin \left( 0 \right) + \cos \left( 0 \right)} \right] \cr
& = 2\sqrt 2 \sin \left( 2 \right) + \sqrt 2 \cos \left( 2 \right) - \sqrt 2 \cr} $$
