Find ${\iint }_D(2x+y)dA$ where $D=\{(x,y)\mid x^2+y^2\le 16,x\ge 0\}$
Round your answer to four decimal places.

Step 1 ：The given region D is a semi-circle with radius 4 in the first quadrant. To solve this double integral, we can convert the Cartesian coordinates to polar coordinates. The conversion is given by $x=r\cos (\theta )$ and $y=r\sin (\theta )$. The double integral in polar coordinates is given by ${\iint }_{D}f(r,\theta )rdrd\theta$. The limits for r are from 0 to 4 and for $\theta$ are from 0 to $\pi /2$.

Step 2 ：Substitute $x=r\cos (\theta )$ and $y=r\sin (\theta )$ into the function $f=2x+y$, we get $f=r\sin (\theta )+2r\cos (\theta )$.

Step 3 ：Calculate the double integral ${\iint }_{D}f(r,\theta )rdrd\theta$ with the limits for r from 0 to 4 and for $\theta$ from 0 to $\pi /2$, we get the integral value is 64.

Step 4 ：Final Answer: The value of the double integral is $\boxed{{\displaystyle 64}}$.