Find $\iint_{D}(2 x+y) d A$ where $D=\left\{(x, y) \mid x^{2}+y^{2} \leq 16, x \geq 0\right\}$
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 \(\iint_{D}f(r, \theta) r dr d\theta\). 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 \(\iint_{D}f(r, \theta) r dr d\theta\) 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 \(\boxed{64}\).