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Evaluate the given integral by changing to polar coordinates.

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\iint_{R} \sqrt{16-x^{2}-y^{2}} \text { dA where } R=\left\{(x, y) \mid x^{2}+y^{2} \leq 16, x \geq 0\right\} \text {, }

Convert to Polar Coordinates: The region $R$ is a semicircle with radius $4$ in the first quadrant. To change to polar coordinates, use $x = r\cos(\theta)$ and $y = r\sin(\theta)$ .
Set Up Integral: The integral in polar coordinates becomes $\iint_{R} \sqrt{16 - r^2} r \, dr \, d\theta$ , with $r$ from $0$ to $4$ and $\theta$ from $0$ to $\frac{\pi}{2}$ .
Integrate with Respect to $r$ : Set up the integral: $\int_{0}^{\pi/2} \int_{0}^{4} \sqrt{16 - r^2} r \, dr \, d\theta$ .
Evaluate Inner Integral: Integrate with respect to $r$ first: $\int_{0}^{\pi/2} \left[\left(-\frac{1}{3}\right)\left(16 - r^2\right)^{\frac{3}{2}}\right]$ from $r=0$ to $r=4$ $d\theta$ .
Simplify Expression: Evaluate the inner integral: \int_{$0$}^{\pi/$2$} \left[\left(-\frac{$1$}{$3$}\right)\left(\(16 - $16$ \right)^{\frac{ $3$ }{ $2$ }} - \left(-\frac{ $1$ }{ $3$ }\right)\left( $16$ - $0$ \right)^{\frac{ $3$ }{ $2$ }}\right] d\theta.
Integrate with Respect to $\theta$ : Simplify the expression: $\int_{0}^{\pi/2} [0 - (-\frac{1}{3})(16)^{\frac{3}{2}}] d\theta$ .
Evaluate Outer Integral: The integral simplifies to $(\frac{16}{3})^{\frac{3}{2}} \int_{0}^{\frac{\pi}{2}} d\theta$ .
Calculate Final Value: Integrate with respect to $\theta$ : $\left(\frac{16}{3}\right)^{\frac{3}{2}} [\theta]$ from $0$ to $\frac{\pi}{2}$ .
Calculate Final Value: Integrate with respect to $\theta$ : $\left(\frac{16}{3}\right)^{\frac{3}{2}} [\theta]$ from $0$ to $\frac{\pi}{2}$ .Evaluate the outer integral: $\left(\frac{16}{3}\right)^{\frac{3}{2}} \left[\frac{\pi}{2} - 0\right]$ .
Calculate Final Value: Integrate with respect to $\theta$ : $(\frac{16}{3})^{\frac{3}{2}} [\theta]$ from $0$ to $\frac{\pi}{2}$ .Evaluate the outer integral: $(\frac{16}{3})^{\frac{3}{2}} [\frac{\pi}{2} - 0]$ .Calculate the final value: $(\frac{16}{3})^{\frac{3}{2}} \times \frac{\pi}{2}$ .