We need to evaluate the double integral $\iint_S \sqrt{4-x^2-y^2} \, dA$, where $S$ is the first quadrant sector of the circle $x^2+y^2 = 4$ between $y=0$ and $y=x$.

1. Problem Description

We need to evaluate the double integral $\iint_S \sqrt{4-x^2-y^2} \, dA$, where $S$ is the first quadrant sector of the circle $x^2+y^2 = 4$ between $y=0$ and $y=x$.

2. Solution Steps

Since we are asked to use polar coordinates, we need to transform the integral into polar coordinates.

First, we have $x^2+y^2 = r^2$. Thus, $\sqrt{4-x^2-y^2} = \sqrt{4-r^2}$. Also, $dA = r \, dr \, d\theta$.

The region $S$ is the first quadrant sector of the circle $x^2+y^2=4$ between $y=0$ and $y=x$.

The equation $x^2+y^2=4$ in polar coordinates is $r^2=4$, so $r=2$.

Since $S$ is in the first quadrant, the angle $\theta$ varies between 0 and $\pi/2$. The region $S$ is bounded by $y=0$ which means $\theta=0$, and $y=x$ which means $\theta=\pi/4$. Thus, $\theta$ varies between $0$ and $\pi/4$ is incorrect.

The condition $y=0$ corresponds to $\theta = 0$ and the condition $y=x$ corresponds to $r \sin \theta = r \cos \theta$, which implies $\sin \theta = \cos \theta$, so $\theta = \pi/4$.

The radius $r$ varies from 0 to

2. The integral becomes:

$$\iint_S \sqrt{4-x^2-y^2} \, dA = \int_0^{\pi/4} \int_0^2 \sqrt{4-r^2} \, r \, dr \, d\theta$$

Let $u = 4-r^2$, so $du = -2r \, dr$, which means $r \, dr = -\frac{1}{2} \, du$.

When $r=0$, $u=4$. When $r=2$, $u=0$.

$$\int_0^2 \sqrt{4-r^2} \, r \, dr = \int_4^0 \sqrt{u} \left( -\frac{1}{2} \, du \right) = \frac{1}{2} \int_0^4 u^{1/2} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_0^4 = \frac{1}{3} \left[ u^{3/2} \right]_0^4 = \frac{1}{3} (4^{3/2} - 0^{3/2}) = \frac{1}{3} (8) = \frac{8}{3}$$

Therefore,

$$\int_0^{\pi/4} \int_0^2 \sqrt{4-r^2} \, r \, dr \, d\theta = \int_0^{\pi/4} \frac{8}{3} \, d\theta = \frac{8}{3} \left[ \theta \right]_0^{\pi/4} = \frac{8}{3} \left( \frac{\pi}{4} - 0 \right) = \frac{8}{3} \cdot \frac{\pi}{4} = \frac{2\pi}{3}$$

3. Final Answer

$\frac{2\pi}{3}$