9. Sketch the region of integration and change the order of integration for the integral \[ \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} f(x, y) d x d y \]

The original integral is \[ \int_{-2}^{2} \int_{0}^{\sqrt{4-y^{2}}} f(x,y) \, dx \, dy. \] **Step 1. Understand the region of integration** For a fixed \( y \) (with \( y \) between \(-2\) and \(2\)), \( x \) varies from \( 0 \) to \( \sqrt{4-y^2} \). The equation \[ x = \sqrt{4-y^2} \] can be squared to yield \[ x^2 = 4-y^2 \quad \Longrightarrow \quad x^2+y^2 = 4. \] This is the equation of a circle with radius 2 centered at the origin. However, because \( x \) goes from \( 0 \) to \(\sqrt{4-y^2}\), we are only considering the portion of the circle where \( x \ge 0 \). Thus, the region of integration is the right half of the circle given by \[ x^2 + y^2 \le 4 \quad \text{with} \quad x \ge 0. \] **Step 2. Sketch the region** Imagine a circle centered at the origin with radius 2. The region of integration is the half of this circle to the right of the \( y \)-axis. **Step 3. Change the order of integration** To reverse the order of integration, we first determine the bounds for \( x \). Notice: - \( x \) ranges from \( 0 \) (since \( x\ge0 \)) to \( 2 \) (the rightmost point of the circle). For a fixed \( x \) between \( 0 \) and \( 2 \), \( y \) must satisfy \[ x^2 + y^2 \le 4 \quad \Longrightarrow \quad y^2 \le 4 - x^2, \] so \[ -\sqrt{4-x^2} \le y \le \sqrt{4-x^2}. \] Thus, the integral with the order switched becomes \[ \int_{x=0}^{2} \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} f(x,y) \, dy \, dx. \] **Final Answer** \[ \int_{0}^{2} \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} f(x,y) \, dy \, dx. \]