Question
Using integration, find the area of the region $\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\leq9,\text{x}+\text{y}\geq3\}.$
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Answer
Given, $\{(\text{x},\text{y}):\text{x}^2+\text{y}^2\leq9,\text{x}+\text{y}\geq3\}$ $\text{x}^2+\text{y}^2\leq9$ $\text{x}^2+\text{y}^2=9$ $\text{y}=\sqrt{9-\text{x}^2}\ \dots(1)$ $\text{x}+\text{y}\geq3$ $\text{x}+\text{y}=3$ $\text{y}=3-\text{x}\ \dots(2)$
By Using Eq. (1) & (2) Required area $=\int\limits^3_0\sqrt{9-\text{x}^2}\text{dx}-\int\limits^3_0(3-\text{x})\text{dx}$ $=\frac{\text{x}}{2}\sqrt{9-\text{x}^2}+\frac{9}{2}\sin^{-1}\Big(\frac{\text{x}}{3}\Big)\Big]^3_0+\frac{1}{2}(3-\text{x})^2\Big]^3_0$ $=\frac{9\pi}{4}-\frac{9}{2}$ or $\frac{9}{4}(\pi-2)$
By Using Eq. (1) & (2) Required area $=\int\limits^3_0\sqrt{9-\text{x}^2}\text{dx}-\int\limits^3_0(3-\text{x})\text{dx}$ $=\frac{\text{x}}{2}\sqrt{9-\text{x}^2}+\frac{9}{2}\sin^{-1}\Big(\frac{\text{x}}{3}\Big)\Big]^3_0+\frac{1}{2}(3-\text{x})^2\Big]^3_0$ $=\frac{9\pi}{4}-\frac{9}{2}$ or $\frac{9}{4}(\pi-2)$
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