Area of the smaller region bounded by the cir | vedclass

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(C) The given equations are the circle $x^2 + y^2 = 2^2$ (radius $r = 2$) and the line $y = x$.To find the intersection points,substitute $y = x$ into

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$\frac{\pi}{2}$

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(C) The given equations are the circle $x^2 + y^2 = 2^2$ (radius $r = 2$) and the line $y = x$.To find the intersection points,substitute $y = x$ into the circle equation: $x^2 + x^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2}$ (since we are in the first quadrant for $y=x$ above the $x$-axis).The area $A$ of the region bounded by the curve $y = f(x)$ and the line $y = x$ above the $x$-axis is given by the integral of the line from $0$ to $\sqrt{2}$ plus the integral of the circle from $\sqrt{2}$ to $2$.$A = \int_{0}^{\sqrt{2}} x \, dx + \int_{\sqrt{2}}^{2} \sqrt{4 - x^2} \, dx$.Evaluating the first part: $[\frac{x^2}{2}]_{0}^{\sqrt{2}} = \frac{2}{2} - 0 = 1$.Evaluating the second part: $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}(\frac{x}{a})$.For $\int_{\sqrt{2}}^{2} \sqrt{4 - x^2} \, dx = [\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}(\frac{x}{2})]_{\sqrt{2}}^{2}$.$= (0 + 2\sin^{-1}(1)) - (\frac{\sqrt{2}}{2}\sqrt{2} + 2\sin^{-1}(\frac{1}{\sqrt{2}})) = (2 	imes \frac{\pi}{2}) - (1 + 2 	imes \frac{\pi}{4}) = \pi - 1 - \frac{\pi}{2} = \frac{\pi}{2} - 1$.Total Area $A = 1 + (\frac{\pi}{2} - 1) = \frac{\pi}{2}$.

Area of the smaller region bounded by the circle $x^2 + y^2 = 4$ and the line $y = x$ above the $x$-axis is:

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