91Ó°ÊÓ

To find the intersection points, we must first simplify the max function given by \(\max(x,y)= \begin{cases}x, & \text { if } x \geq y \\\ y, & \text { if } x<y .\end{cases}\) For x ≥ y, we have max(x, y) = x. The equation \(x^{2}+y^{2}=1\) can be expressed as y = \(\sqrt{1-x^2}\). And for x < y, we have max(x, y) = y. The equation \(x^{2}+y^{2}=1\) can be expressed as x = \(\sqrt{1-y^2}\). Now we have the two cases. For x ≥ y : \(x = \sqrt{1-x^2}\); Squaring both sides \(x^2=1-x^2\) \(2x^2=1\) \(x^2=\frac{1}{2}\) Therefore, \(x=\frac{1}{\sqrt{2}}\) and similarly \(y=\frac{1}{\sqrt{2}}\). For x < y : \(y = \sqrt{1-y^2}\); Squaring both sides \(y^2=1-y^2\) \(2y^2=1\) \(y^2=\frac{1}{2}\) So, as \(x < y\), in this case, \(x = 0\). Thus, the intersection points are \((\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\) and \((0,1)\).

Now we need to set up integral(s) to find the area enclosed by the given lines. We can split the region into two parts - one for x ≥ y and another for x < y. For x ≥ y: Lower limit: 0. Upper limit: \(\frac{1}{\sqrt{2}}\). $$\int_0^{\frac{1}{\sqrt{2}}} (\sqrt{1-x^2}-x) dx$$ For x < y: Lower limit: 0. Upper limit: \(\frac{1}{\sqrt{2}}\). $$\int_0^{\frac{1}{\sqrt{2}}} (y - \sqrt{1-y^2}) dy$$ We have the area enclosed by the two lines as: $$\int_0^{\frac{1}{\sqrt{2}}} (\sqrt{1-x^2}-x) dx + \int_0^{\frac{1}{\sqrt{2}}} (y- \sqrt{1-y^2}) dy$$

To find the solution, we evaluate the integrals: $$\int_0^{\frac{1}{\sqrt{2}}} (\sqrt{1-x^2}-x) dx = \left[\frac{1}{2} \arcsin{x}-\frac{1}{2}x\sqrt{1-x^2}\right]_0^{\frac{1}{\sqrt{2}}} = \frac{\pi}{8} -\frac{1}{4}$$ Similarly, $$\int_0^{\frac{1}{\sqrt{2}}} (y-\sqrt{1-y^2}) dy = \left[\frac{1}{2}y^2 +\frac{1}{2}y\sqrt{1-y^2} -\frac{1}{2}\arcsin{y}\right]_0^{\frac{1}{\sqrt{2}}} = \frac{1}{4}-\frac{\pi}{8}$$ Now, sum up the two integrals to get the area enclosed by the given lines. $$A=\left(\frac{\pi}{8} -\frac{1}{4}\right) +\left(\frac{1}{4}-\frac{\pi}{8}\right) = \frac{1}{2}$$ The area of the plane figure is \(\frac{1}{2}\) square units.