Problem Statement
Find the area of the region in the first quadrant enclosed by the $$x$$-axis, the line $$y = x$$, and the circle $$x^2 + y^2 = 32$$.
Verified Solution & Marking Scheme
Find point of intersection
Substitute $$y = x$$ into $$x^2 + y^2 = 32$$:\n$$x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4$$ (since in first quadrant, $$x > 0$$). Corresponding $$y = 4$$. Intersection point is $$(4, 4)$$. Circle cuts $$x$$-axis at $$x = \sqrt{32} = 4\sqrt{2}$$.
Split area into two regions
$$\text{Area} = \int_{0}^{4} y_{\text{line}} \, dx + \int_{4}^{4\sqrt{2}} y_{\text{circle}} \, dx = \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{32 - x^2} \, dx$$
Evaluate first integral
$$I_1 = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} - 0 = 8$$
Evaluate second integral
Using $$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$$ with $$a = 4\sqrt{2}$$ ($$a^2 = 32$$):\n$$I_2 = \left[ \frac{x}{2}\sqrt{32 - x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}$$\n$$= \left(0 + 16\sin^{-1}(1)\right) - \left(\frac{4}{2}\sqrt{16} + 16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) = 16\left(\frac{\pi}{2}\right) - \left(8 + 16\left(\frac{\pi}{4}\right)\right) = 8\pi - 8 - 4\pi = 4\pi - 8$$
Add areas
$$\text{Total Area} = I_1 + I_2 = 8 + (4\pi - 8) = 4\pi\text{ sq. units}$$