Application of Integrals: Application of Integrals (Circle and Line)

Substituting $y = x$ into $x^2 + y^2 = 32$: $$x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4 \quad (\text{first quadrant, } x > 0)$$ At $x = 4$, $y = 4$. So the point of intersection is $(4, 4)$. The radius of the circle is $r = \sqrt{32} = 4\sqrt{2}$.

From $x = 0$ to $x = 4$, the region is under the line $y = x$. From $x = 4$ to $x = 4\sqrt{2}$, the region is under the circle $y = \sqrt{32 - x^2}$. $$\text{Total Area } A = \int_{0}^{4} x \, dx + \int_{4}^{4\sqrt{2}} \sqrt{(4\sqrt{2})^2 - x^2} \, dx$$

$$A_1 = \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} - 0 = 8\text{ sq. units}$$

Using standard formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$ with $a = 4\sqrt{2}$ and $a^2 = 32$: $$A_2 = \left[ \frac{x}{2}\sqrt{32 - x^2} + 16\sin^{-1}\left(\frac{x}{4\sqrt{2}}\right) \right]_{4}^{4\sqrt{2}}$$ Upper limit $x = 4\sqrt{2}$: $$0 + 16\sin^{-1}(1) = 16 \cdot \frac{\pi}{2} = 8\pi$$ Lower limit $x = 4$: $$\frac{4}{2}\sqrt{32 - 16} + 16\sin^{-1}\left(\frac{4}{4\sqrt{2}}\right) = 2(4) + 16\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 8 + 16\left(\frac{\pi}{4}\right) = 8 + 4\pi$$ $$A_2 = 8\pi - (8 + 4\pi) = 4\pi - 8\text{ sq. units}$$

$$A = A_1 + A_2 = 8 + (4\pi - 8) = 4\pi\text{ sq. units}$$