CBSE 2025, Set 1

Question

A woman discovered a scratch along a straight line on a circular table top of radius 8 cm. She divided the table top into 4 equal quadrants and discovered the scratch passing through the origin inclined at an angle $\dfrac{\pi}{4}$ anticlockwise along the positive direction of the x-axis. Find the area of the region enclosed by the x-axis, the scratch and the circular table top in the first quadrant, using integration.

Edvaya · Accepted Answer

The correct answer is Subjective. **Setup:** - Circle: $x^2 + y^2 = 64$ (radius 8) - Scratch (line): $y = 	an(\pi/4) \cdot x = x$ **Intersection of $y = x$ with the circle:** $x^2 + x^2 = 64 \Rightarrow x = 4\sqrt{2}$ → point $(4\sqrt{2},\ 4\sqrt{2})$ **Required area** = Area under $y = x$ from $0$ to $4\sqrt{2}$ + Area under circle from $4\sqrt{2}$ to $8$: $$A = \int_0^{4\sqrt{2}} x\,dx + \int_{4\sqrt{2}}^{8} \sqrt{64-x^2}\,dx$$ **First integral:** $$\int_0^{4\sqrt{2}} x\,dx = \left[\frac{x^2}{2}ight]_0^{4\sqrt{2}} = \frac{32}{2} = 16$$ **Second integral** (using $\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$ with $a=8$): $$\left[\frac{x}{2}\sqrt{64-x^2} + 32\sin^{-1}\frac{x}{8}ight]_{4\sqrt{2}}^{8}$$ At $x = 8$: $0 + 32\sin^{-1}(1) = 32 \cdot \dfrac{\pi}{2} = 16\pi$ At $x = 4\sqrt{2}$: $\dfrac{4\sqrt{2}}{2} \cdot 4\sqrt{2} + 32\sin^{-1}\!\dfrac{1}{\sqrt{2}} = 16 + 32 \cdot \dfrac{\pi}{4} = 16 + 8\pi$ Second integral $= 16\pi - (16 + 8\pi) = 8\pi - 16$ **Total area:** $$A = 16 + (8\pi - 16) = \boxed{8\pi 	ext{ cm}^2}$$

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Mathematics > Calculus > Application Of Integrals

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