Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #20

Find the first-quadrant area inside x²+y²=4, above the x-axis and below y=√3x, using integration.

✅ Step-by-Step Solution

Final Answer: \(\frac{2 \pi}{3}\) square units

The ray intersects the circle at (1,√3). The upper boundary is the ray up to x=1 and then the circular arc up to x=2.
Use upper minus lower strip widths: \(\sqrt{3} x\) for x∈\(\left[0, 1\right]\); \(\sqrt{4 - x^{2}}\) for x∈\(\left[1, 2\right]\).
Hence A=\(\int_{0}^{1}(\sqrt{3} x)\,dx\) + \(\int_{1}^{2}(\sqrt{4 - x^{2}})\,dx\).
Primitive \(\frac{\sqrt{3} x^{2}}{2}\), endpoint difference \(\frac{\sqrt{3}}{2}\); Primitive \(\frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(- \frac{\sqrt{3}}{2} + \frac{2 \pi}{3}\).
Therefore the area is \(\frac{2 \pi}{3}\) square units.

[1 mark] The ray intersects the circle at (1,√3). The upper boundary is the ray up to x=1 and then the circular arc up to x=2.
[1 mark] Use upper minus lower strip widths: \(\sqrt{3} x\) for x∈\(\left[0, 1\right]\); \(\sqrt{4 - x^{2}}\) for x∈\(\left[1, 2\right]\).
[1 mark] Hence A=\(\int_{0}^{1}(\sqrt{3} x)\,dx\) + \(\int_{1}^{2}(\sqrt{4 - x^{2}})\,dx\).
[1 mark] Primitive \(\frac{\sqrt{3} x^{2}}{2}\), endpoint difference \(\frac{\sqrt{3}}{2}\); Primitive \(\frac{x \sqrt{4 - x^{2}}}{2} + 2 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(- \frac{\sqrt{3}}{2} + \frac{2 \pi}{3}\).
[1 mark] Therefore the area is \(\frac{2 \pi}{3}\) square units.

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