Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #27

Find by integration the first-quadrant region inside x²+y²=32, above the x-axis and below y=x.

✅ Step-by-Step Solution

Final Answer: \(4 \pi\) square units

The line y=x meets the circle at (4,4) in the first quadrant. The x-axis endpoint of the circle is (4√2,0); split the upper boundary at x=4.
Use upper minus lower strip widths: \(x\) for x∈\(\left[0, 4\right]\); \(\sqrt{32 - x^{2}}\) for x∈\(\left[4, 4 \sqrt{2}\right]\).
Hence A=\(\int_{0}^{4}(x)\,dx\) + \(\int_{4}^{4 \sqrt{2}}(\sqrt{32 - x^{2}})\,dx\).
Primitive \(\frac{x^{2}}{2}\), endpoint difference \(8\); Primitive \(\frac{x \sqrt{32 - x^{2}}}{2} + 16 \operatorname{asin}{\left(\frac{\sqrt{2} x}{8} \right)}\), endpoint difference \(-8 + 4 \pi\).
Therefore the area is \(4 \pi\) square units.

[1 mark] The line y=x meets the circle at (4,4) in the first quadrant. The x-axis endpoint of the circle is (4√2,0); split the upper boundary at x=4.
[1 mark] Use upper minus lower strip widths: \(x\) for x∈\(\left[0, 4\right]\); \(\sqrt{32 - x^{2}}\) for x∈\(\left[4, 4 \sqrt{2}\right]\).
[1 mark] Hence A=\(\int_{0}^{4}(x)\,dx\) + \(\int_{4}^{4 \sqrt{2}}(\sqrt{32 - x^{2}})\,dx\).
[1 mark] Primitive \(\frac{x^{2}}{2}\), endpoint difference \(8\); Primitive \(\frac{x \sqrt{32 - x^{2}}}{2} + 16 \operatorname{asin}{\left(\frac{\sqrt{2} x}{8} \right)}\), endpoint difference \(-8 + 4 \pi\).
[1 mark] Therefore the area is \(4 \pi\) square units.

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