Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #21

Using integration, find the area of {(x,y):x²+y²≤9,x+y≥3}.

✅ Step-by-Step Solution

Final Answer: \(- \frac{9}{2} + \frac{9 \pi}{4}\) square units

The line meets the circle at (0,3),(3,0). The desired circular cap lies above y=3−x and below the upper circular arc for 0≤x≤3.
Use upper minus lower strip widths: \(x + \sqrt{9 - x^{2}} - 3\) for x∈\(\left[0, 3\right]\).
Hence A=\(\int_{0}^{3}(x + \sqrt{9 - x^{2}} - 3)\,dx\).
Primitive \(\frac{x^{2}}{2} + \frac{x \sqrt{9 - x^{2}}}{2} - 3 x + \frac{9 \operatorname{asin}{\left(\frac{x}{3} \right)}}{2}\), endpoint difference \(- \frac{9}{2} + \frac{9 \pi}{4}\).
Therefore the area is \(- \frac{9}{2} + \frac{9 \pi}{4}\) square units.

[1 mark] The line meets the circle at (0,3),(3,0). The desired circular cap lies above y=3−x and below the upper circular arc for 0≤x≤3.
[1 mark] Use upper minus lower strip widths: \(x + \sqrt{9 - x^{2}} - 3\) for x∈\(\left[0, 3\right]\).
[1 mark] Hence A=\(\int_{0}^{3}(x + \sqrt{9 - x^{2}} - 3)\,dx\).
[1 mark] Primitive \(\frac{x^{2}}{2} + \frac{x \sqrt{9 - x^{2}}}{2} - 3 x + \frac{9 \operatorname{asin}{\left(\frac{x}{3} \right)}}{2}\), endpoint difference \(- \frac{9}{2} + \frac{9 \pi}{4}\).
[1 mark] Therefore the area is \(- \frac{9}{2} + \frac{9 \pi}{4}\) square units.

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