Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #47

Find by integration the smaller region cut from the ellipse x²/9+y²/4=1 by the line x/3+y/2=1.

✅ Step-by-Step Solution

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

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

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

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