Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_by_integration

Class 12 Maths · Applications Of Integrals · Very Long Answer #42

Using integration, find the area of {(x,y):9x²+y²≤36,3x+y≥6}.

✅ Step-by-Step Solution

Final Answer: \(-6 + 3 \pi\) square units

The ellipse has semiaxes 2 and 6. Substitution y=6−3x gives intersections (0,6),(2,0); the desired cap lies between the line and upper ellipse arc.
Use upper minus lower strip widths: \(3 x + \sqrt{36 - 9 x^{2}} - 6\) for x∈\(\left[0, 2\right]\).
Hence A=\(\int_{0}^{2}(3 x + \sqrt{36 - 9 x^{2}} - 6)\,dx\).
Primitive \(\frac{3 x^{2}}{2} + \frac{3 x \sqrt{4 - x^{2}}}{2} - 6 x + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(-6 + 3 \pi\).
Therefore the area is \(-6 + 3 \pi\) square units.

[1 mark] The ellipse has semiaxes 2 and 6. Substitution y=6−3x gives intersections (0,6),(2,0); the desired cap lies between the line and upper ellipse arc.
[1 mark] Use upper minus lower strip widths: \(3 x + \sqrt{36 - 9 x^{2}} - 6\) for x∈\(\left[0, 2\right]\).
[1 mark] Hence A=\(\int_{0}^{2}(3 x + \sqrt{36 - 9 x^{2}} - 6)\,dx\).
[1 mark] Primitive \(\frac{3 x^{2}}{2} + \frac{3 x \sqrt{4 - x^{2}}}{2} - 6 x + 6 \operatorname{asin}{\left(\frac{x}{2} \right)}\), endpoint difference \(-6 + 3 \pi\).
[1 mark] Therefore the area is \(-6 + 3 \pi\) square units.

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