Class 12 Applications Of Integrals · Very Long Answer5 Marksarea_parameter

Class 12 Maths · Applications Of Integrals · Very Long Answer #15

In the first quadrant, the region inside x²+y²=4, above the x-axis and below the ray y=mx (m>0) has area π/2. Using integration, find m.

✅ Step-by-Step Solution

Final Answer: m=1

The ray meets the circle at (c,mc), where c=2/√(1+m²). Its slope angle θ=arctan m lies in (0,π/2).
Use vertical strips: A=∫₀ᶜmx dx+∫ᶜ²√(4−x²)dx.
A circle primitive is F=x√(4−x²)/2+2arcsin(x/2). Since √(4−c²)=mc, F(c)=mc²/2+2arcsin(c/2), so A=π−2arcsin(c/2).
As c/2=cosθ, arcsin(c/2)=π/2−θ. Therefore A=2θ=2arctan m. Setting A=π/2 gives θ=π/4.
Hence m=tan(π/4)=1.

[1 mark] The ray meets the circle at (c,mc), where c=2/√(1+m²). Its slope angle θ=arctan m lies in (0,π/2).
[1 mark] Use vertical strips: A=∫₀ᶜmx dx+∫ᶜ²√(4−x²)dx.
[1 mark] A circle primitive is F=x√(4−x²)/2+2arcsin(x/2). Since √(4−c²)=mc, F(c)=mc²/2+2arcsin(c/2), so A=π−2arcsin(c/2).
[1 mark] As c/2=cosθ, arcsin(c/2)=π/2−θ. Therefore A=2θ=2arctan m. Setting A=π/2 gives θ=π/4.
[1 mark] Hence m=tan(π/4)=1.

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