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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2021 · 24 February · Morning Shift · Mathematics · Area Under Curves

The area (in sq. units) of the part of the circle x ^ 2 + y ^ 2 =...

The area (in sq. units) of the part of the circle x2+y2=36{x}^{2} + {y}^{2} = 36, which is outside the parabola y2=9x{y}^{2} = 9x, is equal toThe area (in sq. units) of the part of the circle x ^ 2 + y ^ 2 = 36 , which is outside the parabola y ^ 2 = 9x , is

Options

  1. A12π+3312\pi + 3\sqrt{3}
  2. B24π+3324\pi + 3\sqrt{3}
  3. C24π−3324\pi - 3\sqrt{3}
  4. D12π−3312\pi - 3\sqrt{3}

Correct answer

C. 24π−3324\pi - 3\sqrt{3}

Step-by-step solution

The area (in sq. units) of the part of the circle x ^ 2 + y ^ 2 = 36 , which is outside the parabola y ^ 2 = 9x , is

Required area

=π×(6)2−2(∫039xdx+∫3636−x2dx)= \pi \times {\left(6\right)}^{2} - 2\left({∫}_{0}^{3}\sqrt{9x}dx + {∫}_{3}^{6}\sqrt{36 - {x}^{2}}dx\right)

=36π−6[x3/23/2]03−2[x236−x2+18sin⁡−1x6]36= 36\pi - 6{\left[\dfrac{{x}^{3 / 2}}{3 / 2}\right]}_{0}^{3} - 2{\left[\dfrac{x}{2}\sqrt{36 - {x}^{2}} + 18\sin^{- 1}\dfrac{x}{6}\right]}_{3}^{6}

=36π−123−2(0−932+9π−3π)= 36\pi - 12\sqrt{3} - 2\left(0 - \dfrac{9\sqrt{3}}{2} + 9\pi - 3\pi\right)

=24π−33sq.units= 24\pi - 3\sqrt{3}sq . units

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