QJEEM21Ma4fb3: JEE Main 2021 Mathematics - Class 12 JEEM21Ma4fb3 Solution: The area (in sq. units) of the part of the circle x 2 + y 2 =…

Question

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The area (in sq. units) of the part of the circle x² + y² = 36, which is outside the parabola y² = 9x, is :

Accepted Answer

{x^2} + {y^2} = 36

and

{y^2} = 9x

\therefore

{x^2} + 9x - 36 = 0

\Rightarrow x = 3, - 12

Required Area,

A = \pi {{{(6)}^2}} - 2\left[ {{A_1} + {A_2}} \right]

A = \pi {{{(6)}^2} - 2\left[ {\int_0^3 {\sqrt {9x} dx + \int_3^6 {\sqrt {36 - {x^2}} } dx} } \right]}

= 36\pi -

2\left[ {\left[ {3 \times {2 \over 3}{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + {{36} \over 2}{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]

= 36\pi -

2\left[ {\left[ {2{x^{{3 \over 2}}}} \right]_0^3 + \left[ {{x \over 2}\sqrt {36 - {x^2}} + 18{{\sin }^{ - 1}}{x \over 6}} \right]_3^6} \right]

= 36\pi -

2\left[ {\left[ {6\sqrt 3 - 0} \right] + \left[ {\left( {0 + 18{{\sin }^{ - 1}}{6 \over 6}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + 18{{\sin }^{ - 1}}{3 \over 6}} \right)} \right]} \right]

= 36\pi -

2\left[ {\left[ {6\sqrt 3 } \right] + \left[ {\left( {{{18\pi } \over 2}} \right) - \left( {{3 \over 2} \times 3\sqrt 3 + {{18\pi } \over 6}} \right)} \right]} \right]

= 36\pi -

\left( {12\sqrt 3 + 18\pi - 9\sqrt 3 - 6\pi } \right)

=

24\pi - 3\sqrt 3

Note :
(i)

\int {\sqrt {{x^2} + {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} + {a^2}} + {a^2}\log |x + \sqrt {{x^2} + {a^2}} |} \right]} + C

(ii)

\int {\sqrt {{a^2} - {x^2}} dx = {1 \over 2}\left[ {x\sqrt {{a^2} - {x^2}} + {a^2}{{\sin }^{ - 1}}\left( {{x \over a}} \right)} \right]} + C

(iii)

\int {\sqrt {{x^2} - {a^2}} dx = {1 \over 2}\left[ {x\sqrt {{x^2} - {a^2}} - {a^2}\log |x + \sqrt {{x^2} - {a^2}} |} \right]} + C

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