The area (in sq.

Question

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

Solutions · 1