8. Application and integration — ગણિત ધોરણ 12 સાયન્સ — Question

${x^2} + {y^2} = 9$ ....$(i)$

${y^2} = 8x$ ......$(ii)$

Put ${y^2} = 8x$ in $(i),$

${x^2} + 8x - 9 = 0$

==> $(x + 9)(x - 1) = 0$

$i.e.,$  $x = - \,9$ or $1$

$x = - 9$ gives imaginary value of $y$ for equation $(ii)$ hence neglected.

$\therefore A\, \equiv \,(1,\,0)$ and $B \equiv (3,\,0)$

$\therefore $ Required area $= 2 \times$ the hatched areas

$ = 2\left[ {\int_0^1 {y\,dx\,\,} {\rm{for}}\,{\rm{(ii)}}\, + \int_1^3 {y\,dx\,\,{\rm{for}}\,\,{\rm{(i)}}} } \right]\,$

$ = 2\left[ {\int_0^1 2 \sqrt 2 \,\,{{(x)}^{1/2}}\,\,dx + \int_1^3 {\sqrt {{3^2} - {x^2}} \,dx} } \right]$

$ = 2\left[ {2\sqrt 2 \times \left( {\frac{{{x^{3/2}}}}{{3/2}}} \right)_0^1 + \left( {\frac{{x\sqrt {9 - {x^2}} }}{2} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right)_1^3} \right]$

$ = 2\,\left[ {\frac{{4\sqrt 2 }}{3} + \frac{9}{2} \times \frac{\pi }{2} - \frac{{\sqrt 8 }}{2} - \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{1}{3}} \right)} \right]$

$ = \frac{{2\sqrt 2 }}{3} + \frac{{9\pi }}{2} - 9{\sin ^{ - 1}}\left( {\frac{1}{3}} \right)$.