What is the area bounded by the curves {x^2} | vedclass

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Question

(b) Solving the equations,

${x^2} + {y^2} = 9$ ....$(i)$

${y^2} = 8x$ ......$(ii)$

Put ${y^2} = 8x$ in $(i),$

${x^2} + 8x - 9 = 0$

==&g

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Solving the equations,

${x^2} + {y^2} = 9$ ....$(i)$

${y^2} = 8x$ ......$(ii)$

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.

$	herefore A\, \equiv \,(1,\,0)$ and $B \equiv (3,\,0)$

$	herefore $ Required area $= 2 	imes$ the hatched areas

$ = 2\left[ {\int_0^1 {y\,dx\,\,} {m{for}}\,{m{(ii)}}\, + \int_1^3 {y\,dx\,\,{m{for}}\,\,{m{(i)}}} } ight]\,$

$ = 2\left[ {\int_0^1 2 \sqrt 2 \,\,{{(x)}^{1/2}}\,\,dx + \int_1^3 {\sqrt {{3^2} - {x^2}} \,dx} } ight]$

$ = 2\left[ {2\sqrt 2 	imes \left( {\frac{{{x^{3/2}}}}{{3/2}}} ight)_0^1 + \left( {\frac{{x\sqrt {9 - {x^2}} }}{2} + \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{x}{3}} ight)} ight)_1^3} ight]$

$ = 2\,\left[ {\frac{{4\sqrt 2 }}{3} + \frac{9}{2} 	imes \frac{\pi }{2} - \frac{{\sqrt 8 }}{2} - \frac{9}{2}{{\sin }^{ - 1}}\left( {\frac{1}{3}} ight)} ight]$

$ = \frac{{2\sqrt 2 }}{3} + \frac{{9\pi }}{2} - 9{\sin ^{ - 1}}\left( {\frac{1}{3}} ight)$.

What is the area bounded by the curves ${x^2} + {y^2} = 9$ and ${y^2} = 8x$ is

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