Question

Find the area lying above $x-$axis and included between the circle $x^2+y^2=8x$ and inside of the parabola $y^2=4x$.

Answer 1

$\sin \theta$ circle is $\theta$

from $m(x-a{)}^2+(y-b{)}^2=r^2$

$\Rightarrow x^2+y^2=8x$

$x^2-8x+y^2=0$

$x^2-2\times 4\times x+y^2=0$

$x^2-2\times 4\times x+4^2-4^2+y^2=0$

$(x-4{)}^2+y^2=4^2$

So, which has centre $(4,0)$

$r=4$

equation of circle is

$x^2+y^2=8x$

Put $y^2=4x$

$x^2+4x=8x$

$x^2=4x$

$x(x-4)=0$

$x=0$ or $x=4$

For $x=0$

$y^2=4x$

$y=0$

So, $P(0,0)$

For $y=4$

$y^2=4\left(4\right)$

$y^2=16$

$y=4$

So, $P(4,4)$

Since the point $P$ is in $1^{st}$ quad

$P(4,4)$

Now

Area req = Area $opc+$ Area of $peq$

$A\times \left(opc\right)={\int }_0^{4}ydx$

$y^2=4x$

$y=2\sqrt{2}$

${\int }_0^{4}2f\times dx=2{\left[\frac{x^{1/2}}{2/2}\right]}_0^4$

$=\frac{22}{8}$

Area of $pco={\int }_4^{8}4dx$

$x^2+y^2=8x$

$y=pm\sqrt{x-x^2}$

$1st$ quad $\Rightarrow y=\sqrt{2x-x^2}$

${\int }_4^8\sqrt{8x-x^2}dx$

$={\int }_4^8\sqrt{4^2-(5-4{)}^2}dx$

$={[\frac{(x-4)}{2}\sqrt{(x^2-8x)}+8{\sin }^{-1}\frac{(x-4)}{4}]}^2$

$2[2\times 0+8\times \frac{\pi }{2}-\frac{16\sqrt{4}}{3}]$

$=8\pi -\frac{32\sqrt{4}}{3}$