Question

Find the area bounded by the curve $x^2+y^2\le 2ax$ $y^2\ge ax$ $,a>0,x>0,Y>0$.

Answer 1

Given,

$x^2+y^2\le 2ax$

$x^2+y^2+a^2-2ax\le 0+a^2$

${(x-a)}^2+y^2\le a^2........\left(1\right)$

Which is region inside the circle and centre as $(a,0)$

Now ,

$y^2\ge ax$

which is region inside the parabola ……..$\left(2\right)$

$x,y\ge 0$ Region lies in the 1^st quadrant ……….$\left(3\right)$

Let the required area is A ,

$A={\int }_0^a\sqrt{a^2-{(x-a)}^2-\sqrt{ax}}dx$

$A={\int }_0^a\sqrt{a^2-{(x-a)}^2}dx-{\int }_0^a{x}^{\frac{1}{2}}dx$

Since, $\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{2}$

=${{[\frac{(x-a)}{2}\int \sqrt{a^2-{(x-a)}^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x-a}{2}]}^a}_0-\sqrt{a}{{\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]}^a}_0$

$=[\frac{(a-a)\sqrt{a^2-{(a-a)}^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{a-a}{2}-\frac{(0-a)\sqrt{a^2-{(0-a)}^2}}{2}-\frac{a^2}{2}{\sin }^{-1}\left(\frac{0-a}{2}\right)]-\sqrt{a}\frac{2}{3}{a}^{\frac{3}{2}}$

$=0+0-0-\frac{a^2}{2}{\sin }^{-1}\left(1\right)-\frac{2}{3}{a}^2$

$=a^2[\frac{\pi }{4}-\frac{2}{3}]$ sq unit