Question

Using integration, find the area of the following region $\left\{\right(x,y):x^2+y^2\le 2ax,y^2\le ax,y>0\}$

Answer 1

We have,

Given that,

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

We can written that,

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

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

Now,

$y^2\ge ax$

Which is region outside the parabola $y^2=ax$ …… (2)

$x,y\ge 0$ i.e. region lies in the Ist quadrant …… (3)

Therefore the required region,

$A={\int }_0^a[\sqrt{2ax-x^2}-\sqrt{ax}].dx$

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

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

On integrating and we get,

Applying formula,

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

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

Then,

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

$A=0+0-0+\frac{a^2}{2}{\sin }^{-1}1-\frac{2}{3}{a}^2$

$=\frac{a^2}{2}.\frac{\pi }{2}-\frac{2}{3}{a}^2$

$=a^2(\frac{\pi }{4}-\frac{2}{3})$

Hence, this is the answer.