$$Let\space I= \int\limits_0^a \int\limits_{\sqrt{ax-x^2}}^{\sqrt{a^2-x^2}}\dfrac {dxdy}{\sqrt{a^2-x^2-y^2}}$$

The limits of y are

$y=\sqrt{ax-x^2}\space to \space y=\sqrt{a^2-x^2}\\ \therefore 1st \space \lim \space it\\ y^2=ax-x^2\\ \therefore x^2+y^2-ax=0---- (1)$

Which is a circle

Here $2g = -a ,2f = 0,c = 0$

Center is $(-g, -f) = (a/2, 0)$

And radius is $\sqrt{g^2+f^2-c}\\ =\sqrt{\dfrac {a^2}{z^2}}=\dfrac a2$

And 2nd limit

$Y^2= ax – x^2$

$X^2 + y^2 = a^2 ------- (2)$

Which is also circle with center $(0,0)$ and radius a

And the limit of x are $x = 0$ to $x = a$

Which is a straight line

Now on changing to polar co ordinate

We take $x = r \cos θ ,y = r \sin θ$

$dx \space dy = r \space dr \space dθ\\ x^2 + y^2 = r^2 \cos^2 θ + r^2 \sin^2 θ = r^2 $

substituting these values in equation of circles we get

i.e in eq (1) and (2)

Equation (1) becomes

$r^2 – a r \cos θ = 0 \\ r (r – a \cos θ) = 0 \\ r = 0 \space or \space r = a \cos θ$

and equation (2) modifies to

$r^2 = a^2 \\ r = a$

1) The limits of r is

a)Lower limit r is inner circle equation

i.e $r = a \cos θ$

b) Upper limit r is outer circle

i.e $r = a$

2) Limits of θ

To cover bounded region radial strip must slide from 0 to $π/2$

$$\therefore \theta =0\space to \space \theta=\dfrac \pi2$$ $ \therefore I=\int\limits_0^{\frac\pi2}\int\limits_{a\cos \theta}^a\dfrac 1{\sqrt{a^2-r^2}}r\space dr\space d\theta\\ I=\int\limits_0^{\frac \pi2}\dfrac 1{-2}\int\limits^a_{a\cos \theta} \dfrac {-2r}{\sqrt{a^2-r^2}}dr\space d\theta\\ Let \space a^2-r^2=t\\ -2r\space dr=dt\\ when \space r=r\cos \theta, t=a^2(1-\cos^2\theta)\\ When \space r=a,t=0\\ I=\int\limits_0^{\frac \pi2}\dfrac 1{-2}\int\limits^0_{a^2(1-\cos^2\theta)}t^{\frac {-1}2}dt\space d\theta\\ I=\int\limits_0^{\frac \pi2}\dfrac 1{-2}\Bigg[\dfrac {t^{1/2}}{\dfrac 12}\Bigg]^0_{a^2(1-\cos^2\theta)}d\theta\\ I=-\int\limits_0^{\frac \pi2}(0-\sqrt{a^2(1-\cos^2\theta)})d\theta \\ I= \int\limits_0^{\frac \pi2}a\sqrt{1-\cos^2\theta}d\theta\\ I= \int\limits_0^{\frac \pi2}a\sqrt{\sin^2\theta}d\theta \\ I=\int\limits_0^{\frac \pi2}a\sin\theta d\theta\\ I=a[-\cos\theta]_0^{\frac \pi2}\\ I=-a\Big[\cos \dfrac\pi2-\cos 0\Big]\\ I=-a[0-1]\\ I=a$