The limit of y are

The limits of y are $y=0$ to $y=a$

Which is a straight line

The limits of x is $x=(y^2/a) \therefore y^2 = ax$

Which is parabola $x=y$ i.e. straight line

Now intersection of parabola and straight line

$x=(y^2/a)$ and $x=y$

$\therefore xa=x^2 \therefore x(x-a) = 0$

$\therefore x=0$ or $x=a $

When $x=0 ,y=0 $

When $x=a ,y=a.$

Now changing the order

The vertical strip will slide from $x=0$ to $x=a$

1) Outer limit $x: x=0$ to $x=a$

2) Inner limit y

a) Uppe limit is equation of parabola

$ \therefore y^2 = ax\therefore y=\sqrt{ax} $

b) Lower limit is a line

$\therefore y=x$

The integration becomes

$$I=\int\limits_0^a\int\limits_x^{\sqrt {ax}}\dfrac y{(a-x)\sqrt{ax-y^2}}dydx$$

$I=\int\limits_0^a\dfrac 1{a-x}(\dfrac {-1}2)\int\limits_x^{\sqrt{ax}}\dfrac {-2y}{\sqrt{ax-y^2}}dydx$

(Adjusted -2 in numerator and denominator)

$$Let\space I_1= \int\limits_x^{\sqrt{ax}}\dfrac {-2y}{\sqrt{ax-y^2}}dy\\ Put \space ax-y^2 = U \space \text{diff. w.r.t. y }\\ \therefore -2ydy=dU\\ when \space y=\sqrt{ax},U=0\\ when \space y=x,U=x(a-x)\\ \therefore I_1=\int\limits_{x(x-a)}^0\dfrac {dU}{\sqrt U}\\ =\Bigg[\dfrac {\sqrt U}{\frac 12}\Bigg]^0_{x(a-x)}\\ =2\sqrt0-2\sqrt{x(a-x)}\\ =-2\sqrt{x(a-x)}\\ when \space y=\sqrt{ax},U=0\\ when \space y=x,U=x(a-x)\\ \therefore I_1=\int\limits_{x(a-x)}^0\dfrac {dU}{\sqrt U}\\ =\Bigg[\dfrac {\sqrt U}{\frac 12}\Bigg]^0_{x(a-x)}\\ =2\sqrt0-2\sqrt{x(a-x)}dx\\ =-2\sqrt{x(a-x)}\\ \therefore I=\int\limits_0^a\dfrac 1{(a-x)}\times \Bigg(\dfrac 1{-2}\Bigg)\times -2\sqrt{x(a-x)}dx\\ =\int\limits_0^ax^{\frac 12}(a-x)^{\frac {-1}2}dx\\ put\space x=at\\ \therefore dx=adt\\ when\space x=a , t=1 \\When \space x=0 ,t=0\\ I=\int\limits_0^1(at)^{1/2}(a-at)^{-1/2}adt\\ I=a^{1/2}.a^{1/2}\times \int\limits_0^1t^{1/2}(1-t)^{-1/2}dt\\ =\alpha\beta(\frac 32,\frac 12) \\ I=\dfrac {a\Bigg)\overline{\dfrac 32}\Bigg)\overline{\dfrac 12}}{)\overline2}\\ =\dfrac {a.\dfrac12\sqrt{\pi}\times \sqrt{\pi}}1\\ =\dfrac {a\pi}2$$