If the area bounded by the curves y=a x^2 and x=a y^2,(a>0) is 3 sq...

We have given, $y=a x^2$ ....(i)$and x=a y^2$ ......(ii) Put the value of $y$ by Eq. (i) in Eq. (ii), we get $x=a \times a^2 x^4 \Rightarrow x^4 a^3-x=0x\left(x^3 a^3-1\right)=0 \Rightarrow x=0, \frac{1}{2}$ When, $x=0 \Rightarrow y=0$ and $x=\frac{1}{a} \Rightarrow y=\frac{1}{a}$ Here, points of intersection of curves $y=a x^2 and x=a y^2$ are (0,0) and \left(\frac{1}{a}, \frac{1}{a}\right)$.$\therefore \quad$Required area$A=\int_{x=a}^{x=b}\left[f_2(x)-f_1(x)\right] d x3=\int_0^{1 / a}\left(\frac{\sqrt{x}}{\sqrt{a}}-a x^2\right) d x3=\left[\frac{1}{\sqrt{a}} \times \frac{2}{3} x^{3 / 2}-\frac{a x^3}{3}\right]_0^{1 / a}3=\frac{2}{3 \sqrt{a}}\left[\left(\frac{1}{a}\right)^{3 / 2}\right]-\frac{a}{3}\left[\left(\frac{1}{a}\right)^3\right]3=\frac{2}{3 \sqrt{a}} \times \frac{1}{a \sqrt{a}}-\frac{a}{3} \times \frac{1}{a^3}3=\frac{2}{3 a^2}-\frac{1}{3 a^2} \Rightarrow 3=\frac{2-1}{3 a^2} \Rightarrow 9 a^2=1a^2=\frac{1}{9} \Rightarrow a=\frac{1}{3}