Area using double integrals

Question

My question considers double integrals, and calculating the area of cone $z^2=x^2+y^2$ cut out by $(x^2+y^2)^2=2xy$($z\ge 0$). So the second curve is like infinity symbol in first and third quadrant. After using polar coordinates $x=r\cos\phi$, $y=r\sin\phi$ i get $0\le r\le \sqrt{\sin2\phi}$ and $0\le \phi \le \frac{\pi}{4}$. And after i plug that in formula($\iint \sqrt{1+p^2+q^2}$)i get $\frac{\sqrt{2}}{4}$. But correct answer is apparantley 8 times bigger, which doesn't really make sense to me,because we have restriction $z\ge 0$ so that means whe are watching only cone $z=\sqrt{x^2+y^2}$ right?