SOLVED
I have a task: calculate surface area. There is given next expressions: $$z=x^2+y^2\quad z^2\leq xy\quad x\geq 0 \quad y\geq 0$$ This look like this from above: view from above
And like this from below: view from below
I think they ask to calculate surface area of this little green petal below.
So, I've calculated the surface element expression: $$\sqrt{8r^2+1}*dA$$ And limitations in polar coordinates: $$-\sqrt{\cos{\theta}\sin{\theta}} \leq r\leq \sqrt{\cos{\theta}\sin{\theta}}$$ $$r\sin{\theta} \geq 0$$ $$r\cos{\theta} \geq 0$$ How this limitation looks ( turquoise line - less than $r$, blue line - larger than $r$): [limitation][3]
Finally we get: $$\int_{0}^{\frac{\pi}{2}}d\theta \int_{0}^{\sqrt{\cos{\theta}\sin{\theta}}} \sqrt{8r^2+1}*r*dr = \frac{(15^\frac{3}{2}+1)}{6}$$
The surface of the sphere bounded by the hyperbolic sheet.
$dS = (-\frac {\partial z}{\partial x},\frac {\partial z}{\partial y}, 1)\\(-2x,-2y,1)\\ \|dS\| = \sqrt {4x^2+4y^2+1}$
The boundary of the intersection of the two surfaces.
$(x^2 + y^2)^2 = xy$
Convert to cylindrical
$r^4 = r^2(\sin\theta\cos\theta)\\ r = \sqrt {\sin\theta\cos\theta}\\ r = \sqrt {\frac 12 \sin 2\theta}$
$\int_0^{\frac{\pi}2}\int_0^{\sqrt {\frac 12 \sin 2\theta}} r\sqrt{4r^2+1} \ dr\ d\theta$