Area of the surface Compute the area between the surfaces $(x^2+y^2)^2=a^2(x^2-y^2)$ and $x^2+y^2+z^2=a^2.$
My attempt:
I take the first case $z\geq 0$:
$$x=x$$
$$y=y$$
$$z=\sqrt{a^2-(x^2+y^2)}$$
then 
$$T_x=(1,0,\frac {-x}{\sqrt{a^2-(x^2+y^2)}})$$
$$T_x=(0,1,\frac {-y}{\sqrt{a^2-(x^2+y^2)}})$$
Then the area of the surface is:
$$A(S)=\int_S d\sigma=\int\int_D \frac {a}{\sqrt{a^2-(x^2+y^2)}}dxdy$$
By parameterization:
$$x=rcost$$
$$y=rsint$$
then $r=a\sqrt{cos(2t)}\implies t\in [0, \frac {\pi}4]$ and $r\in[0,a\sqrt{cos(2t)}]$
$A(S)=(a^2-\sqrt{2})$. But that's a wrong answer, so what am I doing wrong?
 A: Using parametric form $x = rcost$ and $y = rsint$ the equation of the curve is $r^2 = a^2cos2t$ and the equation of the sphere is $x = rcost$, $y = rsint$, $z = \sqrt{a^2 - r^2}$
The integral of surface area is given by 
$$\int\int_R\sqrt{EG - F^2}drdt$$
$$E = (\frac{\partial x}{\partial r})^2 +(\frac{\partial y}{\partial r})^2 + (\frac{\partial z}{\partial r})^2 = \frac{a^2}{a^2 - r^2}$$
$$G = (\frac{\partial x}{\partial t})^2 + (\frac{\partial y}{\partial t})^2 + (\frac{\partial z}{\partial t})^2 = r^2$$
$$F = (\frac{\partial x}{\partial r})(\frac{\partial x}{\partial t}) + (\frac{\partial y}{\partial r})(\frac{\partial y}{\partial t}) + (\frac{\partial z}{\partial r})(\frac{\partial z}{\partial t}) = 0$$
Therefore the integral for the surface area for $y > 0$ at the upper sphere is
$$\int_0^{\frac{\pi}{4}}\int_0^{a\sqrt{cos2t}}\sqrt{\frac{a^2r^2}{a^2 - r^2}}drdt$$
$$= a^2(\frac{\pi}{4} + 1 - \sqrt{2})$$
Since we have $y < 0$ and the lower sphere plus there are two branches, the total surface area should be 8 times the above.  Hence the total surface area is 
$$a^2(2\pi + 8 - 8\sqrt{2})$$
