Solution verification(double integrals)

Question

I need to find surface area between $z^2=x^2+y^2$ and $(x^2+y^2)^2=2xy$.Using polar coordinates i found bounds for r:$0\le r \le \sqrt{\sin 2\phi}$ and $\phi$ $0\le \phi \le \frac{\pi}{4}$. Then using the formula for surface area i get $\frac{\sqrt{2}}{4}$, but in solution it says $2\sqrt{2}$. Where am I mistaking? Btw for $\phi$ i used this bounds because in $[0,\frac{\pi}{4}]$ is already included every positive value for sine function(because $\sin 2\phi$ needs to be $\ge 0$).

Answer 1

You need to find the surface area of the cone $z^2 = (x^2+y^2) = r^2$ bound by the cylinder $\rho^2 = |\sin2\theta|$.

The cylinder has two petals as shown in the diagram $(0 \leq \theta \leq \frac{\pi}{2} \,$ for the petal in the first octant). The radius of $1$ is maximum at $\theta = \frac{\pi}{4}$ and radius is $0$ at $\theta = 0, \frac{\pi}{2}$.

Please note you need to find surface area in $4$ octants, as the cone $z^2 = x^2 + y^2$ forms both below $z = 0$ and above. So we will find surface area in first octant and then multiply by $4$.

As the cone is a right circular cone with equation $z = \pm r$, we have $\phi = \frac{\pi}{4}$.

Now the cone and cylinder intersect at two points at any given $z$ in the first octant (see the part of the cross section of the cone bound by cylinder, marked in black).

$r^2 = z^2 = \sin 2\theta, \sin (\pi - 2\theta) \implies \theta = \frac{1}{2} \arcsin (r^2), \frac{\pi}{2} - \frac{1}{2} \arcsin (r^2)$.

At $z = 1$, cone touches the cylinder at its maximum radius $1$ and there is no intersection for $z \gt 1$. So $0 \leq r \leq 1$.

$\sqrt {1 + (\frac{\partial z}{\partial x})^2 + (\frac{\partial z}{\partial y)^2})^2} = \sqrt 2$. In cylindrical coordinates, $dx \, dy = r \, d\theta \, dr$.

So surface area $S = \displaystyle 4 \int_0^1 \int_{\arcsin r^2 / 2}^{(\frac{\pi}{2} - \arcsin r^2)/2} \sqrt 2 \, r \, d\theta \, dr = 2 \sqrt2$