Find the area of the upper half of the cone $x^2+y^2=z^2$ above the interior of one loop of $r=cos(2\theta)$.
I know the formula for surface area is $\int_{x_0}^{x_1}\int_{y_0}^{y_1}\sqrt{(f_x)^2+(f_y)^2+1}dxdy$, and in this question it should be $\int_{x_0}^{x_1}\int_{y_0}^{y_1}\sqrt{(\frac{-2x}{2\sqrt{-y^2-x^2}})^2+(\frac{-2y}{2\sqrt{-y^2-x^2}})^2+1}dxdy$, but I am not sure about the bounds. I am guessing that since I'm only taking the area above one loop of $r=cos(2\theta)$, that the bound for $x$ goes from $0$ to $1$. What about the bounds for $y$? Thanks.
In cylindrical coordinates it gets simpler. The surface element for this cone with the parametrization $\vec s=(r\cos\theta,r\sin\theta,r)$ is $\mathbb dS=\sqrt{2}r\,\mathbb dr\,\mathbb d\theta$. So
$$S=\dfrac14\int_0^{2\pi}\int_0^{\cos(2\theta)}\sqrt2r\,\mathbb dr\,\mathbb d\theta$$
The factor 1/4 is because the graph of the curve has four equal lobes.