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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.

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  • $\begingroup$ Hint: Expand and simplify the expression under the root sign, after correcting the negative signs of $y^2$ and $x^2$. $\endgroup$ – random Dec 9 '18 at 0:32
  • $\begingroup$ @random I know how to do the actual integral, but I'm confused as to what the bounds should be. $\endgroup$ – peco Dec 9 '18 at 0:56
  • $\begingroup$ Doing everything in polar coordinates is an option. $\endgroup$ – random Dec 9 '18 at 1:45
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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.

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