# Surface Area of Multiple Integration

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.

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

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
$$S=\dfrac14\int_0^{2\pi}\int_0^{\cos(2\theta)}\sqrt2r\,\mathbb dr\,\mathbb d\theta$$