How do I calculate the following surface area?

Question

I have encountered this problem during my vector calculus studies which I cannot seem to setup correctly. This is a paraphrasing of the question:

Calculate the surface area of the surface $S$, a cone such that $z^2=x^2+y^2$, bounded above the disc $x^2+y^2\le4$.

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What formula would I use to compute this, or what is the process to getting the correct answer?

Answer 1

What I'd do is to plot the region the exercise is talking about (it's just a cone with the vertex in the bottom at (0,0,0) and a it's top at z = 2) and solve for the triple integral that gives you the volume. That is $\displaystyle\int_{\{(x,y,z) \in \mathbb{R}^3 : x^2+y^2 \leq z^2 \leq 4\}}dxdydz$.

Or if you're used to cylindrical coordinates, try those. They're much easier in this case.

Answer 2

Here, given cone $$z = \sqrt{x^2 + y^2}$$ and as it is given that the required region is bounded above the disc $ \Rightarrow z \in (0,2] $

You can solve the problem either by relying on the formula of lateral surface area $ S = \pi r l$ where $r =$ Radius, and $ l = $ slanted height.

Or you can use calculus. Here I will use cylindrical coordinates.

My area element would be $dA = r d\theta dh \sec \frac{\pi}{4}$ ,Here $r = h$, so $$ A = \sec \biggr (\frac{\pi}{4} \biggl )\int_0^2 \int^{2\pi}_0 h dr d \theta $$

Your task is to figure out why there is $\sec \bigr (\frac{\pi}{4} \bigl )$ and $r = h$

Answer 3

The area of $S$ as described is infinite, since $z^2=x^2+y^2$ is a double cone that opens outward to positive and negative infinity. Perhaps you meant to include $z\ge0$, or equivalently $z=\sqrt{x^2+y^2}$? I'll assume that to be the case.

The standard choice of parameterization for $S$ would be one that relies on cylindrical coordinates.

$$\vec s(u,v)=\langle u\cos v,u\sin v,\sqrt{u^2\cos^2v+u^2\sin^2v}\rangle=\langle u\cos v,u\sin v,u\rangle$$

with $u\in[0,2]$ and $v\in[0,2\pi]$. Then the surface element is

$$\mathrm dS=\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\sqrt2\,u\,\mathrm du\,\mathrm dv$$

The area of $S$ is then

$$\iint_S\mathrm dS=\sqrt2\int_0^{2\pi}\int_0^2u\,\mathrm du\,\mathrm dv=\boxed{4\sqrt2\,\pi}$$