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Consider the surface $S=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2=(2-z)^2,0\leq z \leq 1\}$. Parametrize $S$ and calculate the area of $S$.

My parametrization: $ \phi(u,v)=(u,v,2-\sqrt{u^2+v^2}),\ where\ u=r\cos\theta\ ,v=r\sin\theta. \ So\ \phi(u,v)=(u,v,2-r)$

I know $r \leq 1,\ 0\leq\theta\leq2\pi$. How do I calculate the area of $S$? And is the parametrization correct?

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    Do you mean *the surface area* of $\;S\;$ ? This is a 3-dimensional body... – 2017-01-06
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    My teacher asked for area of S – 2017-01-06
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    Your parametrization of $S$ depends on the surface formula want to use – 2017-01-06
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    @MathScientist Perhaps, but you do understand it is **not** the area of a two dimensional body, right? Whatever you or your teacher called it... – 2017-01-06
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    @MathScientist: The component functions of your parametrization are all right, but the bounds on $r$ are not: If $0 \leq z \leq 1$, how large or small can $\sqrt{x^{2} + y^{2}} = |2 - z|$ be? As for the area, do you know about the scalar surface element$$dS = \|\phi_{u} \times \phi_{v}\|\, du\, dv?$$ – 2017-01-06
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    The given $S$ actually **is a surface**, not a body, in $\mathbb{R}^3$. Although it's better and more common to say _surface area_, there's really nothing wrong in asking for the _area_ of a surface. – 2017-01-06
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    It is worth knowing that S is the frustum of a cone, and you should have be able to find the area using classical geometry. – 2017-01-06

2 Answers 2

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No, your parametrization isn't quite right yet, although you do have pretty much all the right ingredients. The main issue with what you wrote is that you didn't quite decide how you're parametrizing it: in terms of $u,v$ or in terms of $r,\theta$? You have kind of a "hybrid" there which is neither this nor that... If you're parametrizing in terms of $u,v$, then your vector function $\phi(u,v)$ must depend on $u,v$ only and can NOT have any other variables, such as $r$. Or if you're parametrizing in terms of $r,\theta$, then it must depend on $r,\theta$ only and can NOT have any other variables, such as $u,v$. And the second issue is that it's not true that $r\le1$.

Note that in your parametrization $u=x$ and $v=y$, so you don't even need these two extra letters. You can parametrize $x=r\cos\theta$ and $y=r\sin\theta$, which gives you $z=2-r$, as you already found correctly. So the entire prametrization is $$\phi(r,\theta)=(r\cos\theta,r\sin\theta,2-r),$$ where $0\le\theta\le2\pi$ and $1\le r\le2$. The bounds for $r$ follow from $r=2-z$ and $0\le z\le1$.

Now you can set up the surface area of $S$ using the integral formula $$\text{Area}=\iint\limits_S dS=\iint\limits_S \|\phi_r\times\phi_{\theta}\|\,dr\,d\theta.$$

With the parametrization above (check all the steps yourself): $$\phi_r=(\cos\theta,\sin\theta,-1);$$ $$\phi_{\theta}=(-r\sin\theta,r\cos\theta,0);$$ $$\phi_r\times\phi_{\theta}=(r\cos\theta,r\sin\theta,r);$$ $$\|\phi_r\times\phi_{\theta}\|=r\sqrt{2}.$$

So the surface area is: $$\text{Area}=\iint\limits_S \|\phi_r\times\phi_{\theta}\|\,dr\,d\theta=\int_0^{2\pi}\int_1^2 r\sqrt{2}\,dr\,d\theta=\cdots=\pi\sqrt{2}.$$

Let us know if you'd like help with the steps in between.

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The easiest way to get the surface area is to put the problem into cylindrical coordinates $(r,\theta,z)$ (a la your parameterization). Then the equations of the surface are

$$ r^2 = (2-z)^2 \implies r = 2-z\\ 0\leq z\leq 1 $$ The surface area of a thin ring from $z = Z$ to $z=Z+\Delta Z$ is $2\pi r$ times $\sqrt{2} \Delta Z$; the $\sqrt{2}$ comes from the fact that hte distance from one point on the ring to the corresponding point on the other plane of the ring is $\sqrt{(\Delta z)^2+(\Delta r)^2} = \sqrt{2(\Delta z)^2}$

So (noting that $r=z$) you need to compute $$ \int_{z=0}^1 (2\pi z) \sqrt{2} \,dz = \sqrt{2}\,\pi $$