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I am trying to calculate the area of the surface $z = x^2 + y^2$, with $x^2 + y^2 \le 1$. By trying to do the surface integral in Cartesian coordinates, I arrive at the following:

$\int_{-1}^{1}dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{4x^2 + 4y^2 + 1} dy$

This is a bit hard to calculate, so I tried to switch to polar coordinates, with

$x=r\cos{\theta}$
$y=r\sin\theta$
$r \in [0, 1]$
$\theta \in [0, 2\pi]$

$f(r, \theta) = (r\cos\theta,\ r\sin\theta,\ r^2)$

$\int_0^{2\pi}d\theta \int_0^1 |J|\ ||\frac{\partial f}{\partial r} \times \frac{\partial f}{\partial \theta}||dr$

But this also results in a square root that is hard to integrate. Am I doing the surface integral wrong, or should I take a different approach?

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Hint: make a further change of variable from $r$ to $u = \sqrt{4 r^2+1}$. One trick to save you some headache is $r dr = \frac12 d r^2 = \frac12 d \frac{u^2-1}{4} = \frac{u du}{4}$... –  achille hui Feb 4 '13 at 12:11
1  
There is no $J$ when you use a parametric representation $f$ of $S$ and compute $|f_r\times f_\theta|$. –  Christian Blatter Feb 4 '13 at 12:57

3 Answers 3

up vote 1 down vote accepted

We have $$\int_{-1}^{1}dx \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \sqrt{4x^2 + 4y^2 + 1} ~dy\to \int_0^{2\pi}\int_0^1\sqrt{4r^2+1}\times r\times drd\theta$$ The $J$, when we convert the Cartesian coordinates into Cylindrical coordinates, is always $r$. To see that just follow the definition of $J$ Jacobian matrix.

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Nice observation! +1 –  amWhy Feb 4 '13 at 14:32

The value of $z$ goes from 0 to 1. And the circle at $z = z_0$ has radius $\sqrt{z_0}$ so the total volume is $$ \int_0^1 2\pi \sqrt{z} dz = \frac{4}{3} \pi$$

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Using the parametrization $$f:\quad (r,\theta)\mapsto(r\cos\theta,r\sin\theta, r^2)$$ with domain $B$ the rectangle $[0,1]\times[0,2\pi]$ in the $(r,\theta)$-plane we get the following expression for the area of $S$: $$\omega(S)=\int_B |f_r\times f_\theta|\ {\rm d}(r,\theta)\ .$$ Now $$f_r=(\cos\theta,\sin\theta, 2r),\quad f_\theta=(-r\sin\theta,r\cos\theta,0),\quad f_r\times f_\theta=(-2r^2\cos\theta,-2r^2\sin\theta,r)$$ and therefore $$|f_r\times f_\theta|^2=4r^4+r^2\ ,$$ independently of $\theta$. It follows that $$\omega(S)=2\pi\int_0^1 \sqrt{4r^2 +1}\ r\ dr=2\pi\>{1\over12}(4r^2+1)^{3/2}\biggr|_0^1={\pi\over6}\bigl(5\sqrt{5}-1\bigr)\ .$$

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