So I have a little doubt about this problem:

Let $\phi(u,v):D\mapsto\mathbb R^3$, where D is the unit disk centered at the origin. Let S be the surface with the parametrization:

$\phi(u,v)=(u-v, u+v, uv)$

Calculate the surface area of S.

What I would try to do to compute the surface area will be:

$\int_{S} \lVert \phi_u\times\phi_v\rVert dudv$

My problem comes with the limits of integration for u and v, and the fact that the domain is a polar region of the form: $D=\{(u,v)\in\mathbb R^2 : u^2 + v^2 \le1\}$. I'm not quite sure if I would need to transform D into a rectangle of the form: $D^*=\{(r,\theta)\in\mathbb R^2 : 0\le r\le 1, 0\le \theta\le 2\pi\}$, and from here calculate the area of the surface via a composition of $\phi(u,v)$ with the transformation to polar coordinates, and if so do I need to include the Jacobian of the polar change of variable when calculating the integral?

Thanks for your help!

  • $\begingroup$ Feel free to change the title if needed. :D $\endgroup$ – Ron Apr 10 '16 at 7:37

The beauty of a given parametrization is that you don't even need to care what the surface S looks like.

$ \phi(u,v)=(u-v, u+v, uv) $ You can imagine that here it says $x,y$ if it makes you feel more at ease.

We need to calculate $\int\int_SdS$ but first we need
$\phi_u'(u,v)=(1,1,v)$ and

Finally we get $n(u,v)=\phi_u'(u,v) \times \phi_v'(u,v)=(u-v,-u-v,2)$

We need $\| n(u,v)\|=\sqrt{(u-v)^2+(u+v)^2+4}=\sqrt{2u^2+2v^2+4}$

Now we can approach to calculating the integral

$$ \int\int_SdS=\int\int_D \sqrt{2u^2+2v^2+4} \space dudv $$

Now we can do a variable change since polar coordinates look as a right thing to do since domain is a unit disk centered at $(0,0)$

$u=r\cos{\gamma}$ , $v=r\sin{\gamma}$ , $r\in[0,1]$ and $\gamma\in[-\pi,\pi]$
Then the jacobian $\left | J \right |=r$

Finally we get

$$ \int\int_SdS=\int\int_D \sqrt{2u^2+2v^2+4} \space dudv=\int_{-\pi}^{\pi}\int_{0}^{1} r\sqrt{2r^2+4} \space drd\gamma $$

$$ =2\pi\sqrt{2} \int_{0}^{1} r\sqrt{r^2+2}dr=2\pi\sqrt{2}\frac{1}{3}(\sqrt{r^2+2})^3|_{0}^{1}=2\pi\sqrt{2}\frac{3\sqrt{3}-2\sqrt{2}}{3} $$

  • $\begingroup$ Thank you! This is exactly what I ended up doing :D $\endgroup$ – Ron Apr 10 '16 at 20:00

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