Calculate the area of the walls of the solid given by $x^2+y^2=1$, $z=0$ and $z=1+xy$. The walls mean the area between $z=0$, and $z=1+xy$.

I know that I can solve this problem by calculating the area of the surface by treating the walls as a toilet paper, unfold the toilet paper on a flat surface, and do a double integral over the toilet paper with the function$f(x,y)$ inside of the double inegral set to 1.

$\int _0^{2\pi}\:\int _0^{1+cos\left(\theta \right)\cdot \:sin\left(\theta \right)}\:1\:dz\:d\theta$

For this problem though I am more interested in using the formula given by:

$\int \:\int _A\left|\frac{\partial r}{\partial u}\times\frac{\partial r}{\partial v}\left(u,v\right)\:\right|du\:dv$

And I am unsure of the parametrization of this problem and then what the jacobian-determinant becomes.

I think the correct parametrization is:

$r\left(\theta ,z\right)=cos\left(\theta \right)i+sin\left(\theta \right)j+\left(z\right)k\:$

Choosing $r(\theta,z)$ as above $\left|\frac{\partial \:r}{\partial \:u}\times \frac{\partial \:r}{\partial \:v}\left(u,v\right)\:\right|$ becomes 1. I generally have a problem with these things when the functions aren't given on the form $z=..$

But then, the question becomes, if that is the correct parametrization, what is the jacobian-determinant?


1 Answer 1


Everything you say is correct. You have the parametrization $${\bf r}(\theta,z)=(\cos\theta,\sin\theta,z)\ ,\tag{1}$$ giving $$|{\bf r}_\theta\times{\bf r}_z|\equiv1\ ,$$ hence the surface element on $S$ is given by ${\rm d}\omega={\rm d}(\theta,z)$. There is no Jacobian determinant here, but a scaling factor from the measure ${\rm d}(\theta,z)$ in the parameter plane to the surface measure ${\rm d}\omega$ in $3$-space. This scaling factor happens to be $1$ in the case at hand. You end up with the integral of your first attempt, obtained using "geometric intuition". Given the parametrization $(1)$ the condition $0\leq z\leq 1+x y$ translates into $0\leq z\leq1+\cos\theta\sin\theta$.

  • $\begingroup$ But why is there no jacobian determinant? If you parametrize with $x=r cos(\theta)$, and $y=rsin(\theta)$, the jacobian determinant will be $r$ for example.I can't calculate the determinant because the jacobian matrix won't be 2x2, or 3x3 (If I am doing it correctly). I tried making it a 3x3 matrix filling in all the blanks with 0, and then the determinant ended up being 0? Maybe that is correct. $\endgroup$
    – David Lund
    Commented Aug 1, 2017 at 16:46
  • $\begingroup$ Do you think you could give me an example where the scaling factor isn't 1? Or point me to some article explaining this maybe. $\endgroup$
    – David Lund
    Commented Aug 1, 2017 at 19:46
  • $\begingroup$ The scaling factor is $|{\bf r}_{.1}\times{\bf r}_{.2}|$, as you knew all along. For a representation of the unit sphere (with $\theta=0$ on the equator) you obtain $|{\bf r}_\phi\times{\bf r}_\theta|=\cos\theta$. – The Jacobian determinant is the scaling factor between volume measures for maps $f:\>{\mathbb R}^n\to{\mathbb R}^n$. $\endgroup$ Commented Aug 2, 2017 at 7:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .