Suppose you have the surface $z=3xy$ and you want to find the area that lies within the cylinder $x^2+y^2\leq 1$.

My homework is forcing me to use the parameterization

$$\textbf{r}_1(s,t)= <s\cos(t), s\sin(t), 3s^2\sin(t)\cos(t)>$$

I am having a difficult time visualizing this parameterization, and I do not have any graphing software to graph the surface, but I want to make sure I understand this concept.

This is quite obvious, but I want to be sure; using the above parameterization, I am not parameterizing the entire surface, right? If I wanted to, I assume the parameterization would be $$\textbf{r}_2(s,t) = <s,t,3st>$$

Instead, is $\textbf{r}_1$ just the parameterization adjusted for the region - the region being the cylinder $x^2+y^2\leq 1$? That is, are we just making a revolution around $z=3xy$?

Any insight would be helpful.


You can see this exercise as a surface integral or you can also see it as a double integral of a function $f(x,y)$ over the unit disc.

$\mathbf r_1$ is the cylindrical parameterization, which, generally, is $$\begin{cases}x=s\cos t\\y=s\sin t\\z=f(x,y)=f(s\cos t,s\sin t)\end{cases}$$

In your case $z=f(x,y)=3xy$. In $\mathbf r_1$, the cylinder becomes $s\le1$ and you have no limitations on $t$. You have to evaluate the integral $$\begin{align}\int_0^1s\times 3s^2\mathrm ds\int_0^{2\pi}\sin(t)\cos(t)\mathrm dt&=0\end{align}$$

I got to this integral by the following way:

$$\iint f(x,y)\mathop{\mathrm dx}\mathop{\mathrm dy}=\iint 3xy\mathop{\mathrm dx}\mathop{\mathrm dy}$$ Using polar coordinates $(x=s\cos t, y=s\sin t)$ and adding the Jacobian: $$\iint3s^3\sin t\cos t\mathop{\mathrm ds}\mathop{\mathrm dt}$$ Because we are in the unit disc, $0<s\le1, \quad0<t<2\pi$ $$\int_0^13s^3\left(\int_0^{2\pi}\sin t\cos t\mathop{\mathrm dt}\right)\mathop{\mathrm ds}$$

  • $\begingroup$ The three upvotes and acceptance notwithstanding this answer is seriously wrong. Why should the area in question be $=0\>$? We have a real "floppy disc" here! $\endgroup$ – Christian Blatter Nov 19 '18 at 11:57
  • $\begingroup$ I think the integrand is right, the limits of integration could be wrong. I tried to evaluate the integral in cartesian coordinates and I got $0$ again: $\oint_{x^2+y^2\le1}3xy\mathop{\mathrm dx\mathrm dy}=3\int_{-1}^1x\mathop{\mathrm dx} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}y\mathop{\mathrm dy}=0$. I honestly do not see where the mistake is $\endgroup$ – Lorenzo B. Nov 19 '18 at 12:21
  • $\begingroup$ I understand your answer (+1). Why mine is wrong? $\endgroup$ – Lorenzo B. Nov 19 '18 at 14:36
  • $\begingroup$ What made you arrive at the integrand $3s^3\cos t\sin t\>$? – That's my last word on this matter. $\endgroup$ – Christian Blatter Nov 19 '18 at 15:00

Since in the end you have to integrate over the unit disc in the $(x,y)$-plane your source proposes to use polar coordinates instead of $x$ and $y$ as parameters. Then $x=s\cos t$, $y=s\sin t$. In this way the idea $z=3xy$ $\>(x^2+y^2\leq1)$ translates into $${\bf r}(s,t)=(s\cos t,s\sin t,3s^2\cos t\sin t)\qquad(0\leq s\leq 1, \ 0\leq t\leq2\pi)\ .$$ In order to find the area of this floppy disc $F$ we have to compute $${\bf r}_s=(\cos t,\sin t, 6s\cos t\sin t),\quad {\bf r}_t=\bigl(-s\sin t,s\cos t ,3s^2\cos(2t)\bigr)$$ and then $${\bf r}_s\times{\bf r}_t=(\ldots,\ldots,\ldots)\ .$$ The area is then finally given as $${\rm area}(F)=\int_0^1\int_0^{2\pi}\bigl|{\bf r}_s\times{\bf r}_t\bigr|\>dt\>ds\ .$$ The resulting integral will be simpler than dreaded.

  • $\begingroup$ Sorry, it's me again. Shouldn't you integrate $|\bf r_s \times \bf r_t|{f}(\bf r)$ instead of just $|\bf r_s \times \bf r_t|$? $\endgroup$ – Lorenzo B. Nov 22 '18 at 18:44

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