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.
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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$$$\int_0^13s^3\left(\int_0^{2\pi}\sin t\cos t\mathop{\mathrm dt}\right)\mathop{\mathrm ds}$$
$\endgroup$ 4
