Csdrhrt

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

My homework is forcing me to use the parameterization

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

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 $x^2+y^2leq 1$? 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
$$begin{cases}x=scos t\y=ssin t\z=f(x,y)=f(scos t,ssin t)end{cases}$$

In your case $z=f(x,y)=3xy$. In $mathbf r_1$, the cylinder becomes $sle1$ and you have no limitations on $t$. You have to evaluate the integral
$$begin{align}int_0^1stimes 3s^2mathrm dsint_0^{2pi}sin(t)cos(t)mathrm dt&=0end{align}$$

I got to this integral by the following way:

$$iint f(x,y)mathop{mathrm dx}mathop{mathrm dy}=iint 3xymathop{mathrm dx}mathop{mathrm dy}$$
Using polar coordinates $(x=scos t, y=ssin t)$ and adding the Jacobian:
$$iint3s^3sin tcos tmathop{mathrm ds}mathop{mathrm dt}$$
Because we are in the unit disc, $0<sle1, quad0<t<2pi$
$$int_0^13s^3left(int_0^{2pi}sin tcos tmathop{mathrm dt}right)mathop{mathrm ds}$$

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 $x=scos t$, $y=ssin t$. In this way the idea $z=3xy$ $>(x^2+y^2leq1)$ translates into
$${bf r}(s,t)=(scos t,ssin t,3s^2cos tsin t)qquad(0leq sleq 1, 0leq tleq2pi) .$$
In order to find the area of this floppy disc $F$ we have to compute
$${bf r}_s=(cos t,sin t, 6scos tsin t),quad {bf r}_t=bigl(-ssin t,scos t ,3s^2cos(2t)bigr)$$
and then
$${bf r}_stimes{bf r}_t=(ldots,ldots,ldots) .$$
The area is then finally given as
$${rm area}(F)=int_0^1int_0^{2pi}bigl|{bf r}_stimes{bf r}_tbigr|>dt>ds .$$
The resulting integral will be simpler than dreaded.