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.
Best Answer
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.