# How to calculate the surface area of parametric surface?

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

$$\textbf{r}_1(s,t)= $$

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 $$\textbf{r}_2(s,t) = $$

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 $$z=3xy$$?

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, $$0 $$\int_0^13s^3\left(\int_0^{2\pi}\sin t\cos t\mathop{\mathrm dt}\right)\mathop{\mathrm ds}$$

• 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! – Christian Blatter Nov 19 '18 at 11:57
• 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 – Lorenzo B. Nov 19 '18 at 12:21
• I understand your answer (+1). Why mine is wrong? – Lorenzo B. Nov 19 '18 at 14:36
• What made you arrive at the integrand $3s^3\cos t\sin t\>$? – That's my last word on this matter. – 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.

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