# Calculate surface area of a F using the surface integral

Given: $$F := \{(x,y,z) \in \mathbb{R}^3 \mid (x,y) \in W,z=f(x,y)\}$$

Calculate the surface area using the surface integral:

$i) \; f(x,y) := x+y \;\; and \;\; W := [12,31] \times [2014,2015]$ $ii) \; f(x,y) := \cos(x) \;\; and \;\; W := [0,\pi] \times [0,2\pi]$



# My Calculations

$i)$

$$\vec{\phi}:[12,31]\times[2014,2015] \rightarrow \mathbb{R}^3, \; \vec{\phi}\left(u,v\right)= \begin{pmatrix} u\\ v\\ u+v \end{pmatrix}$$

$$\left\|\frac{\partial\vec{\phi}}{\partial u} \times \frac{\partial\vec{\phi}}{\partial v}\right\| = \left\| \begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix} \right\| = \left\| \begin{pmatrix} -1\\ -1\\ 1 \end{pmatrix} \right\| = \sqrt{3}$$

$$o(F) = \iint_F 1 \; d\sigma = \iint_F\left\|\frac{\partial \vec{\phi}}{\partial u}\times\frac{\partial \vec{\phi}}{\partial v}\right\|d(u,v)$$ $$= \int_{2014}^{2015} \int_{12}^{31} \sqrt{3} \; dudv = 1 \cdot \sqrt{3} \cdot 19 = 19\sqrt{3}$$

I tried to calculate this roughly via vectors lenghts and the result was $o(F)=\sqrt{2} \cdot 19\sqrt{2} = 38$, one of those two calculations must therefore be wrong, and assuming I did the vector math right it seems I'm doing something wrong in the surface integrals.



$ii)$

$$\vec{\phi}:[0,\pi]\times[0,2\pi] \rightarrow \mathbb{R}^3, \; \vec{\phi}\left(u,v\right)= \begin{pmatrix} u\\ v\\ \cos u \end{pmatrix}$$

$$\left\|\frac{\partial\vec{\phi}}{\partial u} \times \frac{\partial\vec{\phi}}{\partial v}\right\| = \left\| \begin{pmatrix} 1\\ 0\\ -\sin u \end{pmatrix} \times \begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix} \right\| = \left\| \begin{pmatrix} \sin u\\ 0\\ 1 \end{pmatrix} \right\| = \sqrt{\sin^2 u + 1}$$

$$o(F) = \iint_F 1 \; d\sigma = \iint_F\left\|\frac{\partial \vec{\phi}}{\partial u}\times\frac{\partial \vec{\phi}}{\partial v}\right\|d(u,v)$$ $$= \int_{0}^{\pi}\int_{0}^{2\pi} \sqrt{\sin^2 u + 1} \; dvdu = 2\pi \cdot \int_{0}^{\pi}\sqrt{\sin^2 u + 1} \; dvdu = \; ???$$

It seems there is no good solution for this integral, therefore I must be doing something wrong because such tasks will most likely be solvable.

# Question

My suspicion is that my parametric representations of F ($\phi$) are wrong. It seems that one cannot transform them into spherical or polar coordinates (all my previous tasks were transformable, that's why I'm struggling) and thus I'm wondering what to do in such a case.

• Is this still open or have you solved it? – Alec Teal Dec 18 '14 at 17:00
• @AlecTeal It's still open, you found a mistake? – L1me Dec 18 '14 at 18:28
• I have noticed from your recent questions (surface area of a sphere) that you don't really know what's going on, so I have written a huge but detailed answer about how to do questions like this. – Alec Teal Dec 18 '14 at 21:17
• Are you sure it's cos? – Alec Teal Dec 19 '14 at 0:50
• @AlecTeal do you mean $f(x,y) := \cos(x)$ in the given task? Yes I'm sure. I might not have put up the right questions, the problem I'm having is getting a correct parametric representation of F for i) and ii). – L1me Dec 19 '14 at 11:57

Okay it is ABUNDANTLY clear you are not very confident with notation (as it is awful) so I shall teach you all you need to know about doing stuff on surfaces.

Lets tackle this intuitively, I will assume you are happy with the area of a parallelogram being related to the vector cross product, if not it is literally by definition (remember the sin rule for area of a triangle, it's that) so be happy with that, it is important.

We know that: $\|u\times v\|=\|u\|\|v\|\sin(\theta)=$ area of parallelogram with edges the vectors $u$ and $v$

Idea: We want to break the surface, call it $H$ (I choose H because why not) into little chunks, then sum over those chunks, like with all integrals, we shall write this: (where $d\sigma$ is a small chunk and H our surface)$$\iint_Hd\sigma$$
Don't worry about this too much, it's notation, use what helps, some authors write $\int_Hd\sigma$ instead, I use two integrals to emphasise (read: not let me forget) that I am integrating over a surface.

Picture time!

So we take a point $f(x,y)$ and we go $\delta x$ in the x direction, ending up at $f(x+\delta x,y)$ - already you should be thinking of the gradient, we are approximating the surface at $f(x,y)$ with a parallelogram, anyway, this gives us the double-marked line, the single one is by going $\delta y$ in the $y$ direction.

Notice the point $E$, the dotted line shows that this is quite far above the surface at this point, this is an approximation after all (which is why I have used $\delta$s) - anyway using the formula above the area of this parallelogram (which is $\delta\sigma$ - our small chunk of surface) is $\|u\times v\|$ where u and v are the different lines that make the parallelogram, as I have written on the diagram.

$$\delta\sigma = \|(f(x+\delta x,y)-f(x,y))\times(f(x,y+\delta y)-f(x,y))\|$$

Remember the definition of partial-derivatives, $\frac{\partial f}{\partial x}=\lim_{\delta x\rightarrow 0}(\frac{f(x+\delta x,y)-f(x,y)}{\delta x})$ this means $\frac{\partial f}{\partial x}\approx\frac{f(x+\delta x,y)-f(x,y)}{\delta x}$ (is approx. for small $\delta x$) thus:

$\delta x\frac{\partial f}{\partial x}\approx f(x+\delta x,y)-f(x,y)$ and $\delta y\frac{\partial f}{\partial y}\approx f(x,y+\delta y)-f(x,y)$ by the same logic.

So:

$$\delta\sigma = \|(f(x+\delta x,y)-f(x,y))\times(f(x,y+\delta y)-f(x,y))\|\approx\|\delta x\frac{\partial f}{\partial x}\times\delta y\frac{\partial f}{\partial y}\|$$

Recall the definition of cross product, $\|u\times v\|=\|u\|\|v\|\sin(\theta)$, if we have scalars $a,b\ge 0$ (which we do, we have $\delta x$ and $\delta y$) then $\|au\times bv\|=\|au\|\|bv\|\sin(\theta)=|a|\|u\||b|\|v\|\sin(\theta)=ab\|u\|\|v\|\sin(\theta)=ab\|u\times v\|$

So:

$$\delta\sigma \approx\|\delta x\frac{\partial f}{\partial x}\times\delta y\frac{\partial f}{\partial y}\|=\delta x\delta y\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|$$

Taking limits (as $\delta x,\delta y$ tend towards zero, leaving the approximation to get equality) we see that:

$$d\sigma=dxdy\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|$$

Now $\iint_Hd\sigma=\iint_R\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|dA$ where here $dA$ is a small chunk of area, in our case $dA=dxdy=dydx$, here $R$ means "Range" or "Region", it is what we will integrate over.

Why $dA$?

Lets take a function $f$ defined on some region $R$ and the integral $\iint_RfdA$, there are many ways to look at R, we could use cartesian coordinates, $x$ and $y$, or we could use polar coordinates $r,\theta$ say, suppose we take the circle of radius 10, we could define this as $R=\{(x,y)|-10<x<10\text{ and }-\sqrt{100-x^2}<y<\sqrt{100-x^2}\}$ or $R=\{(r,\theta)|r<10\}$

These are the same region, also the function $f(x,y)=x^2+y^2$ is the same as $f(r,\theta)=r^2$, as you can see for the parabaloid $x^2+y^2$ we can use several coordinates, this is what manifolds (in essence) studies, you can pick several coordinates and arrive at the same thing.

Back on topic

Notice that by using $\frac{\partial f}{\partial x}$ and the same for y, I imply $dA$ will be in $dxdy$ form, so if our region is best given in polar coordinates, I'd have to convert it to Cartesian and work with that, this transformation is called "The Jacobian" and will come later for you. So let us give one general statement of surface area:

$$\iint_Hd\sigma=\iint_R\|\frac{\partial f}{\partial u}\times\frac{\partial f}{\partial v}\|dA$$ where $u$ and $v$ are not degenerate, that is they are not vectors of the same direction (as then we wouldn't have a parallelogram)

It is trivial that in the cartesian plane $dA=dxdy=dydx$ - any 5 year old can tell you the area of a rectangle is width times height.

Notice I didn't write $\iint_Hd\sigma=\iint_R\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|dxdy$ earlier, you're probably wondering why, this notation makes it clear that I am looking at R in a cartesian system (I have x and ys) the answer is personal preference, I do not like this because it fixes order (x first in this case) yet we have not given the integral limits in this case. Also the x and y appearing in the partials already imply I am looking at cartesian coordinates, I do not want to say doing the x integral first, hence $dA$

Using the circle example again with the parabaloid (call this parabaloid H) $f(x,y)=x^2+y^2=r^2=f(r,\theta)$, over the circle of radius 10, we can write:

$\iint_Hd\sigma=\iint_R\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|dA$
$=\int^{10}_{-10}\int^{\sqrt{100-x^2}}_{-\sqrt{100-x^2}}\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|dydx$
$=\int^{10}_{-10}\int^{\sqrt{100-y^2}}_{-\sqrt{100-y^2}}\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|dxdy$

Or we can go to polar!

$=\int^{2\pi}_0\int^{10}_{0}\|\frac{\partial f}{\partial r}\times\frac{\partial f}{\partial \theta}\|rdrd\theta$

Now the order of $dx$ and $dy$ matters, which is why I write it!

So where did $rdrd\theta=dA$ come from?

You can see in the first picture that $\delta A$ looks nothing like a rectangle, however lets make our approximation better by looking at a smaller $\delta\theta$, (the second picture), now $\delta A$ looks nice and rectangular, I have used an arc through $\delta\theta$ of radius $r$ has length $r\delta\theta$ to get these, notice though we have two approximations, (we can use $r\delta\theta$ or $(r+\delta r)\delta\theta$ for the "height" of the rectangle), in the latter one we have a $(\delta r)^2$, for tiny $\delta r$ if we square it we will get an even tinier number (think about 0.1, $0.1^2=0.01$ which is so tiny we can discard) meaning the approximations agree with each other, if we take the limit as the deltas tend towards zero we arrive at $dA=rdrd\theta$ as I claimed.

End of explanation

Triple-integrals (where I use $dV$ rather than $dA$ to mean "small chunk of volume") follow this same logic, so this is a good notation to adopt and understanding it makes exams easy!

We can take a chunk of the tangent plane, and map it onto the unit square, notice that the unit square has area 1 and this is a linear transform, so you say "at (x,y) 1 unit of area is worth $\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|$ units of surface, so we can say $dxdy$ units of area is worth $dxdy\|\frac{\partial f}{\partial x}\times\frac{\partial f}{\partial y}\|$ units of surface area, we can integrate over x,y already (this is our region, or range, R)
I've tried to draw part of the surface of a small ball, you can see how the parallelagram who's surface area is worth 1 unit of region area (at x,y) is a crappy approximation to the ball, however take a small chunk of it (I've written $dS$ I should have written $\delta A$) the small chunk that maps to ($\delta\sigma$ (I should have written this, I wrote $d\sigma)) is a good approximation. Please do tell me what you think, also I'll work out the exact answers (without working - just so you can check) after dinner, I've not forgotten, I love being able to check I am right, so I promise to give you exact answers after dinner! I'd love to know what you think, I want to teach and would love feedback on this "lecture" length answer! As promised, solutions I did these last night but I wanted to wait for you to read this first, here is an answer in full. As for the second one, I'll ask a friend today but I'm having a difficult time integrating$\sqrt{\sin^2(x)+1}$and this is backed up by its shape, I suspect it might be a special/nice case because of the$\pi$in the integral limits, but I'll keep you posted. • Well I took those notations right from an accepted answer on a similar question about surface integrals, I wonder how this might be wrong. And beforehand I was told my notation was hard to read, is there any real problem with it being wrong? I think you guys (maybe America vs. Europe) might just have another default notation, but if it means the same there is no problem with any of those notations is there? Also I found it quite offensive that you say I have no idea what I'm talking about. In my only other question about this topic I've made a simple mistake forgetting a square root. – L1me Dec 19 '14 at 12:02 • The one real problem I'm having is getting a correct representation of F to insert into the formula, I already know all of the things you said, maybe it's of use for future readers, but that did not really help me personally. I'll update the question to make this clearer and tidy things up a bit. – L1me Dec 19 '14 at 12:05 • @Alec Teal, there were a few word choices that maybe weren't the best as far as presentation courtesy (any 5 year old can tell you). I'm confident you aren't trying to be rude, it's just hard to tell tone on the internet, so FYI for the future. Mathematical content looked good, so I +1'd it anyway. It's a good post, I like it. :) – FundThmCalculus Dec 19 '14 at 12:44 • @L1me sorry I just assumed it was implicit, H (our surface) is a set of points given by a function$F:\mathbb{R}^2\rightarrow H$, like$F(u,v)=(u,v,f(u,v))$, so to take an$f:\mathbb{R}^2\rightarrow\mathbb{R}$to it's surface you have$F(u,v)=(u,v,f(u,v))$- I did work out the answers but$\sqrt{\sin^2(x)+1}$has no nice solution, so I'm working on it – Alec Teal Dec 19 '14 at 12:48 • @FundThmCalculus I meant it as a way of saying "This is the natural definition of area" because one might ask "why do we define it in terms of Cartesian, why do we turn dxdy into rdrd$\theta\$, why not start in polar, the answer: we define area naturally on the Cartesian plane" - can you think of a better way of saying it? – Alec Teal Dec 19 '14 at 13:02