# Math Help - Area, C of G of a paraboloid section? ...

1. ## Area, C of G of a paraboloid section? ...

Hi All, this is MathWimp here. I posted these queries in less appropriate thread categories here, before I figured out the website structure a bit more - mind you I have an excuse - the welcome E-Mail from the forum just invited me to post a homework question. I believe this to be a bit more involved than most homework problems:

I have an equation for a surface (ie: in 3D). Congratulations are unnecessary ;-) . Wonder if anyone can help with some problems related to it:

In an engineering application, there is a surface:

$z = {\frac{1}{4f}}(x^2+y^2)$

The surface is cut by 4 planes, all "parallel" to the z axis , 2 of them parallel to the y-z plane, 2 parallel to the x-z plane. The planes are $X_{i}$, $X_{f}$ and $Y_{i}$, $Y_{f}$; all 4 to be regarded as constants of the problem (which with ${f}$ makes 5 constants).

(ie: the projection of the cut surface on the x-y plane is a rectangle sides $X_{f}-X_{i}$ parallel to the x axis; and $Y_{f}-Y_{i}$ parallel to the y axis)

What is the area of the section of the surface cut off by these planes?

Note: I found out the hard way that the substitutions $u = {\frac{x}{2f}}$ and $v = {\frac{y}{2f}}$ will save lots of pencil-lead ;-).

I'm hoping for an analytic solution, but I can't find it, and I appreciate there may be no luck that one even exists.

If not, could you suggest a best or at least reasonable approach so that a "user" (say), would have a reasonably short paradigm so that given ${f}$, and the planes $X_{i}$, $X_{f}$, $Y_{i}$, $Y_{f}$, the area of the cut-off section can be found?

And would it be possible to give error bounds on a non-analytical method? A TI 85 program, anybody?

In addition, though I've almost given up this, where is the C of G (ie: the x, y, z thereof) of the cut-off surface? (Assuming it composed of continuous material with uniform area density).

There are a couple of approximate approaches that have occurred to me, but I can't figure a way to tell how approximate they'd be (rather how accurate), so if it could be done more analytically, that'd be great.

Thanks.
Dennis R.

2. Given a surface defined by z=g(x,y) for (x,y) in a region R of the xy plane, then the surface area is given by:
$\iint_R\sqrt{1+\left(\frac{\partial{g}}{\partial{x }}\right)^2+\left(\frac{\partial{g}}{\partial{y}}\ right)^2}dA$.

For your problem, R is the rectangle bound by $x=X_i$, $x=X_f$, $y=Y_i$, and $y=Y_f$; while $\frac{\partial{g}}{\partial{x}}=\frac{x}{2f}$ and $\frac{\partial{g}}{\partial{y}}=\frac{y}{2f}$, and so you have $S=\int_{X_i}^{X_f}\int_{Y_i}^{Y_f}\sqrt{1+\frac{x^ 2}{4f^2}+\frac{y^2}{4f^2}}\,dy\,dx$. Like you said, substitute $u=\frac{x}{2f}$ and $v=\frac{y}{2f}$. Then we get:
$S=4f^2\int_{X_i/2f}^{X_f/2f}\int_{Y_i/2f}^{Y_f/2f}\sqrt{1+u^2+v^2}\,dv\,du$.

Now, a table of integrals will tell you that $\int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$ (see #8 here, for example). Replacing x with v and $a^2$ with $1+u^2$, the above applies to our inner integral, and we have:
$S=4f^2\int_{X_i/2f}^{X_f/2f}\left[\frac{v\sqrt{1+u^2+v^2}}{2}+\frac{1+u^2}{2}\ln(v+\ sqrt{1+u^2+v^2})\right]_{v=Y_i/2f}^{Y_f/2f}\,du$
$S=2f^2\int_{X_i/2f}^{X_f/2f}\left(\frac{Y_f\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}-Y_i\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}}{2f}\right.$
$\;\;\;\;\left.+(1+u^2)\left(\ln\left(\frac{Y_f}{2f }+\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}\right)-\ln\left(\frac{Y_i}{2f}+\sqrt{1+\frac{Y_i^2}{4f^2} +u^2}\right)\right)\right)\,du$.

Now this is a bit more complicated, but the integrand may be broken up into four terms, each of which may be integrated. For the terms $\frac{Y_f}{2f}\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}$ and $-\frac{Y_i}{2f}\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}$, we can use the form $\int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$, this time with x replaced by u, and with either $1+\frac{Y_f^2}{4f^2}$ or $1+\frac{Y_i^2}{4f^2}$ in place of $a^2$.

For the terms $(1+u^2)\ln\left(\frac{Y_f}{2f}+\sqrt{1+\frac{Y_f^2 }{4f^2}+u^2}\right)$ and $-(1+u^2)\ln\left(\frac{Y_i}{2f}+\sqrt{1+\frac{Y_i^2 }{4f^2}+u^2}\right)$, well, those may take more creative substitutions, but I'm pretty sure they can be done analytically as well. Some messy algebra later, you'll have your analytic answer.

--Kevin C.

3. Originally Posted by TwistedOne151
Given a surface defined by z=g(x,y) for (x,y) in a region R of the xy plane, then the surface area is given by:
$\iint_R\sqrt{1+\left(\frac{\partial{g}}{\partial{x }}\right)^2+\left(\frac{\partial{g}}{\partial{y}}\ right)^2}dA$.

For your problem, R is the rectangle bound by $x=X_i$, $x=X_f$, $y=Y_i$, and $y=Y_f$; while $\frac{\partial{g}}{\partial{x}}=\frac{x}{2f}$ and $\frac{\partial{g}}{\partial{y}}=\frac{y}{2f}$, and so you have $S=\int_{X_i}^{X_f}\int_{Y_i}^{Y_f}\sqrt{1+\frac{x^ 2}{4f^2}+\frac{y^2}{4f^2}}\,dy\,dx$. Like you said, substitute $u=\frac{x}{2f}$ and $v=\frac{y}{2f}$. Then we get:
$S=4f^2\int_{X_i/2f}^{X_f/2f}\int_{Y_i/2f}^{Y_f/2f}\sqrt{1+u^2+v^2}\,dv\,du$.

Now, a table of integrals will tell you that $\int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$ (see #8 here, for example). Replacing x with v and $a^2$ with $1+u^2$, the above applies to our inner integral, and we have:
$S=4f^2\int_{X_i/2f}^{X_f/2f}\left[\frac{v\sqrt{1+u^2+v^2}}{2}+\frac{1+u^2}{2}\ln(v+\ sqrt{1+u^2+v^2})\right]_{v=Y_i/2f}^{Y_f/2f}\,du$
$S=2f^2\int_{X_i/2f}^{X_f/2f}\left(\frac{Y_f\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}-Y_i\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}}{2f}+(1+u^2)\le ft(\ln\left(\frac{Y_f}{2f}+sqrt{1+\frac{Y_f^2}{4f^ 2}+u^2}\right)-\ln\left(\frac{Y_i}{2f}+sqrt{1+\frac{Y_i^2}{4f^2}+ u^2}\right)\right)\right)\,du$.

Now this is a bit more complicated, but the integrand may be broken up into four terms, each of which may be integrated. For the terms $\frac{Y_f}{2f}\sqrt{1+\frac{Y_f^2}{4f^2}+u^2}$ and $-\frac{Y_i}{2f}\sqrt{1+\frac{Y_i^2}{4f^2}+u^2}$, we can use the form $\int\sqrt{a^2+x^2}\,dx=\frac{x\sqrt{a^2+x^2}}{2}+\ frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})$, this time with x replaced by u, and with either $1+\frac{Y_f^2}{4f^2}$ or $1+\frac{Y_i^2}{4f^2}$ in place of $a^2$.

For the terms $(1+u^2)\ln\left(\frac{Y_f}{2f}+sqrt{1+\frac{Y_f^2} {4f^2}+u^2}\right)$ and $-(1+u^2)\ln\left(\frac{Y_i}{2f}+sqrt{1+\frac{Y_i^2} {4f^2}+u^2}\right)$, well, those may take more creative substitutions, but I'm pretty sure they can be done analytically as well. Some messy algebra later, you'll have your analytic answer.

--Kevin C.
You could try to make thing a little easier by converting to polar...if possible.

Let $u^2+v^2=r^2$

Thus,
$S=4f^2\int_{X_i/2f}^{X_f/2f}\int_{Y_i/2f}^{Y_f/2f}\sqrt{1+u^2+v^2}\,dv\,du\implies S=4f^2\iint\limits_{R_{r\theta}}\sqrt{1+r^2}r\,dA_ {r\theta}$...

Now, what would the limits of integration be? Does this even work out? I just threw this out there...to see if would simplify things.

--Chris

4. The key to the latter two portions in integration by parts. They're both of the form $(1+u^2)\ln(a+\sqrt{1+a^2+u^2})$, with a being either $\frac{Y_f}{2f}$ or $\frac{Y_i}{2f}$. Now, using integration by parts,
$\int(1+u^2)\ln(a+\sqrt{1+a^2+u^2})\,du$ $=(\frac{u^3}{3}+u)\ln(a+\sqrt{1+a^2+u^2})-\int\frac{u^2(u^2+3)\,du}{3(a+\sqrt{1+a^2+u^2})\sq rt{1+a^2+u^2}}$.
With a bit of work, the latter integral can be done: it's
$\frac{u^3}{9}+\frac{2u}{3}-\frac{au\sqrt{1+a^2+u^2}}{6}-\tfrac{2}{3}\arctan{u}$
$\,\,+\tfrac{2}{3}\arctan\frac{au}{\sqrt{1+a^2+u^2} }+\frac{a(a^2-3)}{6}\ln(u+\sqrt{1+a^2+u^2})+C$.

Giving us that
$\int(1+u^2)\ln(a+\sqrt{1+a^2+u^2})\,du$
$\,\,=\frac{u(u^2+3)}{3}\ln(a+\sqrt{1+a^2+u^2})-\frac{x(x^2+6)}{9}+\frac{au\sqrt{1+a^2+u^2}}{6}$
$\,\,+\tfrac{2}{3}\arctan{u}-\tfrac{2}{3}\arctan\frac{au}{\sqrt{1+a^2+u^2}}-\frac{a(a^2-3)}{6}\ln(u+\sqrt{1+a^2+u^2})+C$

Plug in the right values for a, then plug in the limits for u and subtract; do a whole lot of messy algebra, and you have the answer.

--Kevin C.

5. ## STRIP TEASE!*

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Completely solved! (subject to your approval ;-) ie: any corrections appreciated). ... And ... it can all be done on a SINGLE page ... boo, hoo ...

I never got familiar with the way of writing equations properly here so I refer you to page (ie: slide) #63 (updated, was #61) et seq. in the following Powerpoint document:

http://home.att.net/~gijxixj/Paraboloid_Stuff/Solar_MainlyParaboloidConstrLatest.ppt
Shortforms: http://preview.tinyurl.com/6eopge, http://tinyurl.com/6eopge & http://SPMC.notlong.com

NOTE: this page # can change at any time as I add/delete pages from this document, in case it does change the title of the first page in the applicable sequence is:

"Appendix 2 (1)
COG/Area of "Awkward" Rectangular Cutout from a Paraboloid"

So that an "in-document" search for "Appendix 2 (1)" or may be just "2 (1)" should find it quickly in case the page # changes.

The key is in not making the elemental areas of infinitesimal rectangular form dudv (or dxdy) but in making them of the form of STRIPS of, for example, length Su and width dv: Sudv (Sxdy), Svdu (Sydx) and then summing (integrating) these, and so on. For Zcog ("normalised" as Wcog), you pretty much have to go back to "rectangular" standard "non-strip" elemental areas - but the answer just kind of falls out from previous experience solving Su, Sv & Ucog & Vcog!

I actually figured this out a couple or so days following TwistedOne151's kind replies, unfortuneately had other pressures to bear. I'd gone down the same path - I'm sure it'll lead to the right answers somewhere at the end of a nightmarish long path - but equally certainly unlikely to be in the same extremely simple forms that the method in the Powerpoint document leads to

Moreover the answers given by the STRIP method are unambiguously applicable to any rectangular projection whether involving just +ve or -ve axes or both. One of the reasons I describe the Nightmare method as just that, is that if you go down it far enough, the question of signs becomes unutterably confused.

All the best,
Dennis Revell

P.S: Please feel free to be so kind as to post links to the Powerpoint document wheresover you may feel appropriate (but not to mathworld.wolfram.com - see intro/explanation on page/(slide #63): "Appendix 2 (1)"). Thank you. Any fellow wannabe tree-huggers out there? ;-)

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*Those strips TEASED the Hell out of me!
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