# Maths question concerning three dimensional integration

Discussion in 'General Discussion' started by Drawoh Tesremos, 12 May 2006.

1. Drawoh Tesremos

# Location: UK

A short while ago I was asked to help someone with a three dimensional integration problem.

Specifically, we have a fuction F(x,y) = 16 - x - y^2, and the requirement is,
(a) determine the volume under a defined area, and
(b) determine the surface area within the same limits.

I have no problem with (a), it's a double integral of the function F, and the two sets of limits of integration are relatively easy to work out.

However, with part (b) I just cannot remember what it is that you have to integrate. The limits of integration are ok; they are the same as in part (a).

Can anyone here help? I recall doing this at uni, some 45 years ago, but haven't been able to locate my old text books.

2. Drawoh Tesremos

# Location: UK

For completeness, though not necessary to be able to answer the question I'm asking in the previous post, the defined area is the triangle contained by the three lines
y = 0
x = 2
x = y

3. Stripe

# Posts: 221

I'm in the final year of a maths degree and supposedly meant to be adept with vector calculus. i can't remember for the life of me how to do this though. it'll bug me if i don't do it so i'll get an answer by the end of the night.

ooh friday night with calculus. i am a lucky boy.

4. alexs

# Location: Kent

A long time ago since I last set eyes on such a problem, but I seem to remember that the surface area is given by the integral of sqrt(1 + (dF/dx)^2 + (dF/dy)^2) over the area element dx dy.

5. Stripe

# Posts: 221

ok, here's how i was taught to do it.

f(x,y) = 16 - x - y^2

is the same as

z = 16 - x - y^2

=>

z + x + y^2 -16 = 0

you have to put this into cylindrical polar coordinates with these formulae:

x = RcosA, y = RsinA, z = z (which in this case will be z = 16 - RcosA - R^2 * sin^2{A})

you work out t_R = (dx/dR, dy/dR, dz/dR) = (cosA, sinA, cosA - 2Rsin^2{A})

similarly t_A = (-RsinA, RcosA...

and then some more stuff. i really don't know how to do it anymore. sorry.

but basically, you parameterise the variables and integrate 1 over the surface.

6. Drawoh Tesremos

# Location: UK

Thanks, both of you. You've shown two completely different methods, which is not unusual, as many maths problems can be approached in more than one way. I'll look further at both of your suggestions. Thanks. 7. Stripe

# Posts: 221

sorry i couldn't be of more help. it's really embarassing. i think the other way would probably be better. i think the way i said is only any good if you have x^2 and y^2 as when paramterised, their sum cancels to R or R^2 (i forget which).

sorry again, good luck with it.

8. Ricochet J

# Posts: 12,888

Woah! You can apply Differentiation and Integration to 3 Demensional Graphs? What has the world come to? (Will I learn this in A Level Maths? )

9. spirit

# Posts: 1,885

hehe you dont even learn this crazy stuff in fmaths the wierdest thing i learnt was some dodgy thign with vectors in physics which apparently differentiate or somehting we used in a proof (as you can see i neither cared nor had any idea)

10. Stripe

# Posts: 221

nah, you won't learn that at A-level. but in fact you can integrate into the 4th,5th...nth dimension.

it's weird, at A-level they teach you integration and differentiation but they don't really tell you all the awesome stuff you can do with them.

if you're taking a-level maths, hopefull you'll do P4. there's this result you derive right at the end of it which is like, the best equation ever!!!!

# Location: Cambridge

Who knows if you've solved this or not, but I thought I'd throw in my contribution. Using some (pretty dodgy) geometric reasoning I've convinced myself that alex's element of area is correct, so you need to integrate

sqrt(2+4y^2) dx dy

You can then write y=sinh(z)/sqrt(2), and dy=cosh(z)/sqrt(2) dz, and plug those in:

sqrt(2+2sinh(z)^2) cosh(z)/sqrt(2) dx dz = cosh(z)^2 dx dz

And finally use 2cosh(z)^2 = 1+cosh(2z) to get your answer... I think that works!

12. Psiko

# Posts: 802

Normally, when integrating over a surface, we parametrise the surface. In this case that could be Phi(u,v) = (uv/2, v, 16 - uv/2 - v^2). Then, to integrate a function over the surface, multiply by the surface area element, which is the cross-product of dPhi(u,v)/du x dPhi(u,v))/dv (where the d/du, d/dv are partial derivatives.) To find the surface area we integrate 1 over the surface.

I found that the surface area element was Sqrt(1/2 v^2 + 1), and the parametrisation space is 0<=u<=2, 0<=v<=2.

I'm fairly sure this is right, but not certain. I ended up with 2Sqrt3 + Sqrt2 Arcsinh(Sqrt2).

# Location: Cambridge

Neat choice of parameterisation - I would just have used R=(x,y,F(x,y)) and worked with awkward limits. 14. Psiko

# Posts: 802

Cheers. I'm doing Part IA this year, so it's good revision for vector calculus. I hate working through tripos papers. It's so boring! Are you a mathmo? I've forgotten.

# Location: Cambridge

A Trinity mathmo, no less - fourth year. What is it with all of the mathmos all over the place on this board at the moment?!

Like I said to someone in another thread, feel free to add me on MSN if you want any help. Don't ask me about probability or analysis, but anything else is fair game!

16. Drawoh Tesremos

# Location: UK

Thanks for the further contributions on this. I'm away from home at the moment, but just checked in to see if there was anything further. Before leaving home I passed on the sqrt(1 + (dF/dx)^2 + (dF/dy)^2) suggestion as it seemed more workable than the polar co-ordinate idea. This particular question didn't seem to lend itself to polar co-ordinates due to the region to be integrated being a triangle. Had it been a circle, or similar, I may well have gone with the polar co-cordinates idea.

Thanks for the further comments. I'll look at them when I get back home.