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Surface Integral Help

by EngageEngage
Tags: integral, surface
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EngageEngage
#1
Mar20-08, 08:38 PM
P: 208
1. The problem statement, all variables and given/known data

[tex]\int\int{\frac {x}{\sqrt {1+4\,{x}^{2}+4\,{y}^{2}}}}dS[/tex]
Where S is the parabaloid [tex] z = 25 - x^{2} -y^{2}[/tex] that lies within the cylinder [tex]x^{2}+(y-1)^{2}=1[/tex]

3. The attempt at a solution

First i use the following:

[tex]{\it dS}=\sqrt {1+{\frac {{{\it df}}^{2}}{{{\it dx}}^{2}}}+{\frac {{{
\it df}}^{2}{\it }}{{{\it dy}}^{2}}}}dA[/tex]

(**The above derivatives are not second derivatives, it should be each derivative squared)
to find

[tex]{\it dS}=\sqrt {1+4\,{x}^{2}+4\,{y}^{2}}dA[/tex]

my integral then simplifies to the following, using polar coordinates and the following parametric equations
[tex]x = cos(\theta),and,
y = sin(\theta) + 1, and, dA = rdrd\theta[/tex]

[tex]\int\int cos(\theta) r dr d\theta[/tex]

This is where my trouble starts -- the integral is easy to evaluate, but I don't know how to set up my boundaries so that I am in fact integrating around a cylinder that has been shifted up in the xy plane.
The limits I would use (but which i do not think are right) are as follows:
[tex]2\pi\geq\theta\geq0[/tex]
[tex]1\geq r \geq 0[/tex]
once again, these limits do not take into account the translation of the cylinder on the xy plane. If anyone oculd please tell me where I went wrong or what I should do I would appreciate it greatly.
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tiny-tim
#2
Mar21-08, 05:31 AM
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Quote Quote by EngageEngage View Post
but I don't know how to set up my boundaries so that I am in fact integrating around a cylinder that has been shifted up in the xy plane.
Hi EngageEngage!

You should not be "integrating around a cylinder".

Think of it this way:

Each horizontal slice cuts the parboloid in a circle, and cuts the cylinder in another circle.

You only want the bit of the first circle which is inside the second circle.

My suggestion is to calculate the surface area of each slice, and then integrate over z from the minimum to the maximum values for which the slice is non-zero.
EngageEngage
#3
Mar21-08, 11:31 AM
P: 208
Yeah, i messed that up in the first post, but i think i understand the concept haha. so do you suggest I do this in cylindrical coordinates, where my dA is [tex]rdzd \theta [/tex]? If i did this i could substitute the r in my integral for [tex]sqrt(25 - z)[/tex] and integrate with respect to z and theta. Although I'm still not quite sure how I would incorporate the the translation of the cylinder here. Thanks for the help!

tiny-tim
#4
Mar21-08, 11:50 AM
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Surface Integral Help

Hi EngageEngage!

You're making this too complicated.

Each slice is an arc of a circle (with a slope, of course).

You don't need to integrate to find the length of that arc - it's radius times angle!

Just find the arc-angle of each slice, multiply by the radius of the paraboloid at that height, and something for the slope - that gives you a function of z.

Then integrate over z.

(You have drawn a cross-section of the cylinder and paraboloid at a typical height z, to help you see what's happening, haven't you??)
EngageEngage
#5
Mar21-08, 11:56 AM
P: 208
Actually, if i expand this:
[tex]
x^{2}+(y-1)^{2}=1
[/tex]
to get:
[tex]
x^{2}+y^{2}=2y
r^{2} = 2rsin(\theta)
r = 2sin(\theta)
[/tex]
Which implies the following boundaries:
[tex]
0\leq r \leq 2sin(\theta)
0\leq \theta \leq 2\pi
[/tex]

Though, when i do this my integral reduces to zero which doesn't sound right:
[tex]
\int^{2\pi}_{0}\int^{2sin(\theta)}_{0} cos(\theta)rdrd\theta.......2\int^{2\pi}_{0}cos(\theta)sin^{2}(\theta)d \theta = 0[/tex]
when using simple subsitutions
EngageEngage
#6
Mar21-08, 12:09 PM
P: 208
[tex] r \theta[/tex] will be the arc lenght, and [tex] r = sqrt(25 - z) [/tex]. Im not quite sure what you mean by using the slope though. I've had the picture drawn and thats why i'm not quite sure how this will work because there will always be part of the parabaloid outside of the cylinder. Wont me integrating along z with just the arclength like this include part of the parabaloid thats is not actually within the cylinder? once again, thank you for the help!
tiny-tim
#7
Mar21-08, 01:36 PM
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The slope is the length of the curve between z and z + dz. The paraboloid isn't vertical (like a cylinder), so the length won't be dz, will it?

I can't work out what picture you're working with.

You should have two intersecting circles.

Theta is the arc-angle of the paraboloid that's inside the cylinder. So r x theta x slope = surface area of that slice of the paraboloid inside the cylinder.

Is that your theta?

If so, why are you worried that integrating along z wil include part of the parabaloid that's not inside cylinder?


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