Resolution of a difficult Integral

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Homework Help Overview

The discussion revolves around solving a complex triple integral with specific limits of integration. The integral involves variables x, y, and z, and the participants are exploring different methods to approach the problem, including potential substitutions and coordinate transformations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the possibility of using substitutions, particularly variable changes and trigonometric substitutions, to simplify the integral. There is mention of cylindrical coordinates as a helpful approach, and questions arise regarding how to adjust the limits of integration when changing coordinates. Some participants express uncertainty about the notation used in the discussion.

Discussion Status

The conversation is active, with various suggestions being made regarding the use of cylindrical coordinates and the need for sketches to visualize the limits. Participants are engaging with each other's ideas, but there is no explicit consensus on the best approach yet.

Contextual Notes

Some participants question the setup of the integral and suggest reconsidering the method of dividing the volume for integration. There is also a focus on understanding the notation and terminology used in the discussion.

thonwer
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Homework Statement


I need to solve this integral

[itex]\int[/itex][itex]\int[/itex][itex]\int[/itex]dxdzdy+[itex]\int[/itex][itex]\int[/itex][itex]\int[/itex]dxdzdy

First limits are:

-[itex]\sqrt{z^{2}-y^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{z^{2}-y^{2}}[/itex]
-y[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{y}{2}[/itex]
-[itex]\frac{2}{3}[/itex][itex]\leq[/itex]y[itex]\leq[/itex]0

Second limits are:

-[itex]\sqrt{z^{2}-y^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{z^{2}-y^{2}}[/itex]
y[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{y}{2}[/itex]
0[itex]\leq[/itex]y[itex]\leq[/itex]2

The Attempt at a Solution



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I think it's getting too difficult to solve, is there an easier way to solve it?
 
Last edited:
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Why not try using substitutions?
 
What do you mean with substitutions? Changing variables?
 
What do you mean with substitutions? Changing variables?
Trig substitutions are popular.
The idea is to get rid of that square-root sign when you go to evaluate the dz part.
 
cylindrical coordinates look helpful

$$x\rightarrow r \, \cos(\theta)\\
y\rightarrow r \, \sin(\theta)\\
z\rightarrow z$$
 
If I use cylindrical coordinates x=ρcos(Θ) ; y=ρsin(Θ) ; z=z

I get the integral of ρ but how do I change the limits?
 
I get the integral of ρ but how do I change the limits?
i.e. if a<x<b, then the limits of ρ would be (a/cosθ)<ρ<(b/cosθ)
... and your next integral will be wrt θ

A sketch of the cartesian limits will help you work out what the cylindrical limits should be.
 
Simon Bridge said:
i.e. if a<x<b, then the limits of ρ would be (a/cosθ)<ρ<(b/cosθ)
... and your next integral will be wrt θ

A sketch of the cartesian limits will help you work out what the cylindrical limits should be.

Excuse me, what is wrt?
 
I get:
[itex]\frac{-2}{3}[/itex][itex]\leq[/itex]ρsin(Θ)[itex]\leq[/itex]0

-[itex]\frac{2}{3sin(Θ)}[/itex][itex]\leqρ[/itex][itex]\leq[/itex]0

-ρsin(Θ)[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{ρsin(Θ)}{2}[/itex]

Θ:[0,2π]

And in the other:

0[itex]\leq[/itex]ρsin(Θ)[itex]\leq[/itex]2

0[itex]\leq[/itex]ρ[itex]\leq[/itex][itex]\frac{2}{sin(Θ)}[/itex]

ρsin(Θ)[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{ρsin(Θ)}{2}[/itex]

Is that correct?
 
  • #10
thonwer said:
Excuse me, what is wrt?

wrt = 'with respect to'
 
  • #11
thonwer said:
Excuse me, what is wrt?

"with respect to"

It's just a shorthand notation that we use. Similar to WLOG (without loss of generality) and iff (if and only if).
 
  • #12
Excuse me, what is wrt?
... and everybody leaps into help with the easy question ;)
http://en.wiktionary.org/wiki/WRT
http://www.internetslang.com/WRT-meaning-definition.asp

Is that correct?
- that would require me to do the problem, the idea is that you do it.
I'll look in more detail a bit later - meantime:
It's not always a blind substitution. Which order do you want to do the integration? Like this:
$$\int \int \int \text{d}\theta \text{d}z \rho \text{d}\rho$$

Note: if x^2 = z^2-y^2 then isn't y playing the role of a radius? Or is it z?
Have you sketched out the region of integration?

Are you unfamiliar with integrating in polar coordinates?
http://www.math.dartmouth.edu/~m13w12/notes/class7.pdf

Note: if this is an integral you constructed yourself rather than being handed to you in this form - you should reconsider the way you are dividing the volume up instead. In fact - that may be a good idea anyway.
 
Last edited:

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