Resolution of a difficult Integral

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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?
 
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What do you mean with substitutions? Changing variables?
 
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?
 
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).
 
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.
 
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