Resolution of a difficult Integral

[thonwer]

Homework Statement

I need to solve this integral

\int\int\intdxdzdy+\int\int\intdxdzdy

First limits are:

-\sqrt{z^{2}-y^{2}}\leqx\leq\sqrt{z^{2}-y^{2}}
-y\leqz\leq1+\frac{y}{2}
-\frac{2}{3}\leqy\leq0

Second limits are:

-\sqrt{z^{2}-y^{2}}\leqx\leq\sqrt{z^{2}-y^{2}}
y\leqz\leq1+\frac{y}{2}
0\leqy\leq2

The Attempt at a Solution

I think it's getting too difficult to solve, is there an easier way to solve it?

 

[Simon Bridge]

Why not try using substitutions?

 

[thonwer]

What do you mean with substitutions? Changing variables?

 

[Simon Bridge]

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.

 

[lurflurf]

cylindrical coordinates look helpful

$$x\rightarrow r \, \cos(\theta)\\
y\rightarrow r \, \sin(\theta)\\
z\rightarrow z$$

 

[thonwer]

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?

 

[Simon Bridge]

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.

 

[thonwer]

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?

 

[thonwer]

I get:
\frac{-2}{3}\leqρsin(Θ)\leq0

-\frac{2}{3sin(Θ)}\leqρ\leq0

-ρsin(Θ)\leqz\leq1+\frac{ρsin(Θ)}{2}

Θ:[0,2π]

And in the other:

0\leqρsin(Θ)\leq2

0\leqρ\leq\frac{2}{sin(Θ)}

ρsin(Θ)\leqz\leq1+\frac{ρsin(Θ)}{2}

Is that correct?

 

[SteamKing]

Excuse me, what is wrt?

wrt = 'with respect to'

 

[scurty]

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).

 

[Simon Bridge]

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.