# Resolution of a difficult Integral

1. Apr 27, 2014

### thonwer

1. The problem statement, all variables and given/known data
I need to solve this integral

$\int$$\int$$\int$dxdzdy+$\int$$\int$$\int$dxdzdy

First limits are:

-$\sqrt{z^{2}-y^{2}}$$\leq$x$\leq$$\sqrt{z^{2}-y^{2}}$
-y$\leq$z$\leq$1+$\frac{y}{2}$
-$\frac{2}{3}$$\leq$y$\leq$0

Second limits are:

-$\sqrt{z^{2}-y^{2}}$$\leq$x$\leq$$\sqrt{z^{2}-y^{2}}$
y$\leq$z$\leq$1+$\frac{y}{2}$
0$\leq$y$\leq$2

3. The attempt at a solution

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

Last edited: Apr 27, 2014
2. Apr 27, 2014

### Simon Bridge

Why not try using substitutions?

3. Apr 27, 2014

### thonwer

What do you mean with substitutions? Changing variables?

4. Apr 27, 2014

### Simon Bridge

Trig substitutions are popular.
The idea is to get rid of that square-root sign when you go to evaluate the dz part.

5. Apr 27, 2014

### lurflurf

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

6. Apr 27, 2014

### 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?

7. Apr 27, 2014

### Simon Bridge

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.

8. Apr 27, 2014

### thonwer

Excuse me, what is wrt?

9. Apr 27, 2014

### thonwer

I get:
$\frac{-2}{3}$$\leq$ρsin(Θ)$\leq$0

-$\frac{2}{3sin(Θ)}$$\leqρ$$\leq$0

-ρsin(Θ)$\leq$z$\leq$1+$\frac{ρsin(Θ)}{2}$

Θ:[0,2π]

And in the other:

0$\leq$ρsin(Θ)$\leq$2

0$\leq$ρ$\leq$$\frac{2}{sin(Θ)}$

ρsin(Θ)$\leq$z$\leq$1+$\frac{ρsin(Θ)}{2}$

Is that correct?

10. Apr 27, 2014

### SteamKing

Staff Emeritus
wrt = 'with respect to'

11. Apr 27, 2014

### scurty

"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. Apr 27, 2014

### Simon Bridge

... and everybody leaps in to help with the easy question ;)
http://en.wiktionary.org/wiki/WRT
http://www.internetslang.com/WRT-meaning-definition.asp

- 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: Apr 27, 2014