Help with limits of integration

[2^Oscar]

Hi guys,

I've been doing past paper questions for an exam and I've gotten stuck with the limits of an integral. We have to evaluate

\int\int\int _{\Omega} \frac{1}{(1+z)^2} dx dy dz

where \Omega = \left\{ (x, y, z) : x^2 + y^2 \leq z^2 \leq 1 - x^2 - y^2, z \geq 0 \right\}

using spherical polar coordinates. My problem is finding the limits for r (we use r, theta, phi in lectures), all I get is as far as this inequality r^2 sin^2 (\theta) \leq r^2 cos^2 (\theta) \leq 1- r^2 sin^2 (\theta) and I'm unsure how to go on after this.

I'm sure I'm missing something blindingly obvious, and I'll be fine once I know the limits, but would someone please explain how to proceed and find the limits for r?Cheers,
Oscar

 

[tiny-tim]

Hi Oscar!

… all I get is as far as this inequality r^2 sin(\theta) \leq r^2 cos(\theta) \leq 1- r^2 sin(\theta)

the right side would be correct if you used ²

the left side is just y ≤ x

the best way to do this is to ask yourself what shape we're talking about …

try putting x^2 + y^2 = \rho^2

now what is the shape? ​

 

[2^Oscar]

Sorry about that I missed the squares out on the trig functions. I've corrected them to the inequality I actually got!

Wouldn't \rho^2 = x^2 + y^2 + z^2 though? I tried rearranging the inequality to use that but couldn't get very far. When I sketched the shape it was x^2 + y^2 coming up and then meeting the same curve coming down from one and forming a kind of egg shape...
Thanks for the reply :)
Oscar

 

[tiny-tim]

We usually use r = √(x² + y² + z²) in spherical coordinates, and ρ = √(x² + y²) in cylindrical coordinates.

Try using ρ here, to see what the shape looks like.

 

[2^Oscar]

We usually use r = √(x² + y² + z²) in spherical coordinates, and ρ = √(x² + y²) in cylindrical coordinates.

Try using ρ here, to see what the shape looks like.

Ahh I see, so in the new shape (a cylinder?) we'd have the limits \rho \leq z \leq \sqrt{1 - \rho^2}?

 

[tiny-tim]

… in the new shape (a cylinder?)

No!

Try it with y = 0. ​