1. ## Triple Integral Limits

I don't understand how the limits of integration have been obtained in the following worked example:

So, how did they get the limits $\displaystyle \int^{2 \pi}_0 \int^1_0 \int^1_r$?

I know that for the double integral they've used polar coordinates. The $\displaystyle \int^{2 \pi}_0 \int^1_0$ represents the unit circle. I tried to sketch the region but I'm not sure how the solid is supposed to look like...

Especially I don't know how they got limits "$\displaystyle \int^1_r$"! Does $\displaystyle \sqrt{x^2 + y^2}$ represent a hemisphere?

I really appreciate it if anyone could explain these to me.

2. Originally Posted by demode
[snip]

Especially I don't know how they got limits "$\displaystyle \int^1_r$"! Does $\displaystyle \sqrt{x^2 + y^2}$ represent a hemisphere?
Yes it does. The whole region is, in fact, a cylinder with a hemisphere removed, so that there are sharp edges along the unit circle in the xy plane and there is nothing on the z axis.

Actually, I just figured out a great way to think of this without pictures!

If your only concern is the z bounds, they're actually spelled out in that first condition. For all (x,y) z goes between 1 and sqrt(r^2) = r.

3. Originally Posted by Turiski
Yes it does. The whole region is, in fact, a cylinder with a hemisphere removed, so that there are sharp edges along the unit circle in the xy plane and there is nothing on the z axis.

Actually, I just figured out a great way to think of this without pictures!

If your only concern is the z bounds, they're actually spelled out in that first condition. For all (x,y) z goes between 1 and sqrt(r^2) = r.
$\displaystyle \sqrt{ x^2 + y^2 }$ does NOT represent a hemisphere. It represents a cone.

The reason for the bounds of

$\displaystyle \int_r^1 dz$

is because they are given by $\displaystyle \sqrt{ x^2 + y^2 } \le z \le 1$

In cylindrical co-ordinates $\displaystyle x^2 + y^2 = r^2$ so

$\displaystyle \sqrt{ x^2 + y^2 } \le z \le 1 \to r \le z \le 1$