# Triple integral for cone in cylindrical coordinates.

1. Jun 22, 2012

### Tseliottt

1. The problem statement, all variables and given/known data
Find limits of integration for volume of upside down cone with vertex on origin and base at z=1/sqrt(2). Angle at vertex is pi/2. Do this in cylindrical coordinates.

2. Relevant equations
None.

3. The attempt at a solution
My inner integral conflicts with the books solution. So in my triple integral, the outsides are right: 0<theta<2pi ; 0<r<1/sqrt(2)

But my inner integral is r<z<1/sqrt(2) and the book says its r<z<1.

Where does this 1 come from? I thought the max height was 1/sqrt(2)?

Sorry if my formats messed up. Typing this on my phone.

2. Jun 22, 2012

### hawaiifiver

3. Jun 22, 2012

### SammyS

Staff Emeritus
It looks to me like you did the limits of integration correctly.

4. Jun 22, 2012

### Tseliottt

Yea. After looking at that link, I think the books solution is wrong. Thanks. I spent like an hour trying to figure out what I did wrong. Ill ask my professor to make sure.

5. Jun 22, 2012

### hawaiifiver

I've followed my link above and I think your limits for z are r < z < 1/√2

My triple integral is

int (0 to 1/√2) r dr int (0 to 2∏) dθ int (r to 1/√2) dz