# Triple integral help

1. May 15, 2012

### sevag00

Hello everyone. I need help regarding putting boundaries in triple integrals using spherical coordinates. The figure attached below is an ice cream cone. I want someone to explain to me how to put the boundaries in the order of dr dz dθ? I had understood this in class but unfortunately i forgot the explanation.

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2. May 15, 2012

### sharks

Hi sevag00
Your question contradicts itself. You want to express the integrals in spherical coordinates, but the boundaries in terms of drdθdz are cylindrical coordinates!

What is the whole question? Are you required to find the volume of the cone including the section of paraboloid on top?

Last edited: May 15, 2012
3. May 15, 2012

### sevag00

lol. I meant the volume in cylindrical coordinates.

4. May 15, 2012

### LCKurtz

There's no reason to use spherical coordinates on this problem. You want cylindrical coordinates. It is best not to integrate the dr variable first; better would be $r\, dz dr d\theta$. The dz limits will be z on the lower surface and z on the upper surface, both expressed in terms of r. Once you have those z values in terms of r, you can set them equal to see what r is at their intersection.

5. May 15, 2012

### sevag00

Yeah, i know it's easy integrating in order dzdrdθ . But the question in the book is asking drdzdθ.

6. May 15, 2012

### HallsofIvy

Staff Emeritus
Well, "$drdzd\theta$" is the hard way! I recommend doing it as two separate integrals. For z from 0 to 1, we have the cone, z= r, so, for each z, r goes from 0 to z. For z= 1 to 2, we have the paraboloid, $z= 2- r^2$ so, for each z, $r^2=2-z$ and r, being positive, goes from 0 to $\sqrt{2- z}$.

Last edited: May 15, 2012
7. May 15, 2012

### sharks

The answer will be the sum of two triple integrals; volume of the cone + volume of paraboloid.

8. May 15, 2012

### sevag00

Is this in order dzdrdθ?

9. May 15, 2012

### sharks

For the second triple integral, i think it should be from z=1 to $z=2-r^2$
The radius of the projection is r=1.

10. May 15, 2012

### LCKurtz

Is what in that order? Use the quote button to reply to a message so we know what you are replying to.

11. May 15, 2012

### sevag00

Actually, i know the answer. But i didn't get how r is from 0 to z + 0 to sqrt(2-z). Answer is attached.

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12. May 15, 2012

### LCKurtz

Doing dr first, on the top part (the paraboloid) r goes from 0 to the r on the paraboloid. The equation of the paraboloid is $z = 2-r^2$. Just solve that for $r$ to get $\sqrt{2-z}$.

13. May 15, 2012

### sevag00

Isn't the dr supposed to be from the x-axis to y-axis?

14. May 15, 2012

### LCKurtz

That doesn't make any sense. Think of starting on the z axis up at the level of the paraboloid. r is perpendicular to the z axis and it goes from r = 0 out till it hits the paraboloid. What direction it goes depends on $\theta$.

15. May 15, 2012

### sevag00

Aha. I think i get it.

EDIT: Yes, i get it. Thanks for the help.