Triple integral in cylindrical coordinates

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Homework Help Overview

The discussion revolves around finding the volume of a solid defined by the equations z = x² + y² and x² + y² + z² = 2, utilizing cylindrical coordinates for the setup of a triple integral.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conversion to cylindrical coordinates, questioning the limits of integration for r and z. There is an exploration of splitting the integral into parts and considering the implications of the chosen limits.

Discussion Status

Participants are actively engaging with the problem, raising questions about the setup of the integral and the limits for r. Some guidance has been offered regarding the correct range for r, but no consensus has been reached on the final approach.

Contextual Notes

There is mention of potential confusion regarding the limits of integration, particularly for r, and the need to determine where the curves intersect. The discussion reflects an ongoing examination of the problem setup without definitive conclusions.

Locoism
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Homework Statement



Find the volume of the solid that lies between

z=x2+y2 and
x2+y2+z2=2

Homework Equations



z=r2
z=√(2-r2)


The Attempt at a Solution



So changing this into cylindrical coordinates, I get

z goes from r2 to √(2-r2)
r goes from 0 to √2
theta goes from 0 to 2π

so we get \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt

I dunno... something feels off about this. I think this way it would take into account the negative area on the outside of the curve... How else could I set up the integral?
 
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Locoism said:

Homework Statement



Find the volume of the solid that lies between

z=x2+y2 and
x2+y2+z2=2

Homework Equations



z=r2
z=√(2-r2)


The Attempt at a Solution



So changing this into cylindrical coordinates, I get

z goes from r2 to √(2-r2)
r goes from 0 to √2
theta goes from 0 to 2π

so we get \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt

I dunno... something feels off about this. I think this way it would take into account the negative area on the outside of the curve... How else could I set up the integral?

I would reconsider your statement that r goes from 0 to sqrt(2).
 
Dick said:
I would reconsider your statement that r goes from 0 to sqrt(2).

How so? I was thinking of splitting it into 2 parts and having it go from 0 to the function in terms of z, but then I would have to integrate r first, and I run into the same problem...

ie: z from r^2 to 1, then from 1 to √(2-r^2), and r from 0 to √z and then from 0 to √(2-z^2)...

Is there a way to write it in terms of theta?
 
Locoism said:
How so? I was thinking of splitting it into 2 parts and having it go from 0 to the function in terms of z, but then I would have to integrate r first, and I run into the same problem...

ie: z from r^2 to 1, then from 1 to √(2-r^2), and r from 0 to √z and then from 0 to √(2-z^2)...

Is there a way to write it in terms of theta?

The integral you have is fine. Except that the range of r is off. r^2 is x^2+y^2, not x^2+y^2+z^2. You need to figure out the value of r where r^2=z and r^2+z^2=2 intersect.
 
Awesome, r from 0 to 1, and z from r^2 to √(2-r^2).
Thank you so much, I see it perfectly now.
 

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