Triple integral in cylindrical coordinates

  • Thread starter Locoism
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  • #1
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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 [itex] \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt[/itex]

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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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 [itex] \int_0^{2π} \int_0^{\sqrt{2}} \int_{r^2}^{\sqrt{2-r^2}} r dzdrdt[/itex]

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).
 
  • #3
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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???
 
  • #4
Dick
Science Advisor
Homework Helper
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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.
 
  • #5
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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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