Find volume between parabloid and parabolic sylinder

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SUMMARY

The discussion focuses on calculating the volume of the region bounded by the equations x = y² + z² and x = 2 - y². Participants emphasize the necessity of setting up the correct limits of integration for a triple integral. The suggested method involves slicing the volume with respect to the x-axis and then performing a double integration with respect to y and z. The correct limits of integration are confirmed as y² + z² < x < 2 - y², -√(1 - (z²)/2) < y < √(1 - (z²)/2), and -√2 < z < √2.

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EDIT - well its too late to change the title now, but I can spell corectly... sometimes.

Homework Statement


Find the volume of the region bounded by x = y**2 and z**2 and x = 2 - y**2

Homework Equations


Triple integrals and cylindricals

The Attempt at a Solution


I started by roughly graphing the two figures and I got stuck here for some reason. I tried cylindricals, but I figured there would be a better way. For whatever reason I can't get my head around the limits of integration for this problem. Solving the integral once I have it shouldn't be a problem, but I just need to setup the integral that's why my work here is limited. Thanks for any help
 
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Hi MeMoses! :smile:
MeMoses said:
EDIT - well its too late to change the title now, but I can spell corectly... sometimes.

I think you meant a diabolic sylinder! :biggrin:
I tried cylindricals, but I figured there would be a better way. For whatever reason I can't get my head around the limits of integration for this problem. Solving the integral once I have it shouldn't be a problem, but I just need to setup the integral that's why my work here is limited.

One surface is of revolution, but the other isn't, so there's no short-cuts here. :redface:

You'll have to take slices of width dx perpendicular to the x-axis, and then slice those slices either "horizontally" (with width dz) or "vertically" (with width dy), and double-integrate :wink:
 
tiny-tim said:
Hi MeMoses! :smile:I think you meant a diabolic sylinder! :biggrin:One surface is of revolution, but the other isn't, so there's no short-cuts here. :redface:

You'll have to take slices of width dx perpendicular to the x-axis, and then slice those slices either "horizontally" (with width dz) or "vertically" (with width dy), and double-integrate :wink:
The real question is...
What's a Sylinder?
 
MeMoses said:
EDIT - well its too late to change the title now, but I can spell corectly... sometimes.

Homework Statement


Find the volume of the region bounded by x = y**2 and z**2 and x = 2 - y**2

What does the red highlighted definition mean? There isn't an equation after the word "and".
 
LCKurtz said:
What does the red highlighted definition mean? There isn't an equation after the word "and".

Hi LCKurtz! :smile:

"and" means "+" :wink:
 
tiny-tim said:
Hi LCKurtz! :smile:

"and" means "+" :wink:

How do you know he doesn't mean ##x = y^2## and ##x = z^2##?
 
the title? :rolleyes:
 
tiny-tim said:
Hi LCKurtz! :smile:

"and" means "+" :wink:

LCKurtz said:
How do you know he doesn't mean ##x = y^2## and ##x = z^2##?

tiny-tim said:
the title? :rolleyes:

Well, given that he obviously didn't post the problem word for word, I wouldn't make the assumption that the title is correct. Perhaps the problem is actually stated as I propose and the OP thinks that makes a paraboloid. No way of knowing until we see the correct wording one way or the other.
 
My bad, its x = y**2 + z**2 as thought
 
  • #10
Would the limits
y**2+z**2 < x < 2-y**2,
-sqrt(1-(z**2)/2) < y <sqrt(1-(z**2)/2),
-sqrt(2) < z < sqrt(2)
get the correct volume?
 
  • #11
MeMoses said:
Would the limits
y**2+z**2 < x < 2-y**2,
-sqrt(1-(z**2)/2) < y <sqrt(1-(z**2)/2),
-sqrt(2) < z < sqrt(2)
get the correct volume?

Yes, those are correct.
 

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