Finding Volume of Dome-Shape Confined Between Cylinders and xy-Plane

  • Thread starter Thread starter Inertigratus
  • Start date Start date
  • Tags Tags
    Volume
Click For Summary

Homework Help Overview

The discussion revolves around finding the volume of a dome-shaped region confined between two cylinders and the xy-plane. The original poster describes the shape and provides the equations of the cylinders, expressing uncertainty about how to approach the volume calculation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to derive a function for the height of the dome and explore whether to add the equations of the cylinders or use an average. There are suggestions to consider symmetry and divide the xy-plane into triangles or sections for integration.

Discussion Status

Some participants have offered guidance on using symmetry and breaking the problem into manageable parts. There is an ongoing exploration of how to define the height function and the implications of the intersections of the cylinder equations.

Contextual Notes

Participants note the importance of understanding the geometry of the problem, including the intersections of the cylinder equations and the projection onto the xy-plane. There is mention of the need to clarify the boundaries of integration based on these intersections.

Inertigratus
Messages
123
Reaction score
0
Volume of "dome"-shape

Homework Statement


Find the volume between of the shape that is confined between the two cylinders and the xy-plane. Maybe I have understood it wrong, but after drawing them it looks like the intersection is a dome-looking shape. So I'm guessing the volume that I have to find is between the dome and the ground (xy-plane).


Homework Equations


The two cylinders
[itex]x^2 = 4 - 4z[/itex]
[itex]y^2 = 4 - 4z[/itex]



The Attempt at a Solution


I'm not sure how to do this. I know that [itex]x = y = +- 2[/itex] when [itex]z = 0[/itex] and [itex]x = y = 0[/itex] when [itex]z = 1[/itex].
I thought that maybe adding the two equations can give an equation for how z varies with x and y.
[itex]x^2 + y^2 = 8 - 8z[/itex] and then solve for z and take the integral of z.
In polar coordinates, the radius varies from 0 to 2. That gives me a volume of 3 pi.
The volume is supposed to be 8... any ideas?
 
Physics news on Phys.org


The function for the height of the dome at a given (x,y) is going to have to be piece wise because at some points it's the first cylinder and at some points the second cylinder. That's probably why you're having trouble finding the function to integrate. My advice is to split the xy-plane into 4 triangles for which the volume above them is equal. Then you can find the non-piecewise function above a single triangle, integrate that, and multiply by 4.

Actually, there is even more symmetry to the problem. You can use symmetry to work out that the volume of each triangle based part is 1/8th the volume of a length 4, radius 2 cylinder.
 


Thanks, I understand now that I traced the domain on paper. However, how do I get the function for the height?
Do I add the two equations, solve for z and use that, or do I use one half of the sum of the two equations and solve for z?

Also, could you explain the last part? I'm not that good at understanding symmetry...
 


Inertigratus said:
Thanks, I understand now that I traced the domain on paper. However, how do I get the function for the height?
Do I add the two equations, solve for z and use that, or do I use one half of the sum of the two equations and solve for z?

Also, could you explain the last part? I'm not that good at understanding symmetry...

You should find the intersections of the two functions. This is the line where the dome has a "sharp edge".
To do this you should equal [itex]f_1(x,y)=f_2(x,y)[/itex]
What you obtain are the two lines [itex]y=x, y=-x[/itex].
This lines, together with the other boudaries where [itex]f_1(x,y)=f_2(x,y)=0[/itex], will divide the domain on the xy plane into 4 parts.
You should integrate under one of these areas per time.
Then you should understand that you can is very simple to obtain the whole solid.

Someone else suggested you can further divide that xy plane into 8 areas and so simplify the task more.
 


Ohh, thanks! My math skills are a little rusty hehe... was just looking at an example in my book that is almost identical. I totally forgot about the whole projection-on-the-xy-plane.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K