Polar coordinate to compute the volume

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Homework Help Overview

The problem involves using polar coordinates to compute the volume of a region defined by two surfaces in three-dimensional space, specifically the inequalities 4 - x² - y² ≤ z ≤ 10 - 4x² - 4y².

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the volume integral and question the correctness of the domain. There is an exploration of the intersection of the two surfaces and the implications for the limits of integration. Some participants clarify the distinction between polar and cylindrical coordinates.

Discussion Status

The discussion is active with participants providing insights into the coordinate systems and the setup of the problem. There is a recognition of the need to clarify the coordinate system being used, and some guidance has been offered regarding the volume element and limits of integration.

Contextual Notes

Participants note that the problem involves three-dimensional coordinates, which may lead to confusion between polar and cylindrical coordinates. The intersection points of the surfaces are also under consideration, particularly regarding their implications for the volume calculation.

DrunkApple
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Homework Statement


Use polar coordinates to compute the volume of the region defined by
4 - [itex]x^{2}[/itex] - [itex]y^{2}[/itex] ≤ z ≤ 10 - 4[itex]x^{2}[/itex] - 4[itex]y^{2}[/itex]


Homework Equations





The Attempt at a Solution


I got z = 2 so set up the equation

V = [itex]f^{2pi}_{0}[/itex][itex]f^{5/2}_{2}[/itex][itex]f^{0}_{2}[/itex]r*dzdrdθ

is the domain correct?
 
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DrunkApple said:

Homework Statement


Use polar coordinates to compute the volume of the region defined by
4 - [itex]x^{2}[/itex] - [itex]y^{2}[/itex] ≤ z ≤ 10 - 4[itex]x^{2}[/itex] - 4[itex]y^{2}[/itex]

Homework Equations


The Attempt at a Solution


I got z = 2 so set up the equation

V = [itex]f^{2pi}_{0}[/itex][itex]f^{5/2}_{2}[/itex][itex]f^{0}_{2}[/itex]r*dzdrdθ

is the domain correct?
Those are cylindrical coordinates, not polar. (If you were to do the problem in cylindrical coords, your limits of integration for z would be incorrect.)

The two surfaces intersect at z=2, but that's not particularly important. At what value of r do they intersect?
 
setting 4 - [itex]x^{2}[/itex] - [itex]y^{2}[/itex] and 10 - 4[itex]x^{2}[/itex] - 4[itex]y^{2}[/itex] to equal, r = ±√2, but since r must be greater than 0, it's r = √2
 
The volume element is the area element, r dr dθ, times the height, which you get from zupper - zlower .

z goes from 4 - r2 to 10 - 4r2 .
 
thank you I got it.
But how do I know if it's cylindrical?
 
How do you know if what is cylindrical? If you mean the coordinate system, "cylindrical coordinates" are just polar coordinates for the xy-plane with the z coordinate.
 
I mean just by looking at the equation
Is it because it contains z?
 
DrunkApple said:
thank you I got it.
But how do I know if it's cylindrical?
If you're asking how did I know you were using cylindrical coordinates rather than polar; cylindrical coordinates are in 3 dimensions and use r, θ, and z. Polar coordinates are 2 dimensional using r and θ.
 

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