Polar coordinate to compute the volume

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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
 
thank you I got it.
But how do I know if it's cylindrical?
 
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.