# Polar coordinate to compute the volume

1. Nov 13, 2011

### DrunkApple

1. The problem statement, all variables and given/known data
Use polar coordinates to compute the volume of the region defined by
4 - $x^{2}$ - $y^{2}$ ≤ z ≤ 10 - 4$x^{2}$ - 4$y^{2}$

2. Relevant equations

3. The attempt at a solution
I got z = 2 so set up the equation

V = $f^{2pi}_{0}$$f^{5/2}_{2}$$f^{0}_{2}$r*dzdrdθ

is the domain correct?

2. Nov 13, 2011

### SammyS

Staff Emeritus
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?

3. Nov 13, 2011

### DrunkApple

setting 4 - $x^{2}$ - $y^{2}$ and 10 - 4$x^{2}$ - 4$y^{2}$ to equal, r = ±√2, but since r must be greater than 0, it's r = √2

4. Nov 13, 2011

### SammyS

Staff Emeritus
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 .

5. Nov 14, 2011

### DrunkApple

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

6. Nov 14, 2011

### HallsofIvy

Staff Emeritus
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.

7. Nov 14, 2011

### DrunkApple

I mean just by looking at the equation
Is it because it contains z?

8. Nov 14, 2011

### SammyS

Staff Emeritus
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.