Cylindrical Coordinates Domain

In summary, the problem involves calculating the volume of the region W between two paraboloids. The correct triple integral for this is \int^{2\pi}_{0}\int^{2}_{0}\int^{8-r^2}_{r^2}r dzdrdθ. The integrand should be 1dV and the limits are 0 ≤ z ≤ 8, 0 ≤ r ≤ 2, and 0 ≤ θ ≤ 2\pi.
  • #1
DrunkApple
111
0

Homework Statement


Let W be the region between the paraboloids z = x^2 + y^2 and z = 8 - x^2 - y^2


Homework Equations





The Attempt at a Solution


for each domain,
0 ≤ z ≤ 8
0 ≤ r ≤ 2
0 ≤ θ ≤ 2[itex]\pi[/itex]

So...

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
 
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  • #2
Assuming what you are calculating is the volume, yes.
 
  • #3
wait wait shouldn't it be
[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ
 
  • #4
DrunkApple said:
wait wait shouldn't it be
[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex](8-2[itex]r^{2}[/itex])r drdθ

No. That is wrong on several levels. If you are doing a triple integral for volume the integrand is always 1dV = 1 rdzdrdθ.

1. You are missing the dz.
2. Your z limits are not correct; they would depend on r and maybe θ and you would want the dz on the inside so it is integrated first.
3. Your integrand was zupper-zlower in your double integral, which was correct, and that is what you should get when you integrate the 1dz in the triple integral.
 
  • #5
oh so this is the final one then

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex](8-2[itex]r^{2}[/itex])r dzdrdθ

Yes I forgot dz accidently but I knew that :p
 
  • #6
DrunkApple said:
oh so this is the final one then

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex](8-2[itex]r^{2}[/itex])r dzdrdθ

Yes I forgot dz accidently but I knew that :p

Still wrong. Your integrand should be 1dV.
 
  • #7
How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ
 
  • #8
DrunkApple said:
How about [itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8}_{0}[/itex] rdzdrdθ

No, not correct.
 
  • #9
but isn't dV = dzdrdθ?
 
  • #10
No, it is r dz dr dθ which you had correct. But now you messed up the limits again.
 
  • #11
ARRRGGGG this thing is driving me nuts
ok phew
one more time

[itex]\int^{2\pi}_{0}[/itex][itex]\int^{2}_{0}[/itex][itex]\int^{8-r^2}_{r^2}[/itex]r dzdrdθ
 
  • #12
Finally. Notice that when you work the dz integral what you have left with is exactly the double integral which you had correct in your initial post.
 
  • #13
Yes i got it
 

FAQ: Cylindrical Coordinates Domain

1. What are cylindrical coordinates?

Cylindrical coordinates are a coordinate system used to locate points in three-dimensional space. It consists of a radial distance, an azimuthal angle, and a height or elevation angle.

2. How are cylindrical coordinates defined?

In cylindrical coordinates, the radial distance is measured from the origin to the point, the azimuthal angle is measured from a reference direction in the xy-plane, and the height or elevation angle is measured from the xy-plane to the point.

3. What is the domain of cylindrical coordinates?

The domain of cylindrical coordinates is all points in three-dimensional space that can be uniquely represented by a combination of the radial distance, azimuthal angle, and height or elevation angle.

4. How do cylindrical coordinates differ from Cartesian coordinates?

Cylindrical coordinates use a different set of variables to locate points in three-dimensional space compared to Cartesian coordinates. While Cartesian coordinates use x, y, and z, cylindrical coordinates use r, θ, and z.

5. What are the advantages of using cylindrical coordinates?

Cylindrical coordinates are particularly useful in situations where the geometry of a problem is cylindrical or has circular symmetry. They also simplify certain types of calculations, such as those involving rotational symmetry or cylindrical objects.

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