Change of Variables to find the volume of a part of a sphere in CYLINDRICAL coords

In summary, the bounds for the z-coordinate of a cylinder with a radius of 1 are -sqrt(4-r^2) < z < sqrt(4-r^2), while the bounds for the top cap are 0 < r < 1 and 0 < theta < 2pi.
  • #1
jmm5872
43
0
Make the indicated change of variables (do not evaluate) (Not sure how to write an iterated integral with bounds so I will try and explain by just writing the bounds)
(I also tried using the symbols provided, but everything I tried just put a theta in here so I gave up)

[tex]\int\int\int[/tex]xyz dzdxdy

-1 < z < 1
-sqrt(1-y^2) < x < sqrt(1-y^2) ... circle with radius 1
-sqrt(4-x^2-y^2) < y < sqrt(4-x^2-y^2) ... sphere with radius 4


Relevant equations:
x = rcos[tex]\theta[/tex]
y = rsin[tex]\theta[/tex]
z = z

Attempt:
I understand the bounds, it is a sphere with a radius of 4, however, the bounds constrain this to a cylinder with a radius of 1 along the z-axis. This makes it a cylinder with a rounded top and bottom (the only part of the sphere left).

My initial change of the bounds was this:
-4 < z < 4
0 < r < 1
0 < theta < 2pi

However I know this is wrong because these bounds only make a cylinder with radius 1 and length 8, they do not account for the rounded "caps."

My only idea was that I would have to split into two separate integrals, one for the cylinder and one for the top cap (double that one due to symmetry). But it seems like there should be a better way to account for the endcaps.

Thanks
 
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  • #2


Okay I think I figured it out.

The bound of z would simply be this:

-sqrt(4-r^2) < z < sqrt(4-r^2)

and r and theta:

0 < r < 1
0 < theta < 2pi

A lot easier than I was making it, I guess that's why no one answered me.
 
  • #3


jmm5872 said:
Okay I think I figured it out.

The bound of z would simply be this:

-sqrt(4-r^2) < z < sqrt(4-r^2)

and r and theta:

0 < r < 1
0 < theta < 2pi

A lot easier than I was making it, I guess that's why no one answered me.
For future reference, here's the LaTeX for your iterated integral, showing the limits of integration. To see what I did, click the integral expression.
[tex]\int_{\theta=0}^{2\pi} \int_{r=0}^1 \int_{z=-\sqrt{4 - r^2}}^\sqrt{4 - r^2}}{r^2cos(\theta)sin(\theta) z~ r~dz~dr~d\theta[/tex]
 
  • #4


Thanks, good to see the LaTex. It will help in the future.
 

Related to Change of Variables to find the volume of a part of a sphere in CYLINDRICAL coords

1. How do you convert from spherical to cylindrical coordinates?

To convert from spherical to cylindrical coordinates, you can use the following equations:

x = ρsin(θ)cos(φ)

y = ρsin(θ)sin(φ)

z = ρcos(θ)

Where ρ is the distance from the origin to the point, θ is the angle from the positive z-axis, and φ is the angle from the positive x-axis.

2. What is the formula for finding the volume of a part of a sphere in cylindrical coordinates?

The formula for finding the volume of a part of a sphere in cylindrical coordinates is:

V = ∫∫∫ ρdρdθdz

This is a triple integral that integrates over the desired region to find the volume.

3. How do you set up the limits of integration for the triple integral?

The limits of integration for the triple integral can be determined by the boundaries of the desired region in cylindrical coordinates. ρ will have limits from 0 to the radius of the sphere, θ will have limits from 0 to 2π, and z will have limits from 0 to the height of the desired region.

4. Can this method be used to find the volume of any shape in cylindrical coordinates?

No, this method is specifically for finding the volume of a part of a sphere in cylindrical coordinates. Other shapes may require different methods or formulas for finding their volume.

5. How accurate is the volume calculated using this method?

The accuracy of the volume calculated using this method will depend on the accuracy of the given parameters and the precision of the calculations. As long as the correct limits of integration and equations are used, the calculated volume should be accurate within a reasonable margin of error.

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