What are the Limits for Computing the Volume of a Cylinder Inside a Sphere?

Click For Summary
SUMMARY

The discussion focuses on computing the volume of a cylinder defined by the equation x² + y² = 4, which is situated inside a sphere defined by x² + y² + z² = 9. The correct limits for the integral to compute the volume are established as 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, and -√(9 - r²) ≤ z ≤ √(9 - r²). The final integral for calculating the volume is confirmed as ∫ from 0 to 2π ∫ from 0 to 2 ∫ from -√(9 - r²) to √(9 - r²) r dz dr dθ.

PREREQUISITES
  • Understanding of cylindrical coordinates
  • Familiarity with triple integrals
  • Knowledge of volume calculation in multivariable calculus
  • Basic concepts of geometric shapes: cylinders and spheres
NEXT STEPS
  • Study the application of cylindrical coordinates in volume calculations
  • Learn about triple integrals and their applications in multivariable calculus
  • Explore the geometric properties of spheres and cylinders
  • Investigate the implications of changing limits in integrals
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, as well as engineers and physicists involved in volume calculations of geometric shapes.

boneill3
Messages
126
Reaction score
0
Hi Guys,

I have been given the coordinates of a cylinder inside a sphere and want to convert to Cylindrical coordinates to compute the volume of the cylinder.

Can you please check the limits and integral I have?

The cylinder is x^2+y^2= 4

sphere = x^2+y^2+z^2= 9

As its a cylinder we have

Limits are 0<= theta <= 2\pi 0<= r <= 2 and

Inside a sphere with limits

sphere = x^2+y^2+z^2= 9

z = sqrt{9-r^2}

So would my integral be:


\int{{0}{2\pi} \int{0}{2} \int{0}{sqrt{9-r^2}} r dz dr d(theta)


regards
 
Physics news on Phys.org
Why should the lower limit of z be 0?

Inside a sphere with limits

sphere = x^2+y^2+z^2= 9

z = sqrt{9-r^2}
Are you sure about z=sqrt(9-r^2)) is the only limit set upon z by the above equation?
 
Sorry I want to compute the solid bounded above and below by the sphere and inside the cylinder.

I see your point the sphere can be either side of the z axis .

it should be:

int{-sqrt{9-r^2}} {sqrt{9-r^2}} r d(theta)


Is that alright
 
That's right indeed. :smile:
 
Thanks for your help!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K