Volumes by Slicing and Rotation about an Axis.

Click For Summary
SUMMARY

The volume of the solid generated by revolving the region between the parabola x = y² + 1 and the line x = 3 about the line x = 3 is calculated using the disk method with respect to y, yielding an exact answer of 64π√(2)/15. The incorrect setup for integration with respect to x was identified as using the formula for disks instead of the appropriate shell method. The correct approach involves recognizing the cross-sections as shells when integrating with respect to x.

PREREQUISITES
  • Understanding of the disk and shell methods for volume calculation
  • Familiarity with integration techniques in calculus
  • Knowledge of parabolic equations and their graphical representations
  • Ability to manipulate and evaluate definite integrals
NEXT STEPS
  • Study the shell method for volume calculations in solid of revolution
  • Practice integrating with respect to both x and y for different functions
  • Explore the geometric interpretation of cross-sections in solids of revolution
  • Review the properties of parabolas and their applications in volume problems
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations in solid geometry, as well as educators teaching integration techniques and methods for solids of revolution.

tolove
Messages
164
Reaction score
1

Homework Statement



Find the volume of the solid generated by revolving the region between the parabola x = y^2 + 1 and the line x = 3 about the line x = 3.

Homework Equations



The answer is found by integrating with respect to y with disk method, but I don't understand why my answer is incorrect when I try to integrate it with dx.

The Attempt at a Solution



The exact answer when integrated with respect to y is 64*pi*√(2)/15.

This is the integral I am incorrectly setting up somehow:
2*∫ pi * (√(x-1))^2 dx, x = 1 to 3

Any ideas why this doesn't work?
 
Physics news on Phys.org
tolove said:

Homework Statement



Find the volume of the solid generated by revolving the region between the parabola x = y^2 + 1 and the line x = 3 about the line x = 3.

Homework Equations



The answer is found by integrating with respect to y with disk method, but I don't understand why my answer is incorrect when I try to integrate it with dx.

The Attempt at a Solution



The exact answer when integrated with respect to y is 64*pi*√(2)/15.

This is the integral I am incorrectly setting up somehow:
2*∫ pi * (√(x-1))^2 dx, x = 1 to 3

Any ideas why this doesn't work?

Integrating dy your cross sections are disks. If you want to integrate dx then your cross sections are shells. You need to use a different formula for the integral.
 
Dick said:
Integrating dy your cross sections are disks. If you want to integrate dx then your cross sections are shells. You need to use a different formula for the integral.

The cross-section of 3-d parabola isn't a circle?... ohhh, this isn't a 3d parabola, it's rotated.

Thank you very much!
 

Similar threads

Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K