Integrating in polar coordinates (volume)

Click For Summary
SUMMARY

The discussion focuses on integrating a solid bounded by the parabolas defined by the equations z = 3x^2 + 3y^2 - 7 and z = -x^2 - y^2 + 9 in polar coordinates. The user correctly transformed the equations into polar form, yielding 3r^2 = 7 and r^2 = 9, determining that r ranges from (7/3)^(1/2) to 3 and theta spans from 0 to 2π. However, the user encountered discrepancies between their calculated integral setup and the book's answer, specifically in the integration of the function -4r^2 + 16. The need for assistance in properly setting up the integral is emphasized.

PREREQUISITES
  • Understanding of polar coordinates in three-dimensional space
  • Familiarity with parabolic equations and their graphical representations
  • Knowledge of triple integrals and volume calculations
  • Proficiency in converting Cartesian coordinates to polar coordinates
NEXT STEPS
  • Review the process of converting Cartesian equations to polar coordinates
  • Study the setup and evaluation of triple integrals in cylindrical coordinates
  • Learn about the geometric interpretation of volume bounded by surfaces
  • Explore examples of integrating functions over parabolic regions
USEFUL FOR

Students studying multivariable calculus, particularly those focusing on integration techniques in three-dimensional geometry, as well as educators seeking to clarify concepts related to volume calculations using polar coordinates.

rhyno89
Messages
18
Reaction score
0

Homework Statement



The solid bounded by the parabolids z = 3x^2 + 3y^2 -7 and z = -x^2 -y^2 + 9

Homework Equations





The Attempt at a Solution



Ok so i set the two z equations into polar form and came up with 3r^2 = 7 and r^2 = 9
I thought that r went from (7/3) ^(1/2) to 3 and that theta went from 0 to 2pi.
I figured the equation to integrate was -4r^2 + 16. (Outer bounds -inner bounds) Anyway, the answer the book gives is different than the two different ways I tried. Just need help setting up the integral...thanks
 
Physics news on Phys.org
This is a 3-D problem; you were given z coordinate values in each case. Work with those.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K