How to Evaluate a Triple Integral Bounded by a Cylinder and Planes?

Click For Summary
SUMMARY

The discussion focuses on evaluating a triple integral for the solid bounded by the cylinder defined by the equation y² + z² = 576 and the planes x = 0, y = 4x, and z = 0 in the first octant. The bounds for z are established as 0 to √(576 - y²), while x ranges from 0 to 24. The use of spherical coordinates is recommended due to the symmetry of the region, with ρ ranging from 0 to 24, φ from 0 to π/2, and θ from 0 to arctan(4). This approach simplifies the integration process over the specified volume.

PREREQUISITES
  • Understanding of triple integrals in multivariable calculus
  • Familiarity with cylindrical and spherical coordinate systems
  • Knowledge of the equations of planes and cylinders in three-dimensional space
  • Ability to perform integration over defined regions
NEXT STEPS
  • Study the conversion between Cartesian and spherical coordinates in detail
  • Learn how to set up and evaluate triple integrals in cylindrical coordinates
  • Explore examples of integrating over regions defined by multiple surfaces
  • Practice solving triple integrals involving complex boundaries and shapes
USEFUL FOR

Students in calculus courses, particularly those focusing on multivariable calculus, as well as educators and tutors looking to enhance their understanding of triple integrals and coordinate transformations.

Pete_01
Messages
49
Reaction score
0

Homework Statement


Evaluate the triple integral where E is the solid bounded by the cylinder y^2+ z^2 = 576 and the planes x = 0, y = 4 x and z =0 in the first octant.


Homework Equations





The Attempt at a Solution


I figure that by solving for z I can get the bounds, so between 0 to sqrt(576-y^2) would be the z upper and lower bounds. X bounds would be between 0 and 24? And I am not sure about y.
 
Physics news on Phys.org
I presume you mean "the triple integral for the volume"- i.e. just integrate dV= dxdydz over that region. Otherwise you haven't said what it is you want to integrate! I think, because of the spherical symmetry here, using spherical coordinates would be best. Obviously [itex]\rho[/itex] will go from 0 to [itex]\sqrt{576}= 24[/itex], [itex]\phi[/itex] from 0 to [itex]\pi/2[/itex] and, since y= 4x has slope 4, [itex]\theta[/itex] will go from 0 to arctan(4).
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K