Calculating Volume of A Enclosed by Elliptic Hyperboloid & Planes

  • Thread starter Thread starter sydneyw
  • Start date Start date
  • Tags Tags
    Planes Volume
Click For Summary
SUMMARY

The volume of the region A enclosed by the elliptic hyperboloid defined by the equation x² + y² - z² = a² and the planes z = -h/2 and z = h/2 can be calculated using a triple integral in cylindrical coordinates. The correct integrand is r² cos²(θ) + r² sin²(θ) - z², integrated over the specified limits. The final volume is determined to be (πa⁴h)/2, confirming the calculations are accurate. The symmetry of the figure allows for simplification by calculating only one half of the volume and multiplying by two.

PREREQUISITES
  • Understanding of triple integrals in cylindrical coordinates
  • Familiarity with the equation of an elliptic hyperboloid
  • Knowledge of volume calculation techniques for solids of revolution
  • Ability to perform integration involving trigonometric functions
NEXT STEPS
  • Study the derivation of volume formulas for solids of revolution
  • Learn about the application of cylindrical coordinates in multivariable calculus
  • Explore the properties and applications of elliptic hyperboloids in geometry
  • Practice solving triple integrals with varying limits and integrands
USEFUL FOR

Mathematicians, engineering students, and anyone involved in advanced calculus or geometric analysis will benefit from this discussion, particularly those focused on volume calculations of complex shapes.

sydneyw
Messages
5
Reaction score
0
For positive a and h, let A designate the region of R3 enclosed by the elliptic hyperboloid, x2 +y2 -z2 =a2 and the two planes, z= -h/2 and z=h/2.

Determine the volume of A

So I figure this will be a triple integral in cylindrical coordinates. the first integrand being from -h/2 to h/2, the second from 0 to 2∏ and the third is a transformation to r.

The equation to integrate would be my elliptic hyperboloid equation, correct? So that would look something like r2cos2(θ) +r2sin2(θ)-z2 rdrdθdz...right?

I'm confused on how to go about this problem so I told you what I was able to figure out on my own. Could someone PLEASE explain how I do this?
 
Physics news on Phys.org
i got the triple integral with limits stated above and of r^3 -rz^2 drd(theta)dz..I made my last integrand be from 0 to a..is this correct? I then integrated it all out and ended up with [pi(a^4)h]/2. Is this right?? Could I have possibly solved this?
 
Your surface equation is actually that of a one-sheeted paraboloid. I've attached the graph.

In the plane z=0, r varies from r=a
In the plane z=h/2, r varies to r=\sqrt{a^2+\frac{h^2}{4}}

To get the entire volume, just multiply the triple integral by 2.
 

Attachments

  • para.gif
    para.gif
    10.4 KB · Views: 861
But there is no plane z=0. It's-h/2 and h/2.
 
sydneyw said:
But there is no plane z=0. It's-h/2 and h/2.

You have to evaluate the total volume in two parts. First, from z=0 to z=h/2. Then, from z=-h/2 to z=0.

Note that the figure is symmetrical at z=0, which is at its smallest cross-sectional area, where r=a.

Hence, there is no need to calculate two triple integrals. Just find one and then multiply by 2.
 
okay so 1. do i need to evaluate in cylindrical coords even?
2. am I using my elliptic hyperboloid equation for the integration?
3. it would be a double integral of that elliptic hyperboloid equation or no?

I'm sorry. I'm very confused now.
 
sydneyw said:
For positive a and h, let A designate the region of R3 enclosed by the elliptic hyperboloid, x2 +y2 -z2 =a2 and the two planes, z= -h/2 and z=h/2.

Determine the volume of A

So I figure this will be a triple integral in cylindrical coordinates. the first integrand being from -h/2 to h/2, the second from 0 to 2∏ and the third is a transformation to r.

The equation to integrate would be my elliptic hyperboloid equation, correct? So that would look something like r2cos2(θ) +r2sin2(θ)-z2 rdrdθdz...right?

I'm confused on how to go about this problem so I told you what I was able to figure out on my own. Could someone PLEASE explain how I do this?

This is a solid of revolution. No law says you must use a triple integral. Why not just set it up with circular slabs and integrate in the z direction?
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K