Finding Volume Limits and Integration for Paraboloid of Revolution

Click For Summary
SUMMARY

The discussion focuses on calculating the volume of a solid formed by revolving the area bounded by the parabola \( y = H - \frac{H}{R^{2}}x^{2} \) around the y-axis, resulting in a paraboloid of revolution. The volume is confirmed to be half that of the circumscribing cylinder with height \( H \) and base radius \( R \). The integration process involves substituting \( x^2 \) in terms of \( y \) and \( R \), leading to the final volume calculation of \( \frac{\pi R^2}{H} \int_0^H (H-y)dy \).

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the Disk Method for volume calculations.
  • Knowledge of the properties of paraboloids and their equations.
  • Ability to manipulate algebraic expressions involving variables and constants.
NEXT STEPS
  • Study the Disk Method in detail for volume calculations of solids of revolution.
  • Learn about the properties and applications of paraboloids in geometry.
  • Practice integration techniques, particularly definite integrals involving polynomial functions.
  • Explore the relationship between geometric shapes and their volumes in calculus.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and geometry, as well as educators seeking to enhance their understanding of volume calculations for solids of revolution.

elitespart
Messages
95
Reaction score
0
Volumes: The Disk Method [Resolved]

1. If the area bounded by the parabola y = H - (H/R^{2})x^{2} and the x-axis is revolved about the y-axis, the resulting bullet-shaped solid is a segment of a paraboloid of revolition with height H and radius of base R. Show its volume is half the volume of the circumscribing cylinder

Okay so the thickness of the disk is dy and the area is \pix^{2}. How do I find the limits of intergration and put x in terms of H and R? Thanks.
 
Last edited:
Physics news on Phys.org
Your disks run from y = 0 to y = H.
The incremental volume of each disk is \pi x^2 \Delta y. For any point (x, y) on the parabola, you have a relationship between x and y, so use this relationship to replace x^2.
 
So x^2 = y-H(-R^2/H) and then run that from 0 to H?
 
anybody?
 
elitespart said:
So x^2 = y-H(-R^2/H) and then run that from 0 to H?

sounds good, what do you get when you integrate that?
 
I feel like an idiot but I'm having trouble integrating this. Can someone get me started on it? Thanks.
 
\int_0^H \pi x^2 dy= \int_0^H \pi (y-H)(\frac{-R^2}{H}) dy =\frac{\pi R^2}{H} \int_0^H (H-y)dy

...surely, you know how to integrate this?
 
wooow. there's something wrong with me. So I ended up with the volume of a cylinder over 2. Which is exactly what I needed. Thank you very much.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K