Definite integral distance from volume

Click For Summary

Homework Help Overview

The problem involves determining the height of water in a spherical water tower with a radius of 20 meters when filled to one-quarter of its capacity. The context is centered around the application of definite integrals to find volume.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of definite integrals to calculate the volume of the sphere and how to relate that to the height of the water. There is an exploration of integrating from -20 to a variable height instead of the full range.

Discussion Status

Participants have provided various approaches, including the suggestion to set up a cubic equation based on the volume and to use numerical techniques for solving it. There is acknowledgment of the complexity of the roots involved in the cubic equation.

Contextual Notes

There is mention of the need for numerical methods due to the nature of the cubic equation, and some participants note the presence of complex roots, indicating potential complications in finding real solutions.

poopcaboose
Messages
33
Reaction score
0

Homework Statement

If you have a water tower that is spherical with a radius 20m, how far from the bottom will the water level be if it is filled to 1/4 of it's capacity.



2. Homework Equations [tex]\int_-20^20pi(400-y^2)dy=33510.32164m^3=3,351,032.164Liters[/tex]



The Attempt at a Solution

I can find 1/4 capacity=837,758.041 liters but don't know where to go from here to find the distance from the bottom
 
Physics news on Phys.org
I meant for it to be a definite integral from 20 to -20 not 20 squared
 
Instead of integrating from -20 to 20 to get the full volume, integrate from -20 to h. You'll get a cubic expression in h. Equate that to 1/4 of the full volume and try to solve for h.
 
BTW it looks like you have to use numerical techniques to get an answer. The cubic doesn't factor or anything for me.
 
Thanks for the help, I can use Newtons method to approximate f(y) at zero and solve it from there thanks I would have never thought to set it to y
 
By the way, two of the roots are complex numbers from what I got.. just a warning.
 
I figured it out its 13.054073m from the bottom you either solve for zero with Newtons method or graph it
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
20
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K