Solid Of Revolution Problem - Washer

  • Thread starter Thread starter Bellwether
  • Start date Start date
  • Tags Tags
    Revolution Solid
Click For Summary

Homework Help Overview

The problem involves finding the volume of a solid formed by rotating the area between the function y=x^2+1 and the x-axis from x=0 to x=2 about the line y=-2. Participants are discussing the setup of the integral and the calculation of radii involved in the washer method.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to set up the integral for the volume calculation but express confusion regarding the inner and outer radii. There are discussions about whether to adjust the radius by adding 1 or 3, and some suggest shifting the function to simplify the calculation.

Discussion Status

The discussion is active, with participants providing guidance on how to approach the problem. There are multiple interpretations of the radius calculations being explored, and some participants are clarifying their understanding of the setup.

Contextual Notes

There is uncertainty regarding the correct adjustment of the radii in relation to the axis of rotation, as well as the implications of the function's position relative to the x-axis and the line y=-2.

Bellwether
Messages
5
Reaction score
0

Homework Statement



Find the volume of a solid created when the area between the function y=x^2+1 and the x-axis (for 0<x<2) is rotated about the line y=-2.

Homework Equations



Vs = ∏*r^2*h

The Attempt at a Solution



I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

I think this creates a full solid around y=-2 -- would I just want to add 1 to the radius instead of 3?

Many thanks!
 
Physics news on Phys.org
Welcome to PF!

Hi Bellwether! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
Bellwether said:
I can't seem to set this up correctly and am thrown by the inner radius being the x-axis. I originally set up the integral as : Int {0,2} ∏((x^2+1) + 3)^2 dx

That's for the outer radius …

you now need to subtract the volume for the inner radius. :wink:
 
For problems like this, I used to prefer adding 2 to the function to shift the whole thing up 2 units of y. Then you can simply revolve the function around the line y = 0, and the volume result will be the same.
 
Thanks for the welcome!

So it's, ∫∏((x2+1)+3)2 - 22) ?
 
Hi Bellwether! :smile:

(just got up :zzz:)
Bellwether said:
Thanks for the welcome!

So it's, ∫∏((x2+1)+3)2 - 22) ?

∫ π((x2+1) + 2)2 - 22) dx :wink:
 
I'm confused -- why is the outer radius just +2 , instead of 3? Isn't (x2 + 1) three units away from the axis of revolution?
 
no, you've counted the "1" twice …

(x2 + 1) - (-2 ) = x2 + 3 :wink:
 
Of course, thank you.
 

Similar threads

Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K