Help with yet another solid of revolution question

  • Thread starter Thread starter student93
  • Start date Start date
  • Tags Tags
    Revolution Solid
Click For Summary

Homework Help Overview

The discussion revolves around a solid of revolution problem involving the washer method for calculating volume. Participants are examining the setup of the problem, specifically the limits of integration and the correct application of the method.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the washer method, questioning the limits of integration and the correct bounds for the variable of integration. There is an exploration of the equations involved and their intersections.

Discussion Status

The discussion is active with participants providing insights on the integration process and the necessary adjustments to the limits. Some guidance has been offered regarding the need to ensure bounds correspond to the variable of integration. Multiple interpretations of the limits are being explored.

Contextual Notes

Participants are working with the constraints of the problem as presented, including the need to derive limits from the intersection of curves and the implications of integrating with respect to different variables.

student93
Messages
83
Reaction score
0

Homework Statement



See the attached problem.

Homework Equations



See the attached problem.

The Attempt at a Solution



I used washer method and got an inner radius of x=y^2 and an outer radius of x=y+2, I calculated my upper limit as being 4 and my lower limit as being 0. The answer is 72π/5, but I can't seem to get that answer. Are my limits of integration wrong and/or did I use the wrong method?
 

Attachments

  • Screen shot 2014-01-28 at 5.07.27 AM.png
    Screen shot 2014-01-28 at 5.07.27 AM.png
    6.4 KB · Views: 470
Physics news on Phys.org
That method works for me. Please post your detailed steps.
 
V=π∫(y+2)^2 - (y^2)^2 dy, from 0 to 4
V= -2032π/15, which is obviously not the correct answer since volume can't be negative etc.
 
student93 said:
V=π∫(y+2)^2 - (y^2)^2 dy, from 0 to 4
V= -2032π/15, which is obviously not the correct answer since volume can't be negative etc.
Right integrand, wrong range. x goes from 0 to 4. what's the range for y?
 
haruspex said:
Right integrand, wrong range. x goes from 0 to 4. what's the range for y?

I set √x=x-2, and solved the quadratic and got x=4,1 (I used the 4 as my upper limit and used 0 as my lower limit since it seemed that 0 was the lower limit from how the graph looked). Also how exactly do I go about calculating the limits of integration in regards to y?
 
student93 said:
I set √x=x-2, and solved the quadratic and got x=4,1 (I used the 4 as my upper limit and used 0 as my lower limit since it seemed that 0 was the lower limit from how the graph looked).
Yes, but you are integrating with respect to y, so your bounds must be bounds on y, not x.
Also how exactly do I go about calculating the limits of integration in regards to y?
Sketch the curves and see where they cross.
 
haruspex said:
Yes, but you are integrating with respect to y, so your bounds must be bounds on y, not x.

Sketch the curves and see where they cross.

I set the equations equal and I get my upper limit as 4 with respect to y and my lower limit as 1 with respect to y, however I still don't get the correct answer which is 72π/5.
 
Last edited:
So I finally realized I was supposed to solve y^2=y+2 and get an upper and lower limit of 2 and -1 respectively. After plugging in those two values into the integrand, I finally ended up getting the correct answer. Thanks for the help.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K