• Support PF! Buy your school textbooks, materials and every day products Here!

Revolution about y=-1

  • Thread starter iRaid
  • Start date
  • #1
559
8

Homework Statement


Find the volume of the solid obtained by rotating the region
bounded by the given curves about the specified line. Sketch the
region, the solid, and a typical disk or washer.

y=sinx, y=cosx, 0≤ x≤∏/4, About y=-1


Homework Equations





The Attempt at a Solution


Tried 2 ways, shell method and washer method..

[tex]\pi \int_0^\frac{\pi}{4}(cosx-sinx)^{2}\,dx - \pi \int_0^\frac{\pi}{4}(-1)^{2}\,dx[/tex]
[tex]\pi \int_0^\frac{\pi}{4}cos^{2}x-2sinxcosx+sin^{2}x\,dx - \pi \int_0^\frac{\pi}{4}1\,dx[/tex]
^Hard integral

[tex]2\pi \int_0^\frac{\pi}{4}(x+1)(cosx-sinx)\,dx[/tex]
[tex]2\pi \int_0^\frac{\pi}{4}xcosx+cosx-xsinx-sinx\,dx[/tex]
[tex]2\pi (sinx+cosx)|[/tex] (From 0 to ∏/4, not sure how to do this in latex)
(And so on)


Thanks for any help.
 
Last edited:

Answers and Replies

  • #2
537
0
You might want to relook at the way you set up the problem.

Your first integral is some finite value, let's say 1 (I know that isn't right, but follow the idea). Now imagine, instead of the line y = -1, it is rotated about y = -100. Your volume should be bigger, right? But the way you have the equation, it's getting smaller.

So, for starters, imagine instead you rotate about the line y = 0. What would each individual part look like (the Cos part minus the sin part). Now shift the radius by 1 larger (since you are rotating about y=-1). What is the radius of the outer diameter as you go from 0 to pi/4?
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,224
951

Homework Statement


Find the volume of the solid obtained by rotating the region
bounded by the given curves about the specified line. Sketch the
region, the solid, and a typical disk or washer.

y=sinx, y=cosx, 0≤ x≤∏/4, About y=-1

Homework Equations



The Attempt at a Solution


Tried 2 ways, shell method and washer method..

[tex]\pi \int_0^\frac{\pi}{4}(cosx-sinx)^{2}\,dx - \pi \int_0^\frac{\pi}{4}(-1)^{2}\,dx[/tex]
[tex]\pi \int_0^\frac{\pi}{4}cos^{2}x-2sinxcosx+sin^{2}x\,dx - \pi \int_0^\frac{\pi}{4}1\,dx[/tex]
^Hard integral

[tex]2\pi \int_0^\frac{\pi}{4}(x+1)(cosx-sinx)\,dx[/tex]
[tex]2\pi \int_0^\frac{\pi}{4}xcosx+cosx-xsinx-sinx\,dx[/tex]
[tex]2\pi (sinx+cosx)|[/tex] (From 0 to ∏/4, not sure how to do this in latex)
(And so on)

Thanks for any help.
The result is the same as subtracting,
the volume generated by rotating the region bounded by y=sin(x), 0≤ x≤∏/4, about y=-1
FROM
the volume generated by rotating the region bounded by y=cos(x), 0≤ x≤∏/4, about y=-1 .
 
  • #4
559
8
@SammyS, OK, but it's still a hard integral.
 
Last edited:
  • #5
559
8
You might want to relook at the way you set up the problem.

Your first integral is some finite value, let's say 1 (I know that isn't right, but follow the idea). Now imagine, instead of the line y = -1, it is rotated about y = -100. Your volume should be bigger, right? But the way you have the equation, it's getting smaller.

So, for starters, imagine instead you rotate about the line y = 0. What would each individual part look like (the Cos part minus the sin part). Now shift the radius by 1 larger (since you are rotating about y=-1). What is the radius of the outer diameter as you go from 0 to pi/4?
Ah I see, I thought it was becoming a cylinder..
 

Related Threads for: Revolution about y=-1

  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
9
Views
677
Replies
2
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
1K
Replies
20
Views
945
Replies
14
Views
1K
Replies
4
Views
762
Top