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Volume of a rotated graph?

by tbone413
Tags: graph, rotated, volume
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tbone413
#1
Mar19-08, 12:43 PM
P: 7
1. The problem statement, all variables and given/known data
Find the volume of the solid found by rotating the area bound between
the curves x = 1, y = x^2, y = 0, around x-axis.


2. Relevant equations
I know you have to solve this using the shell method, but I just get extremely confused while trying to do it. Maybe someone could explain it to me?


3. The attempt at a solution
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silicon_hobo
#2
Mar19-08, 01:21 PM
P: 59
This is the formula for the cylindrical shell method:
[tex]\int_a^b2\pi xf(x)dx[/tex]
you can also use [tex]\int_a^bA(x)dx[/tex]
The second one would be easier.
silicon_hobo
#3
Mar19-08, 03:18 PM
P: 59
Sorry! I was mistaken when I said the area was unbounded. The first thing you need to do is find where these curves intersect and determine the leftmost and rightmost boundaries of X.

Soup
#4
Mar19-08, 07:12 PM
P: 2
Volume of a rotated graph?

Normally to find points of intersection you would set the two curves you wish to find the x values that satisythis equation. However they are much more easily found in this particular problem by drawing a quick sketch. It comes out at x=0 to x=1

Also, this does not require shell method at all. Shell method is for when you cannot create an infinite number of lines between between your boundaries that is perpundicular to your axis of revolution such that one of those lines would hit the same curve twice. In this case all you need to is:
[tex]\pi\int_0^1\ (x^2)dx[/tex]
Snazzy
#5
Mar19-08, 07:21 PM
P: 459
Quote Quote by Soup View Post
[tex]\pi\int_0^1\ (x^2)dx[/tex]
Should be (f(x))^2.
paralian
#6
Mar19-08, 09:35 PM
P: 14
Would that end up being...

[tex]\int^{1}_{0}(\pi r^2)dx[/tex]

where r = "Outer radius" (1) minus "inner radius" ([tex]x^2[/tex])

?

(I hope so, I have a test tomorrow)
Snazzy
#7
Mar19-08, 09:42 PM
P: 459
No, because x = 1 is the vertical boundary in your integration sign.
paralian
#8
Mar21-08, 10:36 AM
P: 14
Quote Quote by Snazzy View Post
No, because x = 1 is the vertical boundary in your integration sign.
Ok...I think I get it.

[tex]\int^{1}_{0}\pi (x^2)^2 dx[/tex]

= [tex]\pi*\int^{1}_{0}x^4 dx[/tex]

= [tex]\pi*(1^5/5-0)[/tex]

= [tex]\pi/5[/tex]

Is that right?
Snazzy
#9
Mar21-08, 02:29 PM
P: 459
Looks right.

I should've said x = 1 is the vertical line that limits how far horizontally you should integrate.


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