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Volume of a rotated graph?

  • Thread starter tbone413
  • Start date
7
0
1. Homework Statement
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. Homework 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
 

Answers and Replies

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.
 
Last edited:
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.
 
2
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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]
 
458
0
14
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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)
 
458
0
No, because x = 1 is the vertical boundary in your integration sign.
 
14
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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?
 
458
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Looks right.

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

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