Finding the volume using cylindrical shells

In summary, the problem is asking for the height of a rectangle from the parabola to the x=2 line, but the order of operations is incorrectly being done. The height would be 3+(y-2)2.
  • #1
ThiagoG
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Homework Statement


x=1+(y-2)^2, x=2. Rotating about the x-axis


Homework Equations


Volume=(2∏y)(1+(y-2)2(Δy)

Limits of integration would be from 1 to 3
2∏∫(y)(1+(y-2)2dy
2∏∫y3-4y2+5y dy


The Attempt at a Solution


2∏[y4/4-4y3/3+5y2/2]

Plug in the limits and I get 32∏/3. The answer is 16∏/3. It works if for the circumference you use 2∏ instead of 2∏y.
 
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  • #2
ThiagoG said:

Homework Statement


x=1+(y-2)^2, x=2. Rotating about the x-axis


Homework Equations


Volume=(2∏y)(1+(y-2)2(Δy)

Limits of integration would be from 1 to 3
2∏∫(y)(1+(y-2)2)dy
Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.
 
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  • #3
eumyang said:
Here's where you've gone wrong, I think. The representative rectangles need to be inside the parabola, not outside, as you have set the integral up.

How would you fix this?
 
  • #4
(To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.)

The way you had set it up, your heights of the representative rectangle (or the distance from the y-axis to the parabola) is x (which equals 1 + (y - 2)2). Given that the distance from the y-axis to the line x=2 is 2, can you tell me the height of a rectangle from the parabola to the x=2 line?
 

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  • #5
eumyang said:
(To help visualize what you did earlier, I attached two pics. The red region is what is being rotated around the x-axis. The "Wrong.bmp" file shows what you did, and the "Right.bmp" file shows what the problem is asking.)

The way you had set it up, your heights of the representative rectangle (or the distance from the y-axis to the parabola) is x (which equals 1 + (y - 2)2). Given that the distance from the y-axis to the line x=2 is 2, can you tell me the height of a rectangle from the parabola to the x=2 line?

So it would be 3+(y-2)2?

edit: I tried this and it didn't work
 
  • #6
No, you don't want to add 2 and 1 + (y - 2)2.
 
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  • #7
eumyang said:
No, you don't want to add 2 and 1 + (y - 2)2.

I subtracted 2 from it. So my integral is 2∏∫(y)(-1+(y-1)2)dy. I got a negative number when I did this.
 
  • #8
Almost. Switch the order of subtraction.
 
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  • #9
eumyang said:
Almost. Switch the order of subtraction.

Ok thank you so much for your help. :)
 

1. What is the formula for finding the volume using cylindrical shells?

The formula for finding the volume using cylindrical shells is V = ∫2πrh dx, where r is the radius of the shell, h is the height of the shell, and dx is the thickness of the shell.

2. How is the volume of a solid calculated using cylindrical shells?

The volume of a solid can be calculated using cylindrical shells by breaking the solid into thin cylindrical shells and adding up the volumes of each shell using the formula V = ∫2πrh dx.

3. What is the difference between using cylindrical shells and other methods to find volume?

Cylindrical shells is a method for finding volume that involves integrating the surface area of a cylinder. Other methods, such as the disk method or the washer method, involve integrating the cross-sectional area of a solid.

4. Can the cylindrical shells method be used for any solid?

No, the cylindrical shells method can only be used for solids that have a cylindrical shape or can be broken down into cylindrical shells.

5. Are there any limitations to using the cylindrical shells method?

One limitation of using the cylindrical shells method is that it can be more difficult to set up the integral than other methods. Additionally, it may not be the most efficient method for finding volume in some cases.

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