Volume of Revolution By Shell Method.

In summary: After I drew the region and found the points of intersection, it was a lot easier to pick the appropriate coordinates.Yes, that makes sense. I believe that is what happened for me as well. After I drew the region and found the points of intersection, it was a lot easier to pick the appropriate coordinates.
  • #1
OmniNewton
105
5

Homework Statement


Find the volume of the solid obtained by rotating y=x^2 and x = y^2 about y=2

Homework Equations


V= 2πrh

The Attempt at a Solution


When I had constructed the graph I determined the following:
r= 2-(y(1/2))
h= 1-y2

after converting V into an integral I applied fundamental theorem of calculus.
2-2y^2-y^(1/2) + y^(5/2)
evaluating the expression from lower limit =0 to upper limit = 1

I obtained a value with a denominator of 21.

The final answer I should receive is 31π/30 (I proved this by disk method and this is the stated answer).
 
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  • #2
OmniNewton said:

Homework Statement


Find the volume of the solid obtained by rotating y=x^2 and x = y^2 about y=2

Homework Equations


V= 2πrh

The Attempt at a Solution


When I had constructed the graph I determined the following:
r= 2-(y(1/2))
h= 1-y2
I have r = 2 - y and h = ##y^{1/2} - y^2##.
OmniNewton said:
after converting V into an integral I applied fundamental theorem of calculus.
2-2y^2-y^(1/2) + y^(5/2)
evaluating the expression from lower limit =0 to upper limit = 1

I obtained a value with a denominator of 21.

The final answer I should receive is 31π/30 (I proved this by disk method and this is the stated answer).
I get 31π/30 for the volume.
 
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Likes OmniNewton
  • #3
Mark44 said:
I have r = 2 - y and h = ##y^{1/2} - y^2##.

I get 31π/30 for the volume.

Oh ok I see how you got that. Let me give that a try.
 
  • #4
Thank you so much it worked out! I was at that problem for a good two hours. I erased so much my page nearly ripped. After taking a break to work on another problem and then coming back with your guidance I was able to figure it out.
 
  • #5
OmniNewton said:
Thank you so much it worked out! I was at that problem for a good two hours. I erased so much my page nearly ripped. After taking a break to work on another problem and then coming back with your guidance I was able to figure it out.
I hope you're drawing sketches of 1) the region being revolved, with the two curves and important points labeled, and 2) the solid of revolution. Based on what you showed, I think you did, but sometimes people think they can save time by skipping that step. It's almost always a false economy...
 
  • #6
Mark44 said:
I hope you're drawing sketches of 1) the region being revolved, with the two curves and important points labeled, and 2) the solid of revolution. Based on what you showed, I think you did, but sometimes people think they can save time by skipping that step. It's almost always a false economy...

Yes, my steps for these problems are.

1. draw the region
2. find necessary point of intersections
3. split the axis of rotation for disk method or split the perpendicular axis of rotation for shell method
4. draw atleast one disk or shell (more may be needed)
5. then work from there.

My problem is sometimes I still have a hard time noticing key things like radius and height (getting better with practice)
 
  • #7
Very good. I think the problem you had was not realizing what the appropriate x- or y-coordinates were for writing the radius and the width of the area element. You'll get better at this with practice. Other than that, you seem to be on the right track.
 
  • #8
Mark44 said:
Very good. I think the problem you had was not realizing what the appropriate x- or y-coordinates were for writing the radius and the width of the area element. You'll get better at this with practice. Other than that, you seem to be on the right track.
Yes, that makes sense. I believe that is what happened
 

1. What is the shell method for finding volume of revolution?

The shell method is a mathematical technique used to find the volume of a solid generated by rotating a two-dimensional shape around an axis. It involves slicing the solid into thin cylindrical shells and calculating their volumes.

2. When is the shell method used?

The shell method is typically used when the axis of rotation is parallel to the axis of integration and the shape being rotated is not a perfect disk or washer. It is also useful when the shape being rotated is easier to express in terms of its height rather than its radius.

3. How is the volume of a shell calculated?

To calculate the volume of a shell, we use the formula V = 2πrhΔx, where r is the distance from the axis of rotation to the outer edge of the shell, h is the height of the shell, and Δx is the thickness of the shell. This formula can be derived by considering the volume of a cylinder and subtracting the volume of the hole in the middle created by the axis of rotation.

4. What are the steps for using the shell method to find volume of revolution?

The steps for using the shell method to find volume of revolution are as follows:
1. Determine the axis of rotation and the limits of integration.
2. Express the function defining the shape being rotated in terms of its height, h.
3. Calculate the distance from the axis of rotation to the outer edge of the shell, r.
4. Write the formula V = 2πrhΔx.
5. Integrate the formula with respect to h, using the given limits of integration.
6. Evaluate the integral and simplify the result to find the volume of the solid.

5. Can the shell method be used for any shape?

No, the shell method can only be used for shapes that can be expressed in terms of their height, such as rectangles, triangles, and semicircles. It is not suitable for shapes that are difficult to express in terms of a single variable, such as ellipses or parabolas.

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