Volume of Revolution with different areas
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We have this assignment in Maths class regarding wine glasses. Our task was to determine why two different shaped functions with the same cross-sectional areas will have different volumes of revolution. In the context of our assignment, the two functions represented a wine glass bowl and a champagne flute bowl with equations f(x)=2csc(x+pi/2)-2 and g(x)=10csc(x+pi/2)-2 respectively. The range for the wine glass function is 0<=y<=5 and the range for the champagne flute is 0<=y<=7.36. They must be revolved around the y axis. The problem I am facing is explaining why the volumes of the two functions in the given ranges are different even though they have the same cross sectional area.
Hope someone can help,
Thanks
integration
asked Aug 27 at 0:41
Deep Patel
73
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We have this assignment in Maths class regarding wine glasses. Our task was to determine why two different shaped functions with the same cross-sectional areas will have different volumes of revolution. In the context of our assignment, the two functions represented a wine glass bowl and a champagne flute bowl with equations f(x)=2csc(x+pi/2)-2 and g(x)=10csc(x+pi/2)-2 respectively. The range for the wine glass function is 0<=y<=5 and the range for the champagne flute is 0<=y<=7.36. They must be revolved around the y axis. The problem I am facing is explaining why the volumes of the two functions in the given ranges are different even though they have the same cross sectional area.
Hope someone can help,
Thanks
integration
asked Aug 27 at 0:41
Deep Patel
73
Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We have this assignment in Maths class regarding wine glasses. Our task was to determine why two different shaped functions with the same cross-sectional areas will have different volumes of revolution. In the context of our assignment, the two functions represented a wine glass bowl and a champagne flute bowl with equations f(x)=2csc(x+pi/2)-2 and g(x)=10csc(x+pi/2)-2 respectively. The range for the wine glass function is 0<=y<=5 and the range for the champagne flute is 0<=y<=7.36. They must be revolved around the y axis. The problem I am facing is explaining why the volumes of the two functions in the given ranges are different even though they have the same cross sectional area.
Hope someone can help,
Thanks
integration
asked Aug 27 at 0:41
Deep Patel
73
We have this assignment in Maths class regarding wine glasses. Our task was to determine why two different shaped functions with the same cross-sectional areas will have different volumes of revolution. In the context of our assignment, the two functions represented a wine glass bowl and a champagne flute bowl with equations f(x)=2csc(x+pi/2)-2 and g(x)=10csc(x+pi/2)-2 respectively. The range for the wine glass function is 0<=y<=5 and the range for the champagne flute is 0<=y<=7.36. They must be revolved around the y axis. The problem I am facing is explaining why the volumes of the two functions in the given ranges are different even though they have the same cross sectional area.
Hope someone can help,
Thanks
integration
asked Aug 27 at 0:41
Deep Patel
73
asked Aug 27 at 0:41
Deep Patel
73
asked Aug 27 at 0:41
Deep Patel
73
asked Aug 27 at 0:41
Deep Patel
73
73
Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10
add a comment |Â
Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10
Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10
add a comment |Â
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Revolving an infinitesimal surface near to and around its axis of rotation gives a greater volume than revolving a similar surface further away from it. For a square of area $dA = dx cdot dy$, rotated around the x-axis, $dV = pi (r+dx)^2cdot dy - pi r^2cdot dy approx 2pi rcdot dA$, $r$ the distance from it. This illustrates that the volume of the rotated solid is not a function of the cross-section's area alone, but also its distribution of area away from the axis of rotation (so to speak)
– LPenguin
Aug 27 at 0:54
Thanks for the reply, I understand most of it. However, could you elaborate further about the what you wrote in your first sentence. Why does a larger radius cause a large volume. I did not quite understand the dV part either.
– Deep Patel
Aug 27 at 1:00
I think the first sentence of LPenguin's comment is backwards. A larger radius of revolution will obviously sweep through a longer circumference and produce a greater volume. If you think of two toroids with the same cross section, the one with the larger outer diameter will be larger in volume.
– Phil H
Aug 27 at 5:08
See thats my problem. I can't exactly visualise why this will happen. I don't understand why a larger outer diameter will sweep a larger volume. In the end, the same cross-sectional area is being rotated around the axis and hence both shapes will have the same volume.
– Deep Patel
Aug 27 at 6:10