- #1
Pjennings
- 17
- 0
The volume of a cone =
1
- B H where B is the base of the cone and H is its height.
3
We can think about a cone as the line y = x rotated with respect to the y axis. The volume of a parabaloid =
1
- B H
2
and a parabaloid is the line y = x^2 rotated with respect to the y axis. So what if you wanted to know the what line has the equation 1/4*B*H or in general K*B*H, where K is a constant.
Let y= a*x^n and y=H, which is the height. Then x = (y/a)^1/n.
Then the integral for the volume of revolution becomes
[0]\int[/H](y/a)^1/n dy
Using a substitution and integrating we get that Volume= a*pi*((H/a)^((2+n)/n))/((2+n)/n)
We know that V = K*pi*x^2*H, and therefore we can equate the two equations for volume, and using the relationship that H= ax^n we get
a*pi*((H/a)^((2+n)/n))/((2+n)/n) = K*a*pi*x^2*x^n
Pi and a cancel out on both sides. Using the relationships x^2 =(y/a)^2/n and y=h we get
((H/a)^((2+n)/n))/((2+n)/n))= K*((H/a)^((2+n)/n))
and ((H/a)^((2+n)/n)) cancels out on both sides leaving
K= n/(n+2) where n is the value to which x is raised: x^n. Solving for n we get that n=2k/(1-k). Since we can think of a cone as the line y = x rotated with respect to the y-axis, then we need to let n=1 and see if k=1/3, which it does. You could also use this to determine what shape has the equation 1/4B*H by letting k=1/4, and you get n=2/3. So the line y= x^(2/3) rotated with respect to the y-axis has the equation 1/4*B*H
Having done all of this work, does anyone have any suggestions on where to go from here with this work?
1
- B H where B is the base of the cone and H is its height.
3
We can think about a cone as the line y = x rotated with respect to the y axis. The volume of a parabaloid =
1
- B H
2
and a parabaloid is the line y = x^2 rotated with respect to the y axis. So what if you wanted to know the what line has the equation 1/4*B*H or in general K*B*H, where K is a constant.
Let y= a*x^n and y=H, which is the height. Then x = (y/a)^1/n.
Then the integral for the volume of revolution becomes
[0]\int[/H](y/a)^1/n dy
Using a substitution and integrating we get that Volume= a*pi*((H/a)^((2+n)/n))/((2+n)/n)
We know that V = K*pi*x^2*H, and therefore we can equate the two equations for volume, and using the relationship that H= ax^n we get
a*pi*((H/a)^((2+n)/n))/((2+n)/n) = K*a*pi*x^2*x^n
Pi and a cancel out on both sides. Using the relationships x^2 =(y/a)^2/n and y=h we get
((H/a)^((2+n)/n))/((2+n)/n))= K*((H/a)^((2+n)/n))
and ((H/a)^((2+n)/n)) cancels out on both sides leaving
K= n/(n+2) where n is the value to which x is raised: x^n. Solving for n we get that n=2k/(1-k). Since we can think of a cone as the line y = x rotated with respect to the y-axis, then we need to let n=1 and see if k=1/3, which it does. You could also use this to determine what shape has the equation 1/4B*H by letting k=1/4, and you get n=2/3. So the line y= x^(2/3) rotated with respect to the y-axis has the equation 1/4*B*H
Having done all of this work, does anyone have any suggestions on where to go from here with this work?