Let me think: if the vertical axis of a cone is thought of as time, then evolving up looks like a growing circle until height^2 + (height/slope)^2 = radius^2 where slope is the slope of the cone, and radius is that of the larger sphere. Afterwards, it follows h^2 +r^2 = R^2 This generalizes to n dimensions (a slice of a circle (pie) is the simplest significant case in 2d) When h is the height of the cone alone: That's the integral from 0 to sqrt(R^2 -h^2) of ball(r)dr plus the integral from sqrt(R^2 -h^2) to R of ball(sqrt(R^2-r^2))dr Since the ball function is strictly cx^k, for the ball's volume in k dimensions, the first integral is trivial and returns cx^(k+1)/(k+1), while the second integral is what you get by slicing the top off of (beheading) a ball, and can be solved generally by substituting Rsin(t)=r, you get int(ball(Rcos(t))×cos(t))dt, albeit with new bounds, and ball being cx^k, we get cR^k int(cos(t)^(k+1))dt, which is the limit of my knowledge, please help
Sorry that substitution was supposed to be sin(t)=r/R, having factored R out before any substitution, which gets the R^k in front of the integral instead of R^(k+1)... Or does it? There might be a factor of R in error. But that's a constant here, so it doesn't change the elemental solution relating to x
A related "famous" problem: Assume the ice cream is spherical. Find the radius of the ice cream that maximizes the amount of ice cream inside the cone in terms of the vertex angle of the cone, theta and the height of the cone, h. I get r = h * sin(theta/2)/{[1-sin(theta/2)][1+2sin(theta/2)]}
I didn't understand why the radius rho is bounded by sqrt(8) everywhere. When I tried to describe it, I found out it depended on phi, but apparently it doesn't? So you define the origin at (0,0,0) on your graph. Imagine the center of the hemisphere is C=(1,0,0), so (1+sqrt(8),0,0) is in the hemisphere. (1+sqrt(3),sqrt(5),0) is inside the top hemisphere, and at the top surface of the hemisphere (distance to C is sqrt(8)) The distance to the origin equals sqrt(9+2sqrt(3)) > sqrt(8). I think I missed something, can you help me please?
r^2=x^2+y^2+z^2, that's the equation for a sphere of radius r, so we can rearrange it to get z^2=r^2-x^2-y^2, z=±√[r^2-x^2-y^2], and we choose to take the positive √ because we're using the top side of the sphere. By inspecting the equation in the video we can see that 8 must be r^2, therefore r=√8. The reason it's bounded by √8 is because that's the upper bound of r in the given region between the two curves.
"r^2=x^2+y^2+z^2, that's the equation for a sphere of radius r, so we can rearrange it to get z^2=r^2-x^2-y^2, z=±√[r^2-x^2-y^2], and we choose to take the positive √ because we're using the top side of the sphere. By inspecting the equation in the video we can see that 8 must be r^2, therefore r=√8." I already got that. I started writing a long thing about what I saw as the problem and then I realized what went wrong. For some reason, I thought the hemisphere on the top was centered around C(1,0,0) (I don't know why I got the idea, I think it's because of the drawing), while it's centered around (0,0,0) and we take just the part above sqrt(x²+y²) and it didn't make sense. Then I just thought a little bit about it and now it's clear. Even though you didn't help, thanks for trying to answer.
@@PackSciences You can always take the parameterization and use the equations to derive the bounds.If you substitute x=rsinθcosφ,y=rsinθsinφ,z=rcosθ to the second equation( x^2+y^2+z^2=8) then you get r^2=8 which gives precisely what you want.
Isn't phi the horizontal and theta the vertical? And shouldn't theta be pi/2 to pi/4 since Phi represents the angle starting from z-axis and ending in the xy plane?
Dr. Peyam, your videos are perfect for me. I think you are the sum of your positive divisors.
OMG, best compliment ever ❤️
This video has helped me a lot¡¡¡
Could you do a video solving the jacobian for this?
Let me think: if the vertical axis of a cone is thought of as time, then evolving up looks like a growing circle until height^2 + (height/slope)^2 = radius^2 where slope is the slope of the cone, and radius is that of the larger sphere. Afterwards, it follows h^2 +r^2 = R^2
This generalizes to n dimensions (a slice of a circle (pie) is the simplest significant case in 2d)
When h is the height of the cone alone:
That's the integral from 0 to sqrt(R^2 -h^2) of ball(r)dr plus the integral from sqrt(R^2 -h^2) to R of ball(sqrt(R^2-r^2))dr
Since the ball function is strictly cx^k, for the ball's volume in k dimensions, the first integral is trivial and returns cx^(k+1)/(k+1), while the second integral is what you get by slicing the top off of (beheading) a ball, and can be solved generally by substituting Rsin(t)=r, you get int(ball(Rcos(t))×cos(t))dt, albeit with new bounds, and ball being cx^k, we get cR^k int(cos(t)^(k+1))dt, which is the limit of my knowledge, please help
Sorry that substitution was supposed to be sin(t)=r/R, having factored R out before any substitution, which gets the R^k in front of the integral instead of R^(k+1)... Or does it? There might be a factor of R in error. But that's a constant here, so it doesn't change the elemental solution relating to x
A related "famous" problem: Assume the ice cream is spherical. Find the radius of the ice cream that maximizes the amount of ice cream inside the cone in terms of the vertex angle of the cone, theta and the height of the cone, h. I get r = h * sin(theta/2)/{[1-sin(theta/2)][1+2sin(theta/2)]}
I didn't understand why the radius rho is bounded by sqrt(8) everywhere.
When I tried to describe it, I found out it depended on phi, but apparently it doesn't?
So you define the origin at (0,0,0) on your graph.
Imagine the center of the hemisphere is C=(1,0,0), so (1+sqrt(8),0,0) is in the hemisphere.
(1+sqrt(3),sqrt(5),0) is inside the top hemisphere, and at the top surface of the hemisphere (distance to C is sqrt(8))
The distance to the origin equals sqrt(9+2sqrt(3)) > sqrt(8).
I think I missed something, can you help me please?
r^2=x^2+y^2+z^2, that's the equation for a sphere of radius r, so we can rearrange it to get
z^2=r^2-x^2-y^2,
z=±√[r^2-x^2-y^2], and we choose to take the positive √ because we're using the top side of the sphere. By inspecting the equation in the video we can see that 8 must be r^2, therefore r=√8.
The reason it's bounded by √8 is because that's the upper bound of r in the given region between the two curves.
"r^2=x^2+y^2+z^2, that's the equation for a sphere of radius r, so we can rearrange it to get
z^2=r^2-x^2-y^2,
z=±√[r^2-x^2-y^2], and we choose to take the positive √ because we're using the top side of the sphere. By inspecting the equation in the video we can see that 8 must be r^2, therefore r=√8." I already got that.
I started writing a long thing about what I saw as the problem and then I realized what went wrong.
For some reason, I thought the hemisphere on the top was centered around C(1,0,0) (I don't know why I got the idea, I think it's because of the drawing), while it's centered around (0,0,0) and we take just the part above sqrt(x²+y²) and it didn't make sense. Then I just thought a little bit about it and now it's clear. Even though you didn't help, thanks for trying to answer.
@@PackSciences You can always take the parameterization and use the equations to derive the bounds.If you substitute x=rsinθcosφ,y=rsinθsinφ,z=rcosθ to the second equation( x^2+y^2+z^2=8) then you get r^2=8 which gives precisely what you want.
Isn't phi the horizontal and theta the vertical?
And shouldn't theta be pi/2 to pi/4 since Phi represents the angle starting from z-axis and ending in the xy plane?
I would like you to prove that rho^2 sin ( rho ) is the jacobian.
Inter3sting