Volume of ice cream cone triple integral

In summary: In that case, the problem is not solvable because you need a fourth equation to close the system. You have three equations in four unknowns: ##\rho,~\theta,~\phi##, and the fourth equation would come from the cone. So you need to be given something like the height of the cone or the angle of the cone or something.
  • #1
Panphobia
435
13

Homework Statement


Find the triple integral for the volume between a hemisphere centred at ##z=1## and cone with angle ##\alpha##.

The Attempt at a Solution



What I tried to do first was to get the radius of the hemisphere in terms of the angle ##\alpha##. In this case the radius is ##\tan \alpha##. I already figured out this integral in cylindrical polar coordinates and Cartesian coordinates. I am having a lot of trouble with spherical coordinates. I am trying to get ##\rho## for the hemisphere by drawing the projection of the shape on the xz plane and trying to get a formula for a radial ray that hits the hemisphere. The formula for a circle that is shifted up by one is ##x^2+(z-1)^2=\tan^2 \alpha##, this is the part where I am stuck trying to find the equation for ##\rho## in terms of the angle ##\phi## of the radial ray. The limits for ##\theta## and ##\phi## are really easy, but again ##\rho## I just can't seem to get.
 
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  • #2
Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
 
  • #3
mfb said:
Do you need the integral in spherical coordinates? It is messy, and the other coordinate systems are much easier.

You are missing y2 in your hemisphere formula.
You can find z as function of the two angles, plug it into the equation for the hemisphere (in the right place) and solve for ρ.
So I will have to use the quadratic formula to solve for ##\rho##?
 
  • #5
I disagree with mfb on this problem. In fact spherical coordinates is the natural and easiest way to work this problem. You didn't state it in your problem but I assume you are talking about the upper half of the sphere of radius ##1## centered at ##(0,0,1)##. Its equation is ##x^2+y^2+(z-1)^2 = 1## or, expanding it, ##x^2 + y^2 + z^2 = 2z##. If you change that to spherical coordinates you should get ##\rho = 2\cos\phi##. So the ##\rho,~\theta,~\phi## limits are all very easy.
 
  • #6
How could you fix the radius of the half-sphere to 1?
 
  • #7
mfb said:
How could you fix the radius of the half-sphere to 1?
I couldn't. I scanned it too quickly and assumed it was the "standard" problem you normally see in calculus books.
 

1. What is the formula for calculating the volume of an ice cream cone using a triple integral?

The formula for calculating the volume of an ice cream cone using a triple integral is
V = ∫∫∫ dV = ∫∫∫ r dr dθ dz, where r is the radius of the cone, θ is the angle of the cone, and z is the height of the cone.

2. How do you determine the limits of integration for the triple integral when calculating the volume of an ice cream cone?

The limits of integration for the triple integral depend on the shape and dimensions of the ice cream cone. Generally, the limits for r would be 0 to the radius of the cone, the limits for θ would be 0 to 2π, and the limits for z would be 0 to the height of the cone.

3. What units are used when expressing the volume of an ice cream cone using a triple integral?

The units used for the volume of an ice cream cone using a triple integral would be cubic units, such as cm³ or m³, depending on the units used for the dimensions of the cone.

4. How does the volume of an ice cream cone change if the cone has a different shape?

The volume of an ice cream cone would change if the cone has a different shape, as the dimensions and limits of integration for the triple integral would also change. The formula for calculating the volume would remain the same, however, the values for r, θ, and z would be different.

5. Can the volume of an ice cream cone be calculated using only a single integral instead of a triple integral?

No, the volume of an ice cream cone cannot be calculated using a single integral. A triple integral is necessary because the dimensions of the cone are in three different directions (radius, angle, and height), therefore, a single integral would not take into account all of these dimensions.

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