What shape results from integrating the area of a circle?

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    Circle Integration
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Discussion Overview

The discussion revolves around the integration of the area of a circle and the resulting geometric shape. Participants explore the implications of integrating the circumference to find the area and then integrating the area to derive a volume, questioning the nature of the resulting shape and its relation to known geometric figures like spheres and cones.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant notes that integrating the circumference of a circle yields the area, represented by the formula ∏r².
  • Another participant expresses confusion about integrating the area, arriving at ∏r³/3, and questions what shape this represents, initially suspecting it to be a sphere.
  • A participant suggests that the integration with respect to r results in the volume of a right isosceles cone, describing the process of visualizing this through circular cross sections.
  • There is a mention of the connection between the surface area of a sphere and its volume, with the surface area being 4∏r² and the volume derived from integrating this surface area.
  • One participant seeks clarification on whether a right angle triangle with two 45° angles revolved around an axis could describe the cone, which is affirmed by another participant.

Areas of Agreement / Disagreement

Participants generally agree on the geometric interpretation of the integration process, particularly regarding the cone. However, there is no consensus on the implications of integrating the area of a circle, as some participants question the resulting shape and its relationship to a sphere.

Contextual Notes

Participants express uncertainty regarding the differentiation involved in the integration process, specifically questioning the appropriate variables to use.

Rob K
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Hi there,

I am trying to understand calculus as concerns circles and I can clearly see that the integral of a circumference is an area:
[itex]\int2∏r[/itex] = ∏r[itex]^{2}[/itex]

but what do I get if I integrate the area, I get
∏r[itex]^{3}[/itex]/3

I am confused as to what this shape would be, I kind of was expecting a sphere, but the formula for a sphere is:
4∏r3/3

plus a little technical point: when differentiating this, is it dr/dx or dy/dr, or am I totally off the mark?

Thanks in advance

Rob K
 
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Rob K said:
Hi there,

I am trying to understand calculus as concerns circles and I can clearly see that the integral of a circumference is an area:
[itex]\int2∏r[/itex] = ∏r[itex]^{2}[/itex]

but what do I get if I integrate the area, I get
∏r[itex]^{3}[/itex]/3

I am confused as to what this shape would be, I kind of was expecting a sphere, but the formula for a sphere is:
4∏r3/3

plus a little technical point: when differentiating this, is it dr/dx or dy/dr, or am I totally off the mark?

Thanks in advance

Rob K

The integration is with respect to r. The volume you get is a right isosceles cone with base and height both r. Here's how you can think of it. Take a circle of radius r and think of an r axis perpendicular to the circle through its center. The circle is at distance r from the origin on this axis. If you slide the circle to a larger distance its radius increases accordingly. Sliding the circle in the r direction generates the cone and the integral you are calculating represents calculating the volume of that cone by circular cross sections. And it gets the correct answer of 1/3*Area of base*height for a cone.

The connection between circumference and area of a circle by integration also works for a sphere, but the connection is between surface area and volume. The surface area is ##4\pi r^2## and if you integrate that you get the volume ##\frac 4 3 \pi r^3##. This is the calculation of the volume of a sphere by spherical shells.
 
Thank you, that is very useful as a visualization of integration. Let me get this a little clearer in my head. Is this another way to describe it. a right angle triangle with two 45˚ angles then revolved around the z axis, assuming the z axis is one of the non hypotenuse sides, as you would create it in 3d modelling?
 
Rob K said:
Thank you, that is very useful as a visualization of integration. Let me get this a little clearer in my head. Is this another way to describe it. a right angle triangle with two 45˚ angles then revolved around the z axis, assuming the z axis is one of the non hypotenuse sides, as you would create it in 3d modelling?

Yes, that also describes a right isosceles cone. Both legs of the triangle have length r.
 
The area of the surface of a sphere is 4πr2. Integrate that to get the volume of a sphere.
 
Thank you.
 

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