Editing 2509: Useful Geometry Formulas

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The 3D right circular cone commonly depicted by this drawing would have a volume of πr<sup>2</sup>h/3 where r=a=b.  The area of the "lower" surface would be πr<sup>2</sup>, while the surface area of the upper conical surface would be πr√(h<sup>2</sup> + r<sup>2</sup>).  Neither of these areas can correspond with the caption in the comic, nor does the total surface area (the sum of these two).
 
The 3D right circular cone commonly depicted by this drawing would have a volume of πr<sup>2</sup>h/3 where r=a=b.  The area of the "lower" surface would be πr<sup>2</sup>, while the surface area of the upper conical surface would be πr√(h<sup>2</sup> + r<sup>2</sup>).  Neither of these areas can correspond with the caption in the comic, nor does the total surface area (the sum of these two).
  
If we do not assume that a = b, this drawing could also depict a right elliptic cone.  The volume of the elliptic cone would be <sup>π</sup>/<sub>3</sub>&nbsp;abh.  The area of the lower surface would be πab and the area of the curved upper surface would be <br>2a√(b<sup>2</sup>&nbsp;+&nbsp;h<sup>2</sup>)&nbsp;<sub>0</sub>∫<sup>1</sup>&nbsp;√(<sup>a²h²(t²-1)&nbsp;-&nbsp;b²(a²+h²t²)</sup>/<sub>a²(t²-1)(b²+h²)</sub>)&nbsp;dt.  
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If we do not assume that a = b, this drawing could also depict a right elliptic cone.  The volume of the elliptic cone would be <sup>π</sup>/<sub>3</sub>&nbsp;abh.  The area of the lower surface would be πab and the area of the curved upper surface would be <br>2a√(b<sup>2</sup>&nbsp;+&nbsp;h<sup>2</sup>)&nbsp;<sup>1</sup>∫<sub>0</sub>&nbsp;√(<sup>a²h²(t²-1)&nbsp;-&nbsp;b²(a²+h²t²)</sup>/<sub>a²(t²-1)(b²+h²)</sub>)&nbsp;dt.  
  
 
; Bottom Left - Two ellipses joined vertically, or Cylinder
 
; Bottom Left - Two ellipses joined vertically, or Cylinder
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The non-overlapping parts of the 2D shape are composed of the rectangle formed by the major axes of the two ellipses and the vertical lines, plus half of the top ellipse and half of the bottom ellipse.  The area of the rectangle is dh, and the area of an ellipse with semimajor axis d/2 and semiminor axis r is πrd/2.  The total area is A = d(πr/2 + h), which is captioned below the figure.
 
The non-overlapping parts of the 2D shape are composed of the rectangle formed by the major axes of the two ellipses and the vertical lines, plus half of the top ellipse and half of the bottom ellipse.  The area of the rectangle is dh, and the area of an ellipse with semimajor axis d/2 and semiminor axis r is πrd/2.  The total area is A = d(πr/2 + h), which is captioned below the figure.
  
A 3D right circular prism (cylinder) would have a volume of πr<sup>2</sup>h and a surface area of 2πr<sup>2</sup> + πdh, or 2πr(r + h) since in this case d = 2r.  The area of each flat surface would be πr<sup>2</sup>.  If we do not assume d = 2r, then the lateral surface area of the right elliptic cylinder is 4h&nbsp;<sub>0</sub><sup>1</sup>&nbsp;√(<sup>1&nbsp;-&nbsp;t²(1-4r²/d²)</sup>/<sub>1&nbsp;-&nbsp;t²</sub>)&nbsp;dt. The volume is <sup>π</sup>/<sub>2</sub>&nbsp;rdh.  
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A 3D right circular prism (cylinder) would have a volume of πr<sup>2</sup>h and a surface area of 2πr<sup>2</sup> + πdh, or 2πr(r + h) since in this case d = 2r.  The area of each flat surface would be πr<sup>2</sup>.  If we do not assume d = 2r, then the lateral surface area of the right elliptic cylinder is 4h&nbsp;<sub>0</sub><sup>1</sup>&nbsp;√(<sup>1&nbsp;-&nbsp;t²(1-4r²/d²)</sup>/<sub>1&nbsp;-&nbsp;t²</sub>)&nbsp;dt. The volume is <sup>π</sup>/<sub>2</sub>&nbsp;rdh.  
  
 
; Bottom Right - Parallel Hexagon, or Prism
 
; Bottom Right - Parallel Hexagon, or Prism

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