cones versus cylinders: counterintuitive results re volume

the volume of the cylinder is L × pi × r2,
the volume V of the sphere is given by the formula V = 4/3 × pi ×r3.
the volume of the cone is 1/3 × pi × r2 × h.
-- http://www.mathleague.com/help/geometry/3space.htm

A. lateral surface area of cone

pi times radius times square root of radius squared plus height squared

-- http://mathworld.wolfram.com/SurfaceArea.html

B. The total surface area of a cone is the sum of the base area and lateral surface area:

pi times radius squared plus pi times radius times lateral height

-- http://www.sparknotes.com/testprep/books/sat2/math2c/chapter7section2.rhtml

According to the above, the lateral surface area is pi r l, where l equals the lateral height of the cone, which is pythagorean wise sq root of ht squared plus radius squared

C. surface area of cone
T = Pi r(r+s)
-- http://mathforum.org/dr.math/faq/formulas/faq.cone.html

D. The equation for finding the surface area of a right circular cone:

Surface area = pi × R × (R + (R2 + H2)1/2)

-- http://www.webcalc.net/calc/0040_help.php


Mt. Venus had at its peak, a cylinder, whose width was like a circle 1.75 miles wide, and whose height was 10 miles. Mt. Garage and Mt. Bentley had similar but shorter cylinders at their peaks. Mts Lemon, Myrtle and Heather had at their peak a cone, the height of the cone was 7.5 miles, it's base was 7 miles wide, and the cones depth, relative to what the depth of the mountain would have been if it's slopes continued on under the cone on the peak , at the same angle as before the level of the cone was reached, was 0.5 miles. Which mountain had a more voluminous object at its peak?

The volume of the object at the peak of Mt. Venus, is 10 x 3.14 x (0.87 squared), 31.4 x 0.7569 = 23.8.

The volume of the cone would be, counting the mountain under the cone as part of the cone, 1/3 x 3.14 x (3.5 squared) x 7.5 = 96.25.

The lateral surface area of the cone would be, according to formula A, square root of 3.5 squared + 7.5 squared, times 3.14 x 3.5, or 91, or square root of 68.5 which is 8.28 times 3.14 times 3.5. According to the other formula B it would be, 3.14 x 3.5 x 8.3 = 91. God knows why anyone would use formulas C and D. The volume the cone, not counting the volume of the mountain aside from the cone and underneath the cone would have if the mountain underneath the cone continued on under the cone at the same angles, would be approximately, 0.5 x 91 or 45.5. Thus the ratio between said cylinder and said cone in volume is 23.8 to 45.5.

A question is, has this formula been thrown off due to the squaring of numbers less then 1?

If we get rid of numbers less than one being squared, by multiplying everything by ten, we have:

The volume of the object at the peak of Mt. Venus, is 100 x 3.14 x (0.87 squared), 31.4 x 75.69 = 2377.

The volume of the cone would be, 1/3 x 3.14 x (35 squared) x 75 = 96162.

The lateral surface area of the cone would be, according to formula A, square root of 35 squared + 75 squared, times 3.14 x 35, or 910, 82.8 times 3.14 times 3.5. According to the other formula it would be, 3.14 x 3.5 x 83 = 912.

Thus we see that the shallow cone like structures, are in volume more voluminous than the cylindrical structure. This defies intuition, that a shallow cone, would be more voluminous than a towering cylinder, but, there you have it. If the area of the mountain underneath the cone is counted as part of the cone, the volume of the cone is even greater relative to the cylinder.