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Integrale ridefinito.

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Socratis

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May 27, 2012, 2:28:00 PM5/27/12
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--- dy=o=0,01=cm --Intergrale del cono r=6.o.x, h=8.o.y.

Formula per trovare V = o^3=dy= o --> r^2*pi)/3h
V=o^3---> ((6.o.x)^2*pi)/3*8.o.y = 4.71238898o^3
.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3, Sum.int=301.59..
\/\/\/\/\/\/\/ 13o *o = 61.26105674 cm^3
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3

\/\/\/\/\/ 9o *o = 42.41150082 cm^3 Sum.int=117. 8097..
\/\/\/\/ 7o *o = 32.98672286 cm^3
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3

Naturalmente, la somma integrale del cono, da 0 a dy=dh
si trova per (r^2*pi)/3*h)*(n.dh)^2

es : int \0.5o\ = (4.71..)*(5o)^2 =117. 8097245.. o^3

Socratis.


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