問一道附加數學題(定積分)
回答 (2)
△V
= volume of disc of height △h
= π r^2 △h
= π {R^2 - (R-h)^2 }△h
= π {2Rh - h^2 }△h
Hence,
V= integrate_0_to_h{ π (2Rh - h^2 )dh} = π{Rh^2 - (h^3)/3} = (πh^2)(3R -h)/3
2007-01-15 23:56:49 補充:
由 R=5 及 r=3 , 得 h=5 - 4=1球體體積= 500π/3球缺體積= (π1^2)(3x5 -1)/3 =14π/3圓柱體積= π3^2 x (2x4) =72π所餘體積= 500π/3 - 2x14π/3 - 72π =256π/3
2007-01-16 23:33:23 補充:
好得太誇張啦!!
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