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The volume of the slice is (2*h*r^2)/3.
Make a cross section parallel to the bases. Let the cross section have a center P, intersect line AD at H, arc AB at E, arc AC at F and line AO at G. Let GH=x and AH=y. AGH and AOD are similar triangles, which means AD/DO = AH/HO, which is equivalent to h/r = y/x. PEG is a right triangle with PE=r and PG=r-x, which makes EG=sqrt(r^2-(r-x)^2).
Make a second cross section containing EF and perpendictular to the bases. The cross section is parallel to AD and BC. Let IJ be the line where this cross section intersects base O. A cross section for integration is rectangle EFJI, varying with respect to x over the range 0 to r.
The integral is Integ{0,r}[EF*EI]dx, which after substitutions EF=2*EG and EI=DH=AD-AH becomes:
Integ{0,r}[2*sqrt(r^2-(r-x)^2)*(h-h*x/r)]dx
The integral can be simplified to:
Integ{0,r}[(h/r) * (2*r - 2*x) * sqrt(2*r*x - x^2)]dx
The substitution u=2*r*x - x^2, du = (2*r - 2*x)dx, x=0 -> u=0, x=r -> u=r^2 simplifies the integral to:
Integ{0,r^2}[(h/r) * sqrt(u)]du
This integral easily evaluates to (2*h*r^2)/3. |
