Calculating the Volume of a Spherical Cap - A Quick Guide

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SUMMARY

The volume of a spherical cap can be calculated using calculus, specifically through the use of integrals. The discussion highlights the method of using spherical coordinates and the Jacobian determinant, which is R²sin(T). The volume formula derived is V = (2/3)πhR², where h is the height of the cap and R is the radius of the sphere. The integration domain is defined as D = [0,r] x [0,2π] x [0,T₀], with T₀ determined by the relationship cos(T₀) = (r-h)/r.

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ok i think I've got it. thanks
 
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You need to show your work before you get help. What are your thoughts/ideas on this problem?
 
I do not think that there is any difference between a "spherical cap" and a hemisphere.

Finding a volume using calculus generally involves an integral, there are several methods which can be used to solve such a problem. Do you recall seeing something having to do with "disks"?
 
Let F defined on [0,r]x[0,2\pi]x[0,\pi] by
F(R, S, T)=(R\cos(S)\sin(T), R\sin(S)\sin(T), R\cos(T))
(the change into spherical coordinates).
The absolute value of its Jacobian is R^2\sin(T).

Your domain of integration (cap) is defined by
{(x, y, z) : x^2 + y^2 + z^2 \leq r^2 , r-h \leq z \leq r}
and its pre-image through F is
D=[0,r]x[0,2\pi]x[0,T_0],
where T_0 satisfies \cos(T_0)=(r-h)/r=1-h/r.

Volume = \int_{D}R^2\sin(T)dRdSdT
= (\int_0^rR^2dR).(\int_0^{2\pi}dS).(\int_…
= 1/3r^3.2\pi.(1-\cos(T_0)) = 2\pi.h.r^2/3.
QED.
 
Please do not erase a post after it has been resolved! Other people are sure to have the same question and can still learn from thread.
 

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