Proof for Volume of a Segment of a SphereDate: 11/19/2001 at 02:56:55 From: S. Kozakavich Subject: Proof for Volume of a Segment of a Sphere I am in need of assistance in proving the volume of a truncated spherical cap (or a segment of a sphere I think it is also called). I think I should be subtracting the volume of a smaller top cap from the volume of the whole spherical cap, but am getting lost. I know what the end result should be, but I can't get there. Perhaps my problem is with labelling..... I am unsure. I would greatly appreciate any assistance you could offer. Thank you for your time. Date: 11/19/2001 at 04:04:05 From: Doctor Jeremiah Subject: Re: Proof for Volume of a Segment of a Sphere The shape you're talking about is indeed called a 'spherical cap'. You'll find an illustration at: http://www.mathforum.org/dr.math/faq/formulas/faq.sphere.html (scroll down to Spherical Cap). But I can't answer this question without more information. Do you have the radius of the sphere? If so, do you have the thicknesses of the big cap and little cap? If not, do you have the radius of at least one of the caps? It's important because to generate this proof we need to know what form our answer must take. I could just assume I know what you want and show you how, but if I guess wrong then I will have wasted a lot of time for both of us. Let me know what measurements you want the answer to depend upon and then I will be happy to help you. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 11/19/2001 at 09:42:32 From: Shelaine Kozakavich Subject: Re: Proof for Volume of a Segment of a Sphere Good morning Doctor Jeremiah, Thank you for your quick reply! Sorry about not supplying all the information earlier. Here is what I am given for my question to prove that the volume of the "truncated" spherical cap is: (Pi/6)(h)(3a^2 + 3b^2 + h^2) I am told that the top of a spherical cap is cut off by a plane parallel to the base. In the truncated cap, the base is a circle with radius a and the top of the circle with radius b, and the height is h. I was directed to subtract the volume of the smaller top cap from the volume of the whole spherical cap, but I do not end up with the above formula for the "truncated" spherical cap. Thank-you so much for your assistance here! S. Kozakavich Date: 11/19/2001 at 20:41:23 From: Doctor Jeremiah Subject: Re: Proof for Volume of a Segment of a Sphere Hi Shelaine, It could be done by finding the volume of two spherical caps and subtracting a smaller one from a larger one, and that might be the best way to go, but I am going to do it directly. Unfortunately we don't know the radius of the sphere, but we can figure it out: |-------a--------| |----b---| +++++ +++ +++ +++----b---+ +++ -----------+- + \ | / + | + \ | / + | h +-------a--\-----+ / + ------+- + + \ | / + + | + + R | R + + | Z + R \ | / R + | + + \ | / + + | + + \|/ + + | + + + -+- We have two congruent right angled triangles. b^2 + (Z+h)^2 = R^2 a^2 + Z^2 = R^2 What we need to do is combine these and get rid of z: Z^2 = R^2 - a^2 Z = (R^2 - a^2)^(1/2) b^2 + (Z+h)^2 = R^2 b^2 + Z^2 + 2hZ + h^2 = R^2 b^2 + (R^2 - a^2) + 2h(R^2 - a^2)^(1/2) + h^2 = R^2 2h(R^2 - a^2)^(1/2) = a^2 - b^2 - h^2 (R^2 - a^2)^(1/2) = (a^2 - b^2 - h^2)/(2h) R^2 - a^2 = (a^2 - b^2 - h^2)^2/(4h^2) And that means that: Z = (R^2 - a^2)^(1/2) Z = ( (a^2 - b^2 - h^2)^2/(4h^2) )^(1/2) Z = (a^2 - b^2 - h^2)/(2h) Now we have a Z that does not depend on R at all and we can use R with no reference to Z. To integrate a sphere we need to use a disk of dz thickness that has an area of Pi*x^2 as we move it from z = Z to z = Z+h +++++ +++ +++ +++-----------------+++ -----+- + | + | + | dz +-----+--------/|\--------+-----+ -+- + | + + | + + + + + + + + R^2 = x^2 + z^2 + + + + + + + + + + + + + + + +++ +++ +++ +++ +++++ z=Z+h / Volume = | Pi*x^2 dz <=== x^2 = R^2 - z^2 / z=Z z=Z+h / Volume = | Pi*(R^2 - z^2) dz / z=Z z=Z+h z=Z+h / / Volume = | Pi*R^2 dz - | Pi*z^2 dz / / z=Z z=Z z=Z+h z=Z+h / / Volume = Pi*R^2 | dz - Pi | z^2 dz / / z=Z z=Z z=Z+h z=Z+h | | Volume = Pi*R^2*z | - Pi*z^3/3 | | | z=Z z=Z Let's test the results of our integration. At this point, if the limits were -R and R (the whole sphere) we would have this: z=R z=R | | Volume = Pi*R^2*z | - Pi*z^3/3 | | | z=-R z=-R Volume = (Pi*R^2*(R) - Pi*R^2*(-R)) - (Pi*(R)^3/3 - Pi*(-R)^3/3) Volume = (Pi*R^2*R + Pi*R^2*R) - (Pi*R^3/3 + Pi*R^3/3) Volume = 2*Pi*R^3 - 2*Pi*R^3/3 Volume = 4*Pi*R^3/3 So now let's solve the actual problem. And with limits of Z and Z+h we get: Volume = (Pi*R^2*(Z+h) - Pi*R^2*(Z)) - (Pi*(Z+h)^3/3 - Pi*(Z)^3/3) Volume = Pi*R^2*h - Pi*(Z+h)^3/3 + Pi*Z^3/3 Volume = Pi*R^2*h - Pi*(Z^3+3hZ^2+3Zh^2+h^3)/3 + Pi*Z^3/3 Volume = Pi*R^2*h - Pi*(3hZ^2+3Zh^2+h^3)/3 Volume = Pi*R^2*h - Pi*h*Z^2 - Pi*h^2*Z - Pi*h^3/3 And since Z^2 = R^2 - a^2: Volume = Pi*R^2*h - Pi*h(R^2-a^2) - Pi*h^2*Z - Pi*h^3/3 Volume = Pi*R^2*h - Pi*h*R^2 + Pi*h*a^2 - Pi*h^2*Z - Pi*h^3/3 Volume = Pi*h*a^2 - Pi*h^2*Z - Pi*h^3/3 And since Z = (a^2 - b^2 - h^2)/(2h): Volume = Pi*h*a^2 - Pi*h^2*(a^2 - b^2 - h^2)/(2h) - Pi*h^3/3 Volume = Pi*h*a^2 - Pi*h*(a^2 - b^2 - h^2)/2 - Pi*h^3/3 Volume = Pi*h*a^2 - Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h^3/2 - Pi*h^3/3 Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + 3*Pi*h^3/6 - 2*Pi*h^3/6 Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h^3/6 Volume = Pi*h*a^2/2 + Pi*h*b^2/2 + Pi*h*h^2/6 Volume = 3*Pi*h*a^2/6 + 3*Pi*h*b^2/6 + Pi*h*h^2/6 Volume = Pi*h*(3a^2 + 3b^2 + h^2)/6 Wasn't that messy!? Let me know if you want help doing it with two full spherical caps or if you need more details about the way I did it here. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ Date: 11/23/2001 at 00:17:02 From: Shelaine Kozakavich Subject: Re: Proof for Volume of a Segment of a Sphere I want to thank you for all of your time and expertise. You provide a valuable resource here with this Web site and your assistance with problem questions for those who ask for your help. It is certainly appreciated! Keep up the good work. Thanks again, Shelaine :) |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/