Ann_M Posted March 12, 2004 Share Posted March 12, 2004 hi im trying to prove the volume of a sphere using triple integration and im stuck at how to integrate: ((R^2)- (r^2))^1/2 - {- ((R^2) - (r^2))}^1/2 r dr ive changed my limits to cylndrical limits and R is the radius. ive done the working out but im not getting ne where. please helpppppppp. Link to comment Share on other sites More sharing options...
YT2095 Posted March 12, 2004 Share Posted March 12, 2004 Hi Ann, sorry to kinda butt in, but just out of curiosity, I thought Pi was needed in all calcs to do with circles and stuff, and didn`t see it in your equasion? now I can`t even ballance my own cheque book and so am certainly not a mathematician! I wondered how you can work out your problem without Pi? thnx Link to comment Share on other sites More sharing options...
Sayonara Posted March 12, 2004 Share Posted March 12, 2004 That may not be the entire formula; just the part that needs integration. Link to comment Share on other sites More sharing options...
Dave Posted March 12, 2004 Share Posted March 12, 2004 Ann_M said in post # :hi im trying to prove the volume of a sphere using triple integration and im stuck at how to integrate: ((R^2)- (r^2))^1/2 - {- ((R^2) - (r^2))}^1/2 r dr ive changed my limits to cylndrical limits and R is the radius. ive done the working out but im not getting ne where. please helpppppppp. I think a change of co-ordinates into spherical co-ordinates may be the ones you're looking for, not cylindrical. The triple integration then just becomes trivial. Link to comment Share on other sites More sharing options...
Ann_M Posted March 12, 2004 Author Share Posted March 12, 2004 pi is in der but as the last integral, the integrals can be seen in the attachment these are the cylndrical co-ordinates. Try working it out and then u will c how i have got the integral above. I cant do this using spherical co-ordinates and have to do it either the original way or by changing the co-ordinated to cyclindrical which i find easier, as i have tried to do it normally ie dz dy dx and got very confused. by the way have i got my limits for the second integral correct, im thinking it should be -R to R and not 0 to R, but not sure, please correct me on that as well. thanks Link to comment Share on other sites More sharing options...
Ann_M Posted March 12, 2004 Author Share Posted March 12, 2004 oh and it has to equal the volume of a sphere as the final answer. Link to comment Share on other sites More sharing options...
YT2095 Posted March 12, 2004 Share Posted March 12, 2004 WOW! I`ll not pretend to understand any of that appart from knowing now that Pi`s in there somewhere. but it does look nice I think you`re probably alot smarter than me! Link to comment Share on other sites More sharing options...
Ann_M Posted March 12, 2004 Author Share Posted March 12, 2004 help me den please im sooo confused i get 8/3 pi r ^2 i dont know wat im doing wrong, try it ur self and c wat u get. Link to comment Share on other sites More sharing options...
Dave Posted March 13, 2004 Share Posted March 13, 2004 Your integral is right, you will get the volume of a sphere out, but you need to just work at it a bit more. I went through the entire thing for you, and I'll provide the proof if you really want it. Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted March 13, 2004 Share Posted March 13, 2004 ...if you don't need to show work...go find matlab or maple ...plug it in and POOF Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted March 13, 2004 Share Posted March 13, 2004 oh and where your getting confused is that you lost the r on the first term.... it should look like ([ that big junk you got there ]) *( r ) dr and then you should be able to simplify and work from there its important to keep your brackets... Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted March 13, 2004 Share Posted March 13, 2004 forgot to mention that somewhere you should get the term (R^2-r^2)^(3/2) hopefully that will give you hit. Link to comment Share on other sites More sharing options...
Ann_M Posted March 13, 2004 Author Share Posted March 13, 2004 yep i get that and i also get -2/3 outside the integrals, which i multilply right at the end. but even with that u dont get R^3 in the final answer. this is wat i get wen i have to put the limits of 0 and R in: 1/3 int(0 to 2 pi {(R^2 - r^2)^3/2)} and the limits 0 and R is dis wat ne of u get wen u tried it. Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted March 13, 2004 Share Posted March 13, 2004 The inside R^2-r^2 should get rid of the 1/2 in 3/2 to leave you with 3...when you sub in 0:R Link to comment Share on other sites More sharing options...
Ann_M Posted March 13, 2004 Author Share Posted March 13, 2004 i dont understand, i get (-2R^2)^3/2 - (R^2)^3/2 after i substitued 0:R and then wen i do the final integral i get: 1/3 int 0 to 2pi (-2R^2)^3/2 - (R^2)^3/2 d theta after this im confused, how do u go about subtracting that. Link to comment Share on other sites More sharing options...
Neurocomp2003 Posted March 13, 2004 Share Posted March 13, 2004 ah i see...the problem is that the first term is actaully 0 R^2-r^2--->(r = R) ----> R^2-R^2 = 0 not -2R^2.... and then the 2nd term (R^2)^(3/2) reduces to R^3...read up on your exponential equations. but the 2nd term should also be (2/3)*(R^2)^(3/2)...if your not very table with integration...always differentiate to see if you get the integrall you started with. Link to comment Share on other sites More sharing options...
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