helppppppp

[Ann_M]

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.

[YT2095]

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

[Sayonara]

That may not be the entire formula; just the part that needs integration.

[Dave]

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.

[Ann_M]

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

[Ann_M]

oh and it has to equal the volume of a sphere as the final answer.

[YT2095]

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!

 

[Ann_M]

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.

[Dave]

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.

[Neurocomp2003]

...if you don't need to show work...go find matlab or maple ...plug it in and POOF

[Neurocomp2003]

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...

[Neurocomp2003]

forgot to mention that somewhere you should get the term

(R^2-r^2)^(3/2) hopefully that will give you hit.

[Ann_M]

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.

[Neurocomp2003]

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

[Ann_M]

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.

[Neurocomp2003]

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.