INTEGRATION The inner and outer radii of a half hollow sphere are a and b.Find the center of gravity.Please help

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INTEGRATION The inner and outer radii of a half hollow sphere are a and b.Find the center of gravity.Please help

Mathematics
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Through intergration
You have to consider this as a solid sphere and then integrate from a to b right?
You have to consider this as a solid sphere and then integrate from a to b right?

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Other answers:

Can you help? @ganeshie8 @zepdrix
I know the answer but don't know how to prove
This might help, Zupari. c: http://physics.stackexchange.com/questions/100444/why-is-the-moment-of-inertia-for-a-hollow-sphere-higher-than-a-uniform-sphere
Nope
center of gravity? .... wh... ut? 0_o
Nothing to do with moment of inertia lol
I think it's best that you define for us exactly what "center of gravity" is. @Zupari
Looks more like question for physicist, not mathematician lol.
Sorry i don't know the exact english word but i think thats the word
Centre of gravity of a hollow sphere is a/2 likewise you have to integrate and find the point of this object
the translation is fine :) im just not familiar with the concept. hmm
is it a complete sphere or an hemisphere ?
hemishpere
because if it were a complete sphere, dont u think the center of gravity just lies at the center
ohk..
Complete sphere would be more trivial, but he still have to prove it either way.
answer is 3{(a+b)(a^2 + b^2)}/8( a^2 + ab + b^2)
do you mean (0, 0, 3{(a+b)(a^2 + b^2)}/8( a^2 + ab + b^2))
i think so
https://www.youtube.com/watch?v=UiAH-Ev6PRA watch this if u don't know the concept.
likewise you have to apply that to this boy.
dont know how but i
yoda
|dw:1436562405371:dw|
yep
|dw:1436562485331:dw|
|dw:1436562465745:dw|
oh thats beautiful
i like this simplification, just find it for 2 circles
|dw:1436562542334:dw|
i cant get the given answer
by symmetry the center x is the origin now just determine y center
its not that simple you have to integrate
|dw:1436562726374:dw|
can you think of how to write that with integration
or u can try to find the mean radius too that is fine
i am thinking of this one geometrical way, maybe it works maybe it doesnt take al ook at it
how about finding a point such that the sum of all the vectors from the edge will equal 0
|dw:1436562918280:dw|
\[\int\limits_{a}^bb ^{2}x - x ^{3}/\int\limits_{a}^b a ^{2} - x ^{2}\]
by solving this i dont get the given answer
find this out though because there might be a really nice way if u get this
cause then you can take the mean of the 2 given radii (a+b)/2 and see what the point on that circle radius would be such that all the sum of the vectors = 0
honestly i don't know what u just said maybe its because we are 100s of miles away and our education system is different
im in canada
im in sri lanka if u've heard
yes ofcourse
cool
okay wait, so how did u get those 2 integrals
\[ \int\limits_{a}^bb ^{2}x - x ^{3}/\int\limits_{a}^b a ^{2} - x ^{2} \]
|dw:1436563494387:dw|
m for mass
p for density
okay
well we can ignore that since we are assume its homogenous, we are finding center of volume, but continue
|dw:1436563681149:dw|
ok yes
since the solid part is from a to b integral is from a to b BUT I CANT GET THE ANSWER
okay try this
|dw:1436563948647:dw|
|dw:1436564233941:dw|
|dw:1436564253565:dw|
does this make sense? we are finding the mean height
uhhh...i dont know how to solve that square root inside integral our syllabus doesn;t include that part
oh wait
ya u have to convert to polar coord
here is a better way from polar
|dw:1436564634557:dw|
just forget it
wait why dont give up
|dw:1436565276032:dw|
|dw:1436565312287:dw|
solve that integral
|dw:1436565419033:dw|
|dw:1436565521848:dw|

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