Closed disk of radius limit math problem

ironman2
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Homework Statement


If Dr is a closed disk of radius r centered at (a,b) find lim r->0 (1/pir2) \int\intfdA over Dr.



The Attempt at a Solution


From mean value equality, \int\int fdA = f(x,y)A(D) where A(D) is the area of the region which here is pir2. So the lhs becomes lim r->0 f(x,y). Where do I go from here?
 
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welcome to pf!

hi ironman2! welcome to pf! :smile:
ironman2 said:
… So the lhs becomes lim r->0 f(x,y). Where do I go from here?

hint: where is (x,y) ? :wink:
 


Inside the disk? I thought of using a, b as x,y since a,b were inside the disk and somehow relating a, b to r... but can't seem to do it.
 
ironman2 said:
Inside the disk?

yup! :smile:

(x,y) is inside the disc Dr round (a,b) …

(btw, it would be more mathematical to call it (xr,yr) :wink:)

so limr->0 f(xr,yr) = … ? :smile:
 


Since f is continuous, would lim r->0 f(xr,yr) = (a,b)? I'm thinking a,b = 0,0 since its the center...
 
ironman2 said:
Since f is continuous, would lim r->0 f(xr,yr) = (a,b)?

yes, but you need to prove that, not just say it

hint: sequence :wink:
I'm thinking a,b = 0,0 since its the center...


i've no idea what this means … (a,b) is just (a,b) :confused:
 


Sequence like Taylor-series? I don't quite understand...

Nvm the 0,0 logic, I was assuming it's a disk centered on the origin, which obviously it's not.
 
(x1 , y1) , (x2,y2) , (x3,y3) , … (xn,yn) … :wink:
 


Got it, thanks!
 
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