Closed disk of radius limit math problem

[ironman2]

Homework Statement

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

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?

 

[tiny-tim]

welcome to pf!

hi ironman2! welcome to pf!

… So the lhs becomes lim r->0 f(x,y). Where do I go from here?

hint: where is (x,y) ?

 

[ironman2]

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.

 

[tiny-tim]

Inside the disk?

yup!

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

(btw, it would be more mathematical to call it (x_r,y_r) )

so lim_r->0 f(x_r,y_r) = … ?

 

[ironman2]

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

 

[tiny-tim]

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

I'm thinking a,b = 0,0 since its the center...

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

 

[ironman2]

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.

 

[tiny-tim]

(x₁ , y₁) , (x₂,y₂) , (x₃,y₃) , … (x_n,y_n) …

 

[ironman2]

Got it, thanks!