What is the average value of y=sqrt(64-x^2)

First you have to know its domain, which you can find to be -8 ≤ 0 ≤ 8 as the function neither exists for values of x less than -8 nor for values of x greater than 8.

∫{ √(64-x²)/16 }dx, limits from -8 to 8

= (1/16)[ ∫√(64-x²)dx]

Since the function is even,

= (2/16)[ ∫√(64-x²)dx]

= (1/8) ∫√(64-x²)dx

There is a formula that,

∫√(a²-x²)dx = (x/2)√(a²-x²) + a²/2[sin^-1 (x/a)] + c

Here a=8.
Substituting these in the above formula.

= (1/8){ (x/2)√(64-x²) + 64/2[sin^-1 (x/8)] }

= (1/8){ (x/2)√(64-x²) + 32[sin^-1 (x/8)] }

Applying the limits,

= (1/8){ (8/2)√(64-64) + 32[sin^-1 (8/8)] - (0/2)√(64-0) - 32[sin^-1 (0/8)] }

= (1/8){0+ 32[sin^-1 (1)] - 0 - 32[sin^-1 (0)] }

= (1/8){ 32[sin^-1 (1)] }

= (32/8)(π/2)

= 4π/2

= 2π

Dang i wish i could , good luck (i’m bad at math)