Answer:
(a) ∫∫s₂ f⋅ds = 12
(b) ∫∫s₂ f⋅ds = 3/4
Step-by-step explanation:
From the question we were given s₁ to be = radius of the sphere R₁ = 7
and the Radius of the other sphere S₂ = R₂ = 63
the Flux integral, ∫∫s₁ f⋅ds = 3
(a). we are asked to fine the value of ∫∫s₂ f⋅ds
but first we know that the surface of the sphere = 4πR²
given f is inversely proportional to the square of the distance from the origin;
f ∝ 1/R²
f = K/R where k is a proportionality constant
but recall that surface of the sphere = 4πR² = 196π
∴ ∫∫s₁ f⋅ds = 3 = f(4πR²) = 196πf = 3 ..........(1)
equating gets,
196π(K/R²) = 3 = 196π (K/7₂)
3 = K = 3/4π
considering the second sphere s₂ of radius R₂;
∫∫s₂ f⋅ds = (K/R²)(4π R²) = 4π (3/π ) = 12
∫∫s₂ f⋅ds = 12
(b). already we know, ds ∝ R²
given that f ∝ 1/R³
thus we have ∫∫s f⋅ds = ∫∫s 1/R³*R² = ∫∫s 1/R
if R increases by a factor = 16/4 = R₂/R₁, we have 4
so our surface integral reduces by 1/4
s₂ = s₁/4
s₂ = 3/4
our answer becomes ∫∫s₂ f⋅ds = 3/4
cheers i hope this helps
