(1 pt) suppose f is a radial force field, s1 is a sphere of radius 7 centered at the origin, and the flux integral ∫∫s1f⋅ds=3. let s2 be a sphere of radius 63 centered at the origin, and consider the flux integral ∫∫s2f⋅ds. (a) if the magnitude of f is inversely proportional to the square of the distance from the origin,what is the value of ∫∫s2f⋅ds?

Answer:(a) ∫∫s₂ f⋅ds = 12(b) ∫∫s₂ f⋅ds = 3/4Step-by-step explanation:From the question we were given s₁ to be = radius of the sphere R₁ = 7and the Radius of the other sphere S₂ = R₂ = 63the Flux integral, ∫∫s₁ f⋅ds  = 3(a). we are asked to fine the value of ∫∫s₂ f⋅dsbut first we know that the surface of the sphere = 4πR²given f is inversely proportional to the square of the distance from the origin; f ∝ 1/R² f = K/R      where k is a proportionality constantbut 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 ∝ R²given that f ∝ 1/R³thus we have  ∫∫s f⋅ds =  ∫∫s 1/R³*R² =  ∫∫s 1/Rif R increases by a factor = 16/4 = R₂/R₁, we have 4so our surface integral reduces by 1/4s₂ = s₁/4s₂ = 3/4our answer becomes ∫∫s₂ f⋅ds = 3/4 cheers i hope this helps