The given function is D = ar/r. We are asked to calculate the surface integral of D over a hemispherical surface bounded by r=2 and 0

Question

The given function is D = ar/r. We are asked to calculate the surface integral of D over a hemispherical surface bounded by r=2 and 0< θ < π/2.

The surface integral of a vector field D over a surface S is given by ∫D.ds.

In spherical coordinates, the differential surface area element ds in terms of r, θ, and φ is given by ds = r^2 sin(θ) dθ dφ.

Substituting D = ar/r into the integral, we get ∫D.ds = ∫(ar/r) * r^2 sin(θ) dθ dφ.

Simplifying, we get ∫D.ds = ∫a * r sin(θ) dθ dφ.

We are integrating over a hemispherical surface bounded by r=2 and 0< θ < π/2. So, we substitute these limits into the integral:

∫D.ds = ∫ (from 0 to 2π) ∫ (from 0 to π/2) a * r sin(θ) dθ dφ.

This integral can be separated into two integrals:

∫D.ds = a * r * [∫ (from 0 to 2π) dφ] * [∫ (from 0 to π/2) sin(θ) dθ].

Evaluating these integrals, we get:

∫D.ds = a * r * [2π - 0] * [-cos(π/2) + cos(0)].

Simplifying, we get:

∫D.ds = 2πa * r * [1].

So, ∫D.ds = 2πa * r.

Substituting r = 2, we get:

∫D.ds = 2πa * 2 = 4πa.

So, the surface integral of D over the given hemispherical surface is 4πa. Therefore, the answer is 4π.

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