Answer :
A sphere S with a radius of a and a centre in the centre of the ball will conduct heat flow at a rate of 4πcK
Given In a ball with conductivity K, the temperature at a given place is inversely proportional to the distance from the ball's centre. If the scalar field u(x, y, z) exists and the divergence of u is ∇u = ∂u/∂xi+∂u/∂yj+∂u/∂zk
The surface integral is used to calculate the rate of heat flow over the body's surface S. ∬F⋅dS = F⋅n∬dS
Equation for the radius of a sphere is x^2+y^2+z^2 = a^2
And if the temperature of a ball is u(x, y, z), then u(x, y, z) ∝ 1/√x^2+y^2+z^2
The proportionality constant is denoted by c. So, u(x, y, z) = c/√x^2+y^2+z^2
Then ∇u = ∂(c/√x^2+y^2+z^2)/∂xi+∂(c/√x^2+y^2+z^2)/∂yj+∂(c√x^2+y^2+z^2)/∂zk
∇u = -(cx/√x^2+y^2+z^2)i-(cy/√x^2+y^2+z^2)j-(cz/√x^2+y^2+z^2)k
Hence F = -K.∇u
F = cK(xi+yj+zk)/(x^2+y^2+z^2)^3/2
And the outward unit normal is, n = 1/a(xi+yj+zk)
Thus F.n = cK(xi+yj+zk)/(x^2+y^2+z^2)^3/2 x 1/a(xi+yj+zk)
F.n = cK/a*(x^2+y^2+z^2)^1/2 but x^2+y^2+z^2 = a^2
F.n = cK/a^2
The rate of heat flow = F⋅n∬dS = cK/a^2∬dS
cK/a^2(4πa^2) = 4πcK
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