Test Academy | Given a function ϕ = 1 2 ( x 2 + y 2 + z 2 ) in three-dimensional Cartesian space, the value of the surface integral ∯S n̂ . ∇ϕ dS where S is the surface of a sphere of unit radius and n̂ is the outward unit normal vector on S, is 4π, 3π, 4π/3, 0

Question

Given a function $ϕ = \frac{1}{2} (x^{2} + y^{2} + z^{2})$ in three-dimensional Cartesian space, the value of the surface integral ∯S n̂ . ∇ϕ dS where S is the surface of a sphere of unit radius and n̂ is the outward unit normal vector on S, is

Options :

4π
3π
4π/3
0

Show Answer

Answer :

4π

Solution :

$ϕ = \frac{1}{2} (x^{2} + y^{2} + z^{2})$ , ∯S n̂ . ∇ϕ dS = ?

S = surface of sphere , V = volume of sphere = $\frac{4}{3} π r^{3}$

r = radius of sphere = 1

$\frac{\partial ϕ}{\partial x} \hat{i} + \frac{\partial ϕ}{\partial y} \hat{j} + \frac{\partial ϕ}{\partial z} \hat{k}$

$\frac{1}{2} (2 x \hat{i} + 2 y \hat{j} + 2 z \hat{k})$

$x \hat{i} + y \hat{j} + z \hat{k}$

Using Gauss theorem,

∯S n̂ . ∇ϕ dS = $\int \int \int_{V}$ ∇. ∇ϕ dV

= $\int \int \int_{V}$ (1 + 1 + 1)dV

= $3 \int \int \int_{V}$ dV = 3 x $\frac{4}{3} π r^{3}$

∯S n̂ . ∇ϕ dS = 4π

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