Show, using Gauss divergence theorem, that the surface area of a sphere of radius ‘a’ is 4πa2

To prove that the surface area of a sphere of radius 'a' is 4πa^2 using Gauss divergence theorem, we can follow these steps:

Step 1: Start with the Gauss divergence theorem, which states that for any vector field F, the flux of F across a closed surface S is equal to the triple integral of the divergence of F over the volume V enclosed by S. Mathematically, it can be written as:

∬S F · dA = ∭V ∇ · F dV

Step 2: Consider a sphere of radius 'a' centered at the origin. We want to find the flux of the vector field F = r over the surface of the sphere.

Step 3: Express the vector field F in terms of its Cartesian components. Since F = r, where r = xi + yj + zk is the position vector, we have:

F = xi + yj + zk

Step 4: Calculate the divergence of F. The divergence of F can be found by taking the dot product of the del operator (∇) with F. In Cartesian coordinates, the divergence is given by:

∇ · F = ∂(xi)/∂x + ∂(yj)/∂y + ∂(zk)/∂z = 3

Step 5: Substitute the divergence of F into the Gauss divergence theorem equation:

∬S F · dA = ∭V ∇ · F dV

becomes:

∬S r · dA = ∭V 3 dV

Step 6: Simplify the equation. Since the divergence of F is a constant (3), we can take it out of the integral:

3 ∭V dV = 3V

Step 7: The volume of a sphere of radius 'a' is given by V = (4/3)πa^3. Substitute this value into the equation:

3V = 3(4/3)πa^3 = 4πa^3

Step 8: Finally, equate the flux of F across the surface of the sphere to the volume of the sphere:

∬S r · dA = 4πa^3

Step 9: The left-hand side of the equation represents the surface area of the sphere, denoted as A. Therefore, we have:

A = 4πa^3

Step 10: Simplify the equation to obtain the desired result:

A = 4πa^2

Hence, using Gauss divergence theorem, we have shown that the surface area of a sphere of radius 'a' is 4πa^2.