# Divergence Theorem for a Unit Sphere?

Verify that the Divergence Theorem is true for F(x, y, z) = xi + yj + zj

where S is the unit ball x² + y² + z² ≤ 1

By the Divergence Theorem, ∫ divF dV = ∫∫∫ 3 dV

= ∫ (0 to 2π) ∫ (0 to π) ∫ (0 to 1) 3ρ²sinϕ dρdϕdθ = 4π

However, if I evaluate the surface integral, I got 0

Let g = z - sqrt(1 - x² - y²), ∇g = (x/z)i + (y/z)j + k

∫∫ F•N ds = ∫∫ F•∇g dA = ∫∫ x²/z + y²/z + z²/z dA

= ∫ (0 to 2π) ∫ (0 to 1) r/sqrt(1-r²) drdθ = 2π

Now how do I get the bottom half of the unit sphere?

If I just reverse ∇g so that Nds = -∇g dA, it will just cancel out the 2π

∫∫ F•∇g dA + ∫∫ F•(-∇g) dA = 0

Relevance

The bottom half has N= ∇g too .-

Remember that if z=f(x,y ) , g= z-f(x,y) =0 and

dg= Nabla g . dr =0 , so Nabla g is normal to dr at any point .- dr locates the tangent plane , so Nabla G is normal to the sphere on upper or bottom half .-

∫∫ F•∇g dA + ∫∫ F•(+∇g) dA =2 ∫∫ F•∇g dA

In a sphere you can do this directly using spherical coordinates

dS =<n> dS where <n> is a unit vector normal to dS

If x=RsinZcosT

y=RsinZsinT

z=RcosZ

r= RsinZcosT i+ RsinZsinT j+RcosZk

This vector is normal to dS and <n>= r/IrI , but IrI = R , so

<n> = sinZcosT i+ sinZsinT j+cosZk

dS= RsinZdT ( RdZ)

dS= R^2 sinZ dZdT

Now DIRECTLY ON dS , without any projection

F.<n> dS = (RsinZcosT i+ RsinZsinT j+RcosZk ) .(sinZcosT i+ sinZsinT j+cosZk) R^2 sinZ dZ dT

= R^3 (sin^3 Zcos^2T + sin^3Zsin^2T +sinZcos^2Z) dZ dT

=R^3 ( sin^3Z +sinZcos^2Z)dZdT

=R^3 ( sinZ (1-cos^2Z) +sinZcos^2Z) dZdT

=R^3 sinZ dZ dT

Flux = R^3 INT INT sinZdZ dT

0<T<2pi

0<Z<pi

= 2piR^3 (-cosZ)

=2piR^3 ( -) ( -1-1)

= 4pi

• Login to reply the answers
• Go back to your choice of g. When you chose that g, you had already made the choice of the top half.

• Login to reply the answers