# 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 =...
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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

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

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