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

2 Answers

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  • J
    Lv 7
    8 years ago
    Favorite Answer

    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

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  • Awms A
    Lv 7
    8 years ago

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

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