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Work the previous example for surface S that is a sphere of radius 4 centered at the origin, oriented outward.

6.777 × 10 9

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Key concepts

  • The divergence theorem relates a surface integral across closed surface S to a triple integral over the solid enclosed by S . The divergence theorem is a higher dimensional version of the flux form of Green’s theorem, and is therefore a higher dimensional version of the Fundamental Theorem of Calculus.
  • The divergence theorem can be used to transform a difficult flux integral into an easier triple integral and vice versa.
  • The divergence theorem can be used to derive Gauss’ law, a fundamental law in electrostatics.

Key equations

  • Divergence theorem
    E div F d V = S F · d S

For the following exercises, use a computer algebraic system (CAS) and the divergence theorem to evaluate surface integral S F · n d s for the given choice of F and the boundary surface S. For each closed surface, assume N is the outward unit normal vector.

[T] F ( x , y , z ) = x i + y j + z k ; S is the surface of cube 0 x 1 , 0 y 1 , 0 < z 1 .

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[T] F ( x , y , z ) = ( cos y z ) i + e x z j + 3 z 2 k ; S is the surface of hemisphere z = 4 x 2 y 2 together with disk x 2 + y 2 4 in the xy -plane.

S F · n d s = 75.3982

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[T] F ( x , y , z ) = ( x 2 + y 2 x 2 ) i + x 2 y j + 3 z k ; S is the surface of the five faces of unit cube 0 x 1 , 0 y 1 , 0 < z 1 .

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[T] F ( x , y , z ) = x i + y j + z k ; S is the surface of paraboloid z = x 2 + y 2 for 0 z 9 .

S F · n d s = 127.2345

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[T] F ( x , y , z ) = x 2 i + y 2 j + z 2 k ; S is the surface of sphere x 2 + y 2 + z 2 = 4 .

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[T] F ( x , y , z ) = x i + y j + ( z 2 1 ) k ; S is the surface of the solid bounded by cylinder x 2 + y 2 = 4 and planes z = 0 and z = 1 .

S F · n d s = 37.6991

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[T] F ( x , y , z ) = x y 2 i + y z 2 j + x 2 z k ; S is the surface bounded above by sphere ρ = 2 and below by cone φ = π 4 in spherical coordinates. (Think of S as the surface of an “ice cream cone.”)

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[T] F ( x , y , z ) = x 3 i + y 3 j + 3 a 2 z k (constant a > 0 ) ; S is the surface bounded by cylinder x 2 + y 2 = a 2 and planes z = 0 and z = 1 .

S F · n d s = 9 π a 4 2

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[T] Surface integral S F · d S , where S is the solid bounded by paraboloid z = x 2 + y 2 and plane z = 4 , and F ( x , y , z ) = ( x + y 2 z 2 ) i + ( y + z 2 x 2 ) j + ( z + x 2 y 2 ) k

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Use the divergence theorem to calculate surface integral S F · d S , where F ( x , y , z ) = ( e y 2 ) i + ( y + sin ( z 2 ) ) j + ( z 1 ) k and S is upper hemisphere x 2 + y 2 + z 2 = 1 , z 0 , oriented upward.

S F · d S = π 3

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Use the divergence theorem to calculate surface integral S F · d S , where F ( x , y , z ) = x 4 i x 3 z 2 j + 4 x y 2 z k and S is the surface bounded by cylinder x 2 + y 2 = 1 and planes z = x + 2 and z = 0 .

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Use the divergence theorem to calculate surface integral S F · d S when F ( x , y , z ) = x 2 z 3 i + 2 x y z 3 j + x z 4 k and S is the surface of the box with vertices ( ±1 , ±2 , ±3 ) .

S F · d S = 0

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Use the divergence theorem to calculate surface integral S F · d S when F ( x , y , z ) = z tan −1 ( y 2 ) i + z 3 ln ( x 2 + 1 ) j + z k and S is a part of paraboloid x 2 + y 2 + z = 2 that lies above plane z = 1 and is oriented upward.

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[T] Use a CAS and the divergence theorem to calculate flux S F · d S , where F ( x , y , z ) = ( x 3 + y 3 ) i + ( y 3 + z 3 ) j + ( z 3 + x 3 ) k and S is a sphere with center (0, 0) and radius 2.

S F · d S = 241.2743

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Use the divergence theorem to compute the value of flux integral S F · d S , where F ( x , y , z ) = ( y 3 + 3 x ) i + ( x z + y ) j + [ z + x 4 cos ( x 2 y ) ] k and S is the area of the region bounded by x 2 + y 2 = 1 , x 0 , y 0 , and 0 z 1 .

A vector field in three dimensions, with focus on the area with x > 0, y>0, and z>0. A quarter of a cylinder is drawn with center on the z axis. The arrows have positive x, y, and z components; they point away from the origin.
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Use the divergence theorem to compute flux integral S F · d S , where F ( x , y , z ) = y j z k and S consists of the union of paraboloid y = x 2 + z 2 , 0 y 1 , and disk x 2 + z 2 1 , y = 1 , oriented outward. What is the flux through just the paraboloid?

D F · d S = π

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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