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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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Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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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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