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z=x2+y2z = \sqrt { x ^ { 2 } + y ^ { 2 } }

Question 8

Multiple Choice

 Let F(x,y,z) =(xy2+z) i+(x2y+z) j+ek and let S be the surface bounded by \text { Let } \mathbf { F } ( x , y , z ) = \left( x y ^ { 2 } + z \right) \mathbf { i } + \left( x ^ { 2 } y + z \right) \mathbf { j } + e \mathbf { k } \text { and let } S \text { be the surface bounded by } z=x2+y2z = \sqrt { x ^ { 2 } + y ^ { 2 } } and z=4z = 4 . Verify the Divergence Theorem by evaluating SFNds\iint_{S} \mathbf{F} \cdot \mathbf{N} d s as a surface integral and as a triple integral. Round your answer to two decimal places.


A) 80.4280.42
B) 321.70321.70
C) 402.12402.12
D) 1,286.801,286.80
E) 337.02337.02

Correct Answer:

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