$z = \sqrt { x ^ { 2 } + y ^ { 2 } }$

question 8

Multiple Choice

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

• A) \(80.42\)
• B) \(321.70\)
• C) \(402.12\)
• D) \(1,286.80\)
• E) \(337.02\)

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