$\text { Let } \mathbf { F } ( x , y , z ) = \left( x y ^ { 2 } + z \right) \mathbf { i } + \left( x ^ { 2 } y + z \right) \mathbf { j } + e \mathbf { k } \text { and let } S \text { be the surface bounded by }$ $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$ 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) $337.02$

Question

$\text { Let } \mathbf { F } ( x , y , z ) = \left( x y ^ { 2 } + z \right) \mathbf { i } + \left( x ^ { 2 } y + z \right) \mathbf { j } + e \mathbf { k } \text { and let } S \text { be the surface bounded by }$ $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$ 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) $337.02$

Quizplus · Accepted Answer

The Answer of $\text { Let } \mathbf { F } ( x , y , z ) = \left( x y ^ { 2 } + z \right) \mathbf { i } + \left( x ^ { 2 } y + z \right) \mathbf { j } + e \mathbf { k } \text { and let } S \text { be the surface bounded by }$ $z = \sqrt { x ^ { 2 } + y ^ { 2 } }$ 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) $337.02$

z=x2+y2z = \sqrt { x ^ { 2 } + y ^ { 2 } }z=x2+y2​

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