Let S be the surface of the upper part of the cylinder 4x²+ z² = 1, z $\ge$ 0, between the planes y = -1, y = 1, with an upward-pointing normal.
(a)Evaluate the flux integral $\int _ { S } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) \cdot \vec { d A }$ (b)Consider W, the solid region described by -1 $\le$ y $\le$ 1, 4x² + z² $\le$ 1, z $\ge$ 0.Evaluate $\int _ { W } \operatorname { div } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) d V$ Does this contradict the Divergence Theorem? Explain.

Question

Let S be the surface of the upper part of the cylinder 4x²+ z² = 1, z

\ge

0, between the planes y = -1, y = 1, with an upward-pointing normal.
(a)Evaluate the flux integral

\int _ { S } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) \cdot \vec { d A }

(b)Consider W, the solid region described by -1

\le

y

\le

1, 4x² + z²

\le

1, z

\ge

0.Evaluate

\int _ { W } \operatorname { div } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) d V

Does this contradict the Divergence Theorem? Explain.

Quizplus · Accepted Answer

The Answer of Let S be the surface of the upper part of the cylinder 4x²+ z² = 1, z $\ge$ 0, between the planes y = -1, y = 1, with an upward-pointing normal. (a)Evaluate the flux integral $\int _ { S } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) \cdot \vec { d A }$ (b)Consider W, the solid region described by -1 $\le$y $\le$ 1, 4x² + z² $\le$ 1, z $\ge$ 0.Evaluate $\int _ { W } \operatorname { div } \left( - 3 x y ^ { 2 } z \vec { i } + 7 \cos z \vec { j } + 3 \vec { k } \right) d V$ Does this contradict the Divergence Theorem? Explain.

Let S Be the Surface of the Upper Part of the Cylinder