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On an Exam, Students Are Asked to Find the Line F=2xyi+yzj+y2k { \vec { F } } = 2 x y \vec { i } + y z \vec { j } + y ^ { 2 } \vec { k }

question 7

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On an exam, students are asked to find the line integral of F=2xyi+yzj+y2k { \vec { F } } = 2 x y \vec { i } + y z \vec { j } + y ^ { 2 } \vec { k } over the curve C which is the boundary of the upper hemisphere x2+y2+z2=22,z0x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 ^ { 2 } , z \geq 0 oriented in a counter-clockwise direction when viewed from above.One student wrote: " curlF=2xj+2k\operatorname { curl } { \vec { F } } = - 2 x \vec { j } + 2 \vec { k } By Stokes' Theorem CFdr=S(2xj+2k) dA\int _ { C } \vec { F } \cdot \vec{ d r } = \int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A } where S is the hemisphere.Since div(2xj+2k) =0\operatorname { div } ( - 2 x \vec { j } + 2 \vec { k } ) = 0 by the Divergence Theorem S(2xj+2k) dA=W0dV=0\int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A } = \int _ { W } 0 d V = 0 where W is the solid hemisphere.Hence we have CFdr=0\int _ { C } \vec { F } \cdot \vec{ d r } = 0 "
This answer is wrong.Which part of the student's argument is wrong? Select all that apply.


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