Solved

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

Multiple Choice

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.


A) The student has been careless with the orientations of the curve and surface.
B) The student has to be careful with the orientations of the curve and surface.However, Stokes' Theorem has been applied correctly.
C) The student has used Divergence Theorem incorrectly.The upper hemisphere S does not include the bottom, hence it does not enclose any region in space, and we cannot apply the Divergence Theorem to this surface.
D) The student has to be careful with the orientations of the curve and surface.However, the Divergence Theorem has been applied correctly.
E) The student has the correct orientations of the curve and surface but the student has used Divergence Theorem for the hemisphere incorrectly.The upper hemisphere S does not include the bottom, hence it does not enclose any region in space, and we cannot apply the Divergence Theorem to this surface.

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions