On an exam, students are asked to find the line integral of $ { \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 $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 ^ { 2 } , z \geq 0$ oriented in a counter-clockwise direction when viewed from above.One student wrote: " $\operatorname { curl } { \vec { F } } = - 2 x \vec { j } + 2 \vec { k }$ By Stokes' Theorem $\int _ { C } \vec { F } \cdot \vec{ d r } = \int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A }$ where S is the hemisphere.Since $\operatorname { div } ( - 2 x \vec { j } + 2 \vec { k } ) = 0$ by the Divergence Theorem $\int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A } = \int _ { W } 0 d V = 0$ where W is the solid hemisphere.Hence we have $\int _ { C } \vec { F } \cdot \vec{ d r } = 0$ "
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.

Question

On an exam, students are asked to find the line integral of $  { \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 $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 ^ { 2 } , z \geq 0$ oriented in a counter-clockwise direction when viewed from above.One student wrote: " $\operatorname { curl }  { \vec { F } } = - 2 x \vec { j } + 2 \vec { k }$ By Stokes' Theorem $\int _ { C } \vec { F } \cdot \vec{ d r } = \int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A }$ where S is the hemisphere.Since $\operatorname { div } ( - 2 x \vec { j } + 2 \vec { k } ) = 0$ by the Divergence Theorem $\int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A } = \int _ { W } 0 d V = 0$ where W is the solid hemisphere.Hence we have $\int _ { C } \vec { F } \cdot \vec{ d r } = 0$ "
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.

Quizplus · Accepted Answer

The Answer of On an exam, students are asked to find the line integral of $  { \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 $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 2 ^ { 2 } , z \geq 0$ oriented in a counter-clockwise direction when viewed from above.One student wrote: " $\operatorname { curl }  { \vec { F } } = - 2 x \vec { j } + 2 \vec { k }$ By Stokes' Theorem $\int _ { C } \vec { F } \cdot \vec{ d r } = \int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A }$ where S is the hemisphere.Since $\operatorname { div } ( - 2 x \vec { j } + 2 \vec { k } ) = 0$ by the Divergence Theorem $\int _ { S } ( - 2 x \vec { j } + 2 \vec { k } ) \cdot \vec { d A } = \int _ { W } 0 d V = 0$ where W is the solid hemisphere.Hence we have $\int _ { C } \vec { F } \cdot \vec{ d r } = 0$ "
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.

On an Exam, Students Are Asked to Find the Line F⃗=2xyi⃗+yzj⃗+y2k⃗ { \vec { F } } = 2 x y \vec { i } + y z \vec { j } + y ^ { 2 } \vec { k }F=2xyi+yzj​+y2k

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