On an exam, students were asked to evaluate $\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r }$ , where C is the circle centered at the origin of radius r: $x = r \cos t , y = r \sin t , 0 \leq t \leq 2 \pi$ .One student wrote:
"Since $\frac { \partial } { \partial y } \left( \frac { - y } { x ^ { 2 } + y ^ { 2 } } \right) = \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } , \frac { \partial } { \partial x } \left( \frac { x } { x ^ { 2 } + y ^ { 2 } } \right) = \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }$ Using Green's Theorem, $\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r } = \int _ { D } 0 d A = 0$ ."
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The Answer of On an exam, students were asked to evaluate $\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r }$ , where C is the circle centered at the origin of radius r: $x = r \cos t , y = r \sin t , 0 \leq t \leq 2 \pi$ .One student wrote:
"Since $\frac { \partial } { \partial y } \left( \frac { - y } { x ^ { 2 } + y ^ { 2 } } \right) = \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } + y ^ { 2 } } , \frac { \partial } { \partial x } \left( \frac { x } { x ^ { 2 } + y ^ { 2 } } \right) = \frac { y ^ { 2 } - x ^ { 2 } } { x ^ { 2 } + y ^ { 2 } }$ Using Green's Theorem, $\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r } = \int _ { D } 0 d A = 0$ ."
Do you agree with the student?

On an Exam, Students Were Asked to Evaluate ∫C(−yi⃗+xj⃗x2+y2)⋅dr⃗\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r }∫C​(x2+y2−yi+xj​​)⋅dr

On an Exam, Students Were Asked to Evaluate $\int _ { C } \left( \frac { - y \vec { i } + x \vec { j } } { x ^ { 2 } + y ^ { 2 } } \right) \cdot d \vec { r }$