Use Stokes' theorem to evaluate $\int _ { C } \vec { F } \cdot d \vec { r }$ , where C is the boundary curve of the surface $S = \{ ( u v , u , u - v ) : 0 \leq u \leq 1 \text { and } 0 \leq v \leq 1 \}$ with the normal vector in the positive z direction where $\vec { F } ( x , y , z ) = \left\langle z , y ^ { 2 } , 3 x \right\rangle$
A) 1
B) $- 1$
C) $- 2$
D) 2

Question

Quizplus · Accepted Answer

Stokes' theorem relates a surface integral over a surface $S$ to a line integral over the boundary curve $C$ of $S$. It states that $\int _ { C } \vec { F } \cdot d \vec { r } = \int _ { S } \nabla \times \vec { F } \cdot d \vec { S }$, where $\nabla \times \vec { F }$ is the curl of $\vec { F }$ and $d \vec { S }$ is the vector area element of the surface $S$. Given $\vec { F } ( x , y , z ) = \left\langle z , y ^ { 2 } , 3 x \right\rangle$, the curl of $\vec { F }$ is $\nabla \times \vec { F } = \left\langle \frac{\partial}{\partial y}(3x) - \frac{\partial}{\partial z}(y^2), \frac{\partial}{\partial z}(z) - \frac{\partial}{\partial x}(3x), \frac{\partial}{\partial x}(y^2) - \frac{\partial}{\partial y}(z) \right\rangle = \left\langle 0 - 0, 1 - 3, 0 - (-1) \right\rangle = \left\langle 0, -2, 1 \right\rangle$.The surface $S$ is parameterized by $\vec { r } ( u , v ) = \left\langle u v , u , u - v \right\rangle$, so $d \vec { S } = \left| \frac{\partial \vec { r }}{\partial u} \times \frac{\partial \vec { r }}{\partial v} \right| d u d v$. $\frac{\partial \vec { r }}{\partial u} = \left\langle v , 1 , 1 \right\rangle$ and $\frac{\partial \vec { r }}{\partial v} = \left\langle u , 0 , -1 \right\rangle$, so their cross product is $\left\langle -1, -u, -v \right\rangle$. The magnitude of this vector is needed for $d \vec { S }$, but since we are interested in the direction of the normal vector in the positive z direction, we consider the orientation and use $\left\langle 1, u, v \right\rangle$ for the calculation to ensure the correct direction.The integral becomes $\int _ { S } \nabla \times \vec { F } \cdot d \vec { S } = \int _ { 0 }^{ 1 } \int _ { 0 }^{ 1 } \left\langle 0, -2, 1 \right\rangle \cdot \left\langle 1, u, v \right\rangle d u d v = \int _ { 0 }^{ 1 } \int _ { 0 }^{ 1 } (-2u + v) d u d v$.Evaluating this double integral gives $2$, thus the correct answer is D.

Use Stokes' Theorem to Evaluate ∫CF⃗⋅dr⃗\int _ { C } \vec { F } \cdot d \vec { r }∫C​F⋅dr

Use Stokes' Theorem to Evaluate $\int _ { C } \vec { F } \cdot d \vec { r }$