Let $\mathbf { F } ( x , y , z ) = 4 x z \mathbf { i } + y \mathbf { j } + 4 x y \mathbf { k }$ and C be the boundary of the surface $9 - x ^ { 2 } - y ^ { 2 } = z$ above z = 0. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

A)3
B)2
C)1
D)0
E)-1

Question

Let

\mathbf { F } ( x , y , z ) = 4 x z \mathbf { i } + y \mathbf { j } + 4 x y \mathbf { k }

and C be the boundary of the surface

9 - x ^ { 2 } - y ^ { 2 } = z

above z = 0. Using Stokes' Theorem,

\int _ { C } \mathbf { F } \cdot \mathbf { T } d s

is

A)3
B)2
C)1
D)0
E)-1

Quizplus · Accepted Answer

Stokes' Theorem relates a surface integral over a vector field to a line integral over the boundary of the surface. Specifically, it states that the line integral of a vector field $\mathbf{F}$ around a closed curve $C$ is equal to the surface integral of the curl of $\mathbf{F}$ over the surface $S$ bounded by $C$. Mathematically, $\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.Given $\mathbf{F}(x, y, z) = 4xz\mathbf{i} + y\mathbf{j} + 4xy\mathbf{k}$, we first find the curl of $\mathbf{F}$, $\nabla \times \mathbf{F}$.$\nabla \times \mathbf{F} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ 4xz & y & 4xy \end{array} \right| = (0 - 0)\mathbf{i} - (0 - 4x)\mathbf{j} + (4z - 0)\mathbf{k} = -4x\mathbf{j} + 4z\mathbf{k}$.The surface $S$ is defined by $9 - x^2 - y^2 = z$ for $z \geq 0$, which is a paraboloid. The boundary $C$ is the circle at $z = 0$, or $x^2 + y^2 = 9$.To apply Stokes' Theorem, we need to integrate $\nabla \times \mathbf{F}$ over the surface $S$. However, since the question asks for the line integral around $C$, we can directly use the result of Stokes' Theorem without performing the surface integral explicitly.The key observation is that the curl of $\mathbf{F}$ does not depend on $y$, and its $j$ component will integrate to zero over a symmetric surface like the disk bounded by $C$. Similarly, the $k$ component will also result in zero when integrated over the surface because the surface is symmetric about the $z$-axis. Therefore, the integral of $\nabla \times \mathbf{F}$ over the surface $S$ is zero, which means the line integral of $\mathbf{F}$ around $C$ is also zero.

Let F(x,y,z)=4xzi+yj+4xyk\mathbf { F } ( x , y , z ) = 4 x z \mathbf { i } + y \mathbf { j } + 4 x y \mathbf { k }F(x,y,z)=4xzi+yj+4xyk

Let $\mathbf { F } ( x , y , z ) = 4 x z \mathbf { i } + y \mathbf { j } + 4 x y \mathbf { k }$