Let $\mathbf { F } ( x , y , z ) = - 4 y \mathbf { i } + 2 z \mathbf { j } + 3 x \mathbf { k }$ , C be the boundary of the region $z = 10 - x ^ { 2 } - y ^ { 2 }$ above z = 1. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \bullet \mathbf { T } d s$ is

A)18?
B)22?
C)24?
D)28?
E)36?

Question

Let

\mathbf { F } ( x , y , z ) = - 4 y \mathbf { i } + 2 z \mathbf { j } + 3 x \mathbf { k }

, C be the boundary of the region

z = 10 - x ^ { 2 } - y ^ { 2 }

above z = 1. Using Stokes' Theorem,

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

is

A)18?
B)22?
C)24?
D)28?
E)36?

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 $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s = \int _ { S } \nabla \times \mathbf { F } \cdot d \mathbf { S }$, where $\mathbf { F }$ is a vector field, $\mathbf { T }$ is the unit tangent vector to $C$, $ds$ is the differential arc length along $C$, $\nabla \times \mathbf { F }$ is the curl of $\mathbf { F }$, and $d \mathbf { S }$ is the vector area element of $S$.Given $\mathbf { F } ( x , y , z ) = - 4 y \mathbf { i } + 2 z \mathbf { j } + 3 x \mathbf { k }$, we first find the curl of $\mathbf { F }$, $\nabla \times \mathbf { F }$.$\nabla \times \mathbf { F } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \ \partial / \partial x & \partial / \partial y & \partial / \partial z \ -4y & 2z & 3x \end{array} \right| = (3 - 0) \mathbf { i } - (-4 - 0) \mathbf { j } + (0 - (-4)) \mathbf { k } = 3 \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k }$The surface $S$ is the disk in the plane $z = 10 - x^2 - y^2$ above $z = 1$, which corresponds to $10 - x^2 - y^2 = 1$, or $x^2 + y^2 = 9$, a circle of radius 3 in the $xy$-plane.The vector area element $d \mathbf { S }$ can be expressed as $\hat{n} dS$, where $\hat{n}$ is the unit normal vector to the surface. For a flat disk in the $xy$-plane, $\hat{n} = \mathbf { k }$, and $dS = dx dy$.The integral over the surface $S$ becomes $\int _ { S } (3 \mathbf { i } + 4 \mathbf { j } + 4 \mathbf { k }) \cdot \mathbf { k } dS = \int _ { S } 4 dS$, since only the $k$-component of $\nabla \times \mathbf { F }$ contributes when dotted with $\mathbf { k }$.The area of the disk is $\pi r^2 = \pi (3)^2 = 9\pi$, so the integral is $4 \times 9\pi = 36\pi$.Therefore, the correct answer is $36\pi$, which is closest to choice E) 36, assuming a misunderstanding in the question's phrasing regarding the exact numerical answer versus the presence of $\pi$.

Let F(x,y,z)=−4yi+2zj+3xk\mathbf { F } ( x , y , z ) = - 4 y \mathbf { i } + 2 z \mathbf { j } + 3 x \mathbf { k }F(x,y,z)=−4yi+2zj+3xk

Let $\mathbf { F } ( x , y , z ) = - 4 y \mathbf { i } + 2 z \mathbf { j } + 3 x \mathbf { k }$