Exam 16: Vector Calculus

If $\mathbf { F } ( x , y ) = e ^ { y ^ { 2 } } \cos x \mathbf { i } + 2 y e ^ { y ^ { 2 } } \sin x \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

Let $\mathbf { F } ( x , y ) = 9 x ^ { 2 } y \mathbf { i } + \left( 5 x ^ { 2 } - y \right) \mathbf { j }$ , where C is the $y = x ^ { 3 } + 1$ from (1,2) to (3,28). Then $\int$ C $\mathbf { F } \bullet d \mathbf { r }$ is

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { 1 - x } \mathbf { i } + \ln y \mathbf { j }$ is

A parameterization of $y = 6 x ^ { 2 }$ is

The domain of the vector field $\mathbf { F } ( x , y , z ) = \frac { y z } { e ^ { y z } } \mathbf { i } + \frac { x z } { e ^ { x z } } \mathbf { j } + \frac { x y } { e ^ { z y } } \mathbf { k }$ is

The work done by the force $\mathbf { F } ( x , y ) = - ( x + y ) \mathbf { i } - ( y - x ) \mathbf { j }$ moving along $\mathbf { r } ( t ) = t ^ { 3 } \mathbf { i } + t ^ { 2 } \mathbf { j }$ from (8, 4) to (1, 1) is

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

The rectangular equation for the parametric surface and $x ( u , v ) = 2 \cos v \cos u$ is

Let $\mathbf { F } ( x , y , z ) = ( 2 x - y ) \mathbf { i } - ( 2 x - z ) \mathbf { j } + z \mathbf { k }$ and S is the region bounded by $2 x + 4 y + 2 z = 12$ and the coordinate planes. Using the Divergence Theorem, $\iint S$$\mathbf { F } \cdot \mathbf { n } d S$ is

The domain of the vector field $\mathbf { F } ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j }$ is

If $\mathbf { F } ( x , y ) = 2 x \sec 2 y \mathbf { i } + 2 x ^ { 2 } \sec 2 y \tan 2 y \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

An equation of the tangent plane to the surfaceat (1, -1, 1) is

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

Let $\mathbf { F } ( x , y , z ) = x y \mathbf { i } + y ^ { 2 } \mathbf { j } + 2 \mathbf { k }$ and C be the boundary of $z = x ^ { 2 } + y ^ { 2 }$ below z = 1.Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

The work done by the force $\mathbf { F } ( x , y ) = - x ^ { 2 } y \mathbf { i } + 2 y \mathbf { j }$ moving along $\mathbf { r } ( t ) = 2 \cos t \mathbf { i } + 2 \sin t \mathbf { j }$ with $0 \leq t \leq \frac { \pi } { 2 }$ is

Using Green's Theorem, the line integral $\int$ C $\left( x ^ { 2 } + y \right) d x$ where C is the parabola $y = 4 - x ^ { 2 }$ from (-2, 0) to (0, 4), from (0, 4) to (-2, 0), along the parabola, the line segment from (2, 0) to (-2, 0), is

The work done by the force $\mathbf { F } ( x , y ) = - x ^ { 2 } y \mathbf { i } + 2 y \mathbf { j }$ moving along the line segment from (2, 0) to (0, 2) is

Let $\mathbf { F } ( x , y , z ) = 2 x y \mathbf { i } + \left( 6 y ^ { 2 } - x z \right) \mathbf { j } + 10 z \mathbf { k } , \mathbf { r } ( t ) = t \mathbf { i } + t ^ { 2 } \mathbf { j } + t ^ { 3 } \mathbf { k } , 0 \leq t \leq 1$ . Then $\int$ C $\mathbf { F } \bullet d \mathbf { r }$ is

The area of the surface $\mathbf { r } ( u , v ) = u \mathbf { i } - \frac { v } { 2 } \mathbf { j } + \frac { v } { 2 } \mathbf { k }$ , where $0 \leq u \leq 2,0 \leq v \leq 1$ is

Let $\mathbf { F } ( x , y , z ) = y ^ { 2 } \mathbf { i } + x \mathbf { j } + z ^ { 2 } \mathbf { k }$ and C be the boundary of $z = x ^ { 2 } + y ^ { 2 }$ below z = 1. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is