Question 1

A parameterization of $2 x - 3 y + z = 5$ is

Accepted Answer

A)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 - 2 u + 3 v$ 
B)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 - 2 u - 3 v$ 
C)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 + 2 u + 3 v$ 
D)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = - 5 - 2 u + 3 v$ 
E)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = - 5 + 2 u + 3 v$ 
A)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 - 2 u + 3 v$ 
B)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 - 2 u - 3 v$ 
C)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = 5 + 2 u + 3 v$ 
D)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = - 5 - 2 u + 3 v$ 
E)  $x ( u , v ) = u$ $y ( u , v ) = v$ $z ( u , v ) = - 5 + 2 u + 3 v$

Question 2

Let $\mathbf { F } ( x , y , z ) = 3 z \mathbf { i } - 4 \mathbf { j } + y \mathbf { k }$ and S is $x + y + z = 1$ in the first octant. Then the flux of $\text { F }$ through S is

Accepted Answer

A)  $- \frac { 4 } { 3 }$ 
B)  $\frac { 4 } { 3 }$ 
C)  $\frac { 4 } { 5 }$ 
D)  $- \frac { 4 } { 5 }$ 
E)  $- \frac { 4 } { 7 }$ 
A)  $- \frac { 4 } { 3 }$ 
B)  $\frac { 4 } { 3 }$ 
C)  $\frac { 4 } { 5 }$ 
D)  $- \frac { 4 } { 5 }$ 
E)  $- \frac { 4 } { 7 }$

Question 3

Let $\mathbf { F } ( x , y , z ) = 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$ . Using the Divergence Theorem,$\iint S$ $\mathbf { F } \bullet \mathbf { n } d S$ is

Accepted Answer

A)  $\frac { 8 \pi } { 3 }$ 
B)  $\frac { 4 \pi } { 3 }$ 
C)  $\frac { 4 \pi } { 5 }$ 
D)  $\frac { 8 \pi } { 5 }$ 
E)  $\frac { 4 \pi } { 7 }$ 
A)  $\frac { 8 \pi } { 3 }$ 
B)  $\frac { 4 \pi } { 3 }$ 
C)  $\frac { 4 \pi } { 5 }$ 
D)  $\frac { 8 \pi } { 5 }$ 
E)  $\frac { 4 \pi } { 7 }$

Question 4

Let $\mathbf { F } ( x , y , z ) = z \mathbf { i } + x \mathbf { j } + y \mathbf { k } , \mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k } , 0 \leq t \leq 2 \pi$ . Then $\int $C $\mathbf { F } \bullet d \mathbf { r }$ is

Accepted Answer

A)  $4 \pi$ 
B)  $3 \pi$ 
C)  $2 \pi$ 
D)  $\pi$ 
E)  $\frac { \pi } { 2 }$ 
A)  $4 \pi$ 
B)  $3 \pi$ 
C)  $2 \pi$ 
D)  $\pi$ 
E)  $\frac { \pi } { 2 }$

Question 5

Let $f ( x , y ) = 2 x ^ { 3 } y + 7 x ^ { 2 } y ^ { 2 } - 5 x y ^ { 3 }$ . Its gradient vector field is

Accepted Answer

A)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y - 14 x y ^ { 2 } + 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
B)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } - 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
C)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y - 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
D)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
E)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } - \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
A)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y - 14 x y ^ { 2 } + 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
B)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } - 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
C)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y - 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
D)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } + \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$ 
E)  $\nabla f ( x , y ) = \left( 6 x ^ { 2 } y + 14 x y ^ { 2 } - 5 y ^ { 3 } \right) \mathbf { i } - \left( 2 x ^ { 3 } + 14 x ^ { 2 } y - 15 x y ^ { 2 } \right) \mathbf { j }$

Question 6

The line integral $\int $C $( x + y ) d s$ , where C is the curve $x = t , y = 1 - t , 0 \leq t \leq 1 ,$ is

Accepted Answer

A)  $2 \sqrt { 2 }$ 
B)  $\sqrt { 2 }$ 
C)  $3 \sqrt { 2 }$ 
D)  $\sqrt { 3 }$ 
E)  $\frac { 1 } { \sqrt { 2 } }$ 
A)  $2 \sqrt { 2 }$ 
B)  $\sqrt { 2 }$ 
C)  $3 \sqrt { 2 }$ 
D)  $\sqrt { 3 }$ 
E)  $\frac { 1 } { \sqrt { 2 } }$

Question 7

The divergence of $\mathbf { F } ( x , y , z ) = \left( y ^ { 2 } + z ^ { 2 } \right) \mathbf { i } + x e ^ { y } \mathbf { j } - x e ^ { y } \cos z \mathbf { k }$ is

Accepted Answer

A)  $- x e ^ { y } ( 1 + \sin z )$ 
B)  $x e ^ { y } ( 1 - \cos z )$ 
C)  $x e ^ { y } ( 1 + \cos z )$ 
D)  $x e ^ { y } ( 1 + \sin z )$ 
E)  $x e ^ { y } ( 1 - \sin z )$ 
A)  $- x e ^ { y } ( 1 + \sin z )$ 
B)  $x e ^ { y } ( 1 - \cos z )$ 
C)  $x e ^ { y } ( 1 + \cos z )$ 
D)  $x e ^ { y } ( 1 + \sin z )$ 
E)  $x e ^ { y } ( 1 - \sin z )$

Question 8

The surface integral $\int \int$S $x d S$ where S is $z = x ^ { 2 }$ in the first octant bounded by $x = 1 , y = 2$ and the coordinate planes, is

Accepted Answer

A)  $\frac { 5 \sqrt { 5 } - 1 } { 6 }$ 
B)  $\frac { 5 \sqrt { 5 } + 1 } { 6 }$ 
C)  $\frac { 5 \sqrt { 5 } - 1 } { 3 }$ 
D)  $\frac { 5 \sqrt { 5 } + 1 } { 3 }$ 
E)  $\frac { 5 \sqrt { 5 } - 1 } { 2 }$ 
A)  $\frac { 5 \sqrt { 5 } - 1 } { 6 }$ 
B)  $\frac { 5 \sqrt { 5 } + 1 } { 6 }$ 
C)  $\frac { 5 \sqrt { 5 } - 1 } { 3 }$ 
D)  $\frac { 5 \sqrt { 5 } + 1 } { 3 }$ 
E)  $\frac { 5 \sqrt { 5 } - 1 } { 2 }$

Question 9

Let $f ( x , y , z ) = \ln \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)$ . Its gradient vector field is

Accepted Answer

A)  $\nabla f ( x , y , z ) = - \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
B)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } - \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } - \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
C)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } - \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
D)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } - \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
E)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
A)  $\nabla f ( x , y , z ) = - \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
B)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } - \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } - \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
C)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } - \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
D)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } - \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$ 
E)  $\nabla f ( x , y , z ) = \frac { 2 x } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { i } + \frac { 2 y } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { j } + \frac { 2 z } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } \mathbf { k }$

Question 10

The line integral $\int $C $( x + y ) d x + ( y + z ) d y + ( x + z ) d z$ , where C is the line segment from (0,0,0) to (1,2,4), is

Accepted Answer

A)  $\frac { 39 } { 2 }$ 
B)  $\frac { 37 } { 2 }$ 
C)  $\frac { 35 } { 2 }$ 
D)  $\frac { 33 } { 2 }$ 
E)  $\frac { 31 } { 2 }$ 
A)  $\frac { 39 } { 2 }$ 
B)  $\frac { 37 } { 2 }$ 
C)  $\frac { 35 } { 2 }$ 
D)  $\frac { 33 } { 2 }$ 
E)  $\frac { 31 } { 2 }$

Question 11

Let $\mathbf { F } ( x , y , z ) = \frac { x } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathbf { i } + \frac { y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathbf { j } + \mathbf { k }$ . Then curl F is

Accepted Answer

A) -k 
B) 0 
C) i 
D) j 
E) k 
A) -k 
B) 0 
C) i 
D) j 
E) k

Question 12

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

Accepted Answer

A)  $2 x ^ { 2 } y + y \cos x + C$ 
B)  $- x ^ { 2 } y - y \cos x + C$ 
C)  $- x ^ { 2 } y + y \cos x + C$ 
D)  $x ^ { 2 } y - y \cos x + C$ 
E)  $x ^ { 2 } y + y \cos x + C$ 
A)  $2 x ^ { 2 } y + y \cos x + C$ 
B)  $- x ^ { 2 } y - y \cos x + C$ 
C)  $- x ^ { 2 } y + y \cos x + C$ 
D)  $x ^ { 2 } y - y \cos x + C$ 
E)  $x ^ { 2 } y + y \cos x + C$

Question 13

Using Green's Theorem, the line integral $\int $C $\cos y d x + \cos x d y$ where C is the rectangle with vertices $( 0,0 ) , \left( \frac { \pi } { 3 } , 0 \right) , \left( \frac { \pi } { 3 } , \frac { \pi } { 4 } \right)$ and $\left( 0 , \frac { \pi } { 4 } \right)$ is

Accepted Answer

A)  $\frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 12 }$ 
B)  $\frac { \pi ( 5 + 4 \sqrt { 2 } ) } { 12 }$ 
C)  $\frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 24 }$ 
D)  $\frac { \pi ( 5 + 4 \sqrt { 2 } ) } { 24 }$ 
E)  $- \frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 24 }$ 
A)  $\frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 12 }$ 
B)  $\frac { \pi ( 5 + 4 \sqrt { 2 } ) } { 12 }$ 
C)  $\frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 24 }$ 
D)  $\frac { \pi ( 5 + 4 \sqrt { 2 } ) } { 24 }$ 
E)  $- \frac { \pi ( 5 - 4 \sqrt { 2 } ) } { 24 }$

Question 14

Let $f ( x , y ) = \sin x + x y + \cos y .$ Its gradient vector field is

Accepted Answer

A)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } - ( x + \sin y ) \mathbf { j }$ 
B)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } + ( x + \sin y ) \mathbf { j }$ 
C)  $\nabla f ( x , y ) = ( \cos x - y ) \mathbf { i } + ( x - \sin y ) \mathbf { j }$ 
D)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } + ( x - \sin y ) \mathbf { j }$ 
E)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } - ( x - \sin y ) \mathbf { j }$ 
A)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } - ( x + \sin y ) \mathbf { j }$ 
B)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } + ( x + \sin y ) \mathbf { j }$ 
C)  $\nabla f ( x , y ) = ( \cos x - y ) \mathbf { i } + ( x - \sin y ) \mathbf { j }$ 
D)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } + ( x - \sin y ) \mathbf { j }$ 
E)  $\nabla f ( x , y ) = ( \cos x + y ) \mathbf { i } - ( x - \sin y ) \mathbf { j }$

Question 15

Let $\mathbf { F } ( x , y , z ) = x z \mathbf { i } + x y \mathbf { j } + y ^ { 2 } \mathbf { k }$ s and C be the boundary of the surface consisting of $z = 4 - x ^ { 2 }$ in the first octant cut by y = 3, and the coordinate planes. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

Accepted Answer

A) 36 
B) 40 
C) 42 
D) 45 
E) 48 
A) 36 
B) 40 
C) 42 
D) 45 
E) 48

Question 16

Let $\mathbf { F } ( x , y , z ) = 4 x \mathbf { i } - 3 y \mathbf { j } + 5 \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = z$ such that $x ^ { 2 } + y ^ { 2 } \leq 4$ Then the flux of $\text { F }$ through S is

Accepted Answer

A) 24? 
B) 20? 
C) 18? 
D) 16? 
E) 12? 
A) 24? 
B) 20? 
C) 18? 
D) 16? 
E) 12?

Question 17

Using Green's Theorem, the line integral $\int $C $4 y d x + 3 x d y$ where C is the quadrilateral with vertices at (0, 0), (1, 0), (1, 1), and (0, 1), is

Accepted Answer

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

Question 18

If $\mathbf { F } ( x , y ) = ( 6 x - 5 y ) \mathbf { i } - \left( 5 x - 6 y ^ { 2 } \right) \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

Accepted Answer

A)  $3 x ^ { 2 } - 5 x y + 2 y ^ { 3 } + C$ 
B)  $3 x ^ { 2 } + 5 x y + 2 y ^ { 3 } + C$ 
C)  $3 x ^ { 2 } - 5 x y - 2 y ^ { 3 } + C$ 
D)  $3 x ^ { 2 } + 5 x y - 2 y ^ { 3 } + C$ 
E)  $- 3 x ^ { 2 } - 5 x y + 2 y ^ { 3 } + C$ 
A)  $3 x ^ { 2 } - 5 x y + 2 y ^ { 3 } + C$ 
B)  $3 x ^ { 2 } + 5 x y + 2 y ^ { 3 } + C$ 
C)  $3 x ^ { 2 } - 5 x y - 2 y ^ { 3 } + C$ 
D)  $3 x ^ { 2 } + 5 x y - 2 y ^ { 3 } + C$ 
E)  $- 3 x ^ { 2 } - 5 x y + 2 y ^ { 3 } + C$

Question 19

The surface integral $\int \int$S $( x + y ) d S$ where S is $4 x + 3 y + 6 z = 12$ in the first octant, is

Accepted Answer

A)  $\frac { 5 \sqrt { 5 } - 1 } { 6 }$ 
B)  $\frac { 15 \pi } { \sqrt { 2 } }$ 
C)  $\frac { 7 \sqrt { 61 } } { 3 }$ 
D)  $\frac { \sqrt { 3 } } { 6 }$ 
E) 0 
A)  $\frac { 5 \sqrt { 5 } - 1 } { 6 }$ 
B)  $\frac { 15 \pi } { \sqrt { 2 } }$ 
C)  $\frac { 7 \sqrt { 61 } } { 3 }$ 
D)  $\frac { \sqrt { 3 } } { 6 }$ 
E) 0

Question 20

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

Accepted Answer

A)  $\{ ( x , y , z ) : x > y > z > 0 \}$ 
B)  $\{ ( x , y , z ) : x y > 0 \}$ 
C)  $\{ ( x , y , z ) : x z > 0 \}$ 
D)  $\{ ( x , y , z ) : y z > 0 \}$ 
E)  $\{ ( x , y , z ) : x > 0 , y > 0 , z > 0 \}$ 
A)  $\{ ( x , y , z ) : x > y > z > 0 \}$ 
B)  $\{ ( x , y , z ) : x y > 0 \}$ 
C)  $\{ ( x , y , z ) : x z > 0 \}$ 
D)  $\{ ( x , y , z ) : y z > 0 \}$ 
E)  $\{ ( x , y , z ) : x > 0 , y > 0 , z > 0 \}$

A parameterization of $2 x - 3 y + z = 5$ is

Let $\mathbf { F } ( x , y , z ) = 3 z \mathbf { i } - 4 \mathbf { j } + y \mathbf { k }$ and S is $x + y + z = 1$ in the first octant. Then the flux of $\text { F }$ through S is

Let $\mathbf { F } ( x , y , z ) = 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$ . Using the Divergence Theorem, $\iint S$ $\mathbf { F } \bullet \mathbf { n } d S$ is

Let $\mathbf { F } ( x , y , z ) = z \mathbf { i } + x \mathbf { j } + y \mathbf { k } , \mathbf { r } ( t ) = \cos t \mathbf { i } + \sin t \mathbf { j } + t \mathbf { k } , 0 \leq t \leq 2 \pi$ . Then $\int$ C $\mathbf { F } \bullet d \mathbf { r }$ is

Let $f ( x , y ) = 2 x ^ { 3 } y + 7 x ^ { 2 } y ^ { 2 } - 5 x y ^ { 3 }$ . Its gradient vector field is

The line integral $\int$ C $( x + y ) d s$ , where C is the curve $x = t , y = 1 - t , 0 \leq t \leq 1 ,$ is

The divergence of $\mathbf { F } ( x , y , z ) = \left( y ^ { 2 } + z ^ { 2 } \right) \mathbf { i } + x e ^ { y } \mathbf { j } - x e ^ { y } \cos z \mathbf { k }$ is

The surface integral $\int \int$ S $x d S$ where S is $z = x ^ { 2 }$ in the first octant bounded by $x = 1 , y = 2$ and the coordinate planes, is

Let $f ( x , y , z ) = \ln \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)$ . Its gradient vector field is

The line integral $\int$ C $( x + y ) d x + ( y + z ) d y + ( x + z ) d z$ , where C is the line segment from (0,0,0) to (1,2,4), is

Let $\mathbf { F } ( x , y , z ) = \frac { x } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathbf { i } + \frac { y } { \left( x ^ { 2 } + y ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathbf { j } + \mathbf { k }$ . Then curl F is

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

Using Green's Theorem, the line integral $\int$ C $\cos y d x + \cos x d y$ where C is the rectangle with vertices $( 0,0 ) , \left( \frac { \pi } { 3 } , 0 \right) , \left( \frac { \pi } { 3 } , \frac { \pi } { 4 } \right)$ and $\left( 0 , \frac { \pi } { 4 } \right)$ is

Let $f ( x , y ) = \sin x + x y + \cos y .$ Its gradient vector field is

Let $\mathbf { F } ( x , y , z ) = 4 x \mathbf { i } - 3 y \mathbf { j } + 5 \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = z$ such that $x ^ { 2 } + y ^ { 2 } \leq 4$ Then the flux of $\text { F }$ through S is

Using Green's Theorem, the line integral $\int$ C $4 y d x + 3 x d y$ where C is the quadrilateral with vertices at (0, 0), (1, 0), (1, 1), and (0, 1), is

If $\mathbf { F } ( x , y ) = ( 6 x - 5 y ) \mathbf { i } - \left( 5 x - 6 y ^ { 2 } \right) \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

The surface integral $\int \int$ S $( x + y ) d S$ where S is $4 x + 3 y + 6 z = 12$ in the first octant, is

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

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Exam 16: Vector Calculus

A parameterization of 2x−3y+z=52 x - 3 y + z = 52x−3y+z=5 is

Let F(x,y,z)=3zi−4j+yk\mathbf { F } ( x , y , z ) = 3 z \mathbf { i } - 4 \mathbf { j } + y \mathbf { k }F(x,y,z)=3zi−4j+yk and S is x+y+z=1x + y + z = 1x+y+z=1 in the first octant. Then the flux of F \text { F } F through S is

Let F(x,y,z)=2zk\mathbf { F } ( x , y , z ) = 2 z \mathbf { k }F(x,y,z)=2zk and S is x2+y2+z2=1x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1x2+y2+z2=1 . Using the Divergence Theorem, ∬S\iint S∬S F∙ndS\mathbf { F } \bullet \mathbf { n } d SF∙ndS is

Let f(x,y)=2x3y+7x2y2−5xy3f ( x , y ) = 2 x ^ { 3 } y + 7 x ^ { 2 } y ^ { 2 } - 5 x y ^ { 3 }f(x,y)=2x3y+7x2y2−5xy3 . Its gradient vector field is

The line integral ∫\int ∫ C (x+y)ds( x + y ) d s(x+y)ds , where C is the curve x=t,y=1−t,0≤t≤1,x = t , y = 1 - t , 0 \leq t \leq 1 ,x=t,y=1−t,0≤t≤1, is

The divergence of F(x,y,z)=(y2+z2)i+xeyj−xeycos⁡zk\mathbf { F } ( x , y , z ) = \left( y ^ { 2 } + z ^ { 2 } \right) \mathbf { i } + x e ^ { y } \mathbf { j } - x e ^ { y } \cos z \mathbf { k }F(x,y,z)=(y2+z2)i+xeyj−xeycoszk is

The surface integral ∫∫\int \int∫∫ S xdSx d SxdS where S is z=x2z = x ^ { 2 }z=x2 in the first octant bounded by x=1,y=2x = 1 , y = 2x=1,y=2 and the coordinate planes, is

Let f(x,y,z)=ln⁡(x2+y2+z2)f ( x , y , z ) = \ln \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)f(x,y,z)=ln(x2+y2+z2) . Its gradient vector field is

The line integral ∫\int ∫ C (x+y)dx+(y+z)dy+(x+z)dz( x + y ) d x + ( y + z ) d y + ( x + z ) d z(x+y)dx+(y+z)dy+(x+z)dz , where C is the line segment from (0,0,0) to (1,2,4), is

If F(x,y)=(2xy−ysin⁡x)i+(x2+cos⁡x)j\mathbf { F } ( x , y ) = ( 2 x y - y \sin x ) \mathbf { i } + \left( x ^ { 2 } + \cos x \right) \mathbf { j }F(x,y)=(2xy−ysinx)i+(x2+cosx)j is a conservative field, then its potential function f(x,y)f ( x , y )f(x,y) is

Let f(x,y)=sin⁡x+xy+cos⁡y.f ( x , y ) = \sin x + x y + \cos y .f(x,y)=sinx+xy+cosy. Its gradient vector field is

Let F(x,y,z)=4xi−3yj+5k\mathbf { F } ( x , y , z ) = 4 x \mathbf { i } - 3 y \mathbf { j } + 5 \mathbf { k }F(x,y,z)=4xi−3yj+5k and S is x2+y2=zx ^ { 2 } + y ^ { 2 } = zx2+y2=z such that x2+y2≤4x ^ { 2 } + y ^ { 2 } \leq 4x2+y2≤4 Then the flux of F \text { F } F through S is

Using Green's Theorem, the line integral ∫\int ∫ C 4ydx+3xdy4 y d x + 3 x d y4ydx+3xdy where C is the quadrilateral with vertices at (0, 0), (1, 0), (1, 1), and (0, 1), is

If F(x,y)=(6x−5y)i−(5x−6y2)j\mathbf { F } ( x , y ) = ( 6 x - 5 y ) \mathbf { i } - \left( 5 x - 6 y ^ { 2 } \right) \mathbf { j }F(x,y)=(6x−5y)i−(5x−6y2)j is a conservative field, then its potential function f(x,y)f ( x , y )f(x,y) is

The surface integral ∫∫\int \int∫∫ S (x+y)dS( x + y ) d S(x+y)dS where S is 4x+3y+6z=124 x + 3 y + 6 z = 124x+3y+6z=12 in the first octant, is

The domain of the vector field F(x,y,z)=ln⁡xi+ln⁡yj+ln⁡zk\mathbf { F } ( x , y , z ) = \ln x \mathbf { i } + \ln y \mathbf { j } + \ln z \mathbf { k }F(x,y,z)=lnxi+lnyj+lnzk is

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A parameterization of $2 x - 3 y + z = 5$ is

Let $\mathbf { F } ( x , y , z ) = 3 z \mathbf { i } - 4 \mathbf { j } + y \mathbf { k }$ and S is $x + y + z = 1$ in the first octant. Then the flux of $\text { F }$ through S is

Let $\mathbf { F } ( x , y , z ) = 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$ . Using the Divergence Theorem, $\iint S$ $\mathbf { F } \bullet \mathbf { n } d S$ is

Let $f ( x , y ) = 2 x ^ { 3 } y + 7 x ^ { 2 } y ^ { 2 } - 5 x y ^ { 3 }$ . Its gradient vector field is

The line integral $\int$ C $( x + y ) d s$ , where C is the curve $x = t , y = 1 - t , 0 \leq t \leq 1 ,$ is

The divergence of $\mathbf { F } ( x , y , z ) = \left( y ^ { 2 } + z ^ { 2 } \right) \mathbf { i } + x e ^ { y } \mathbf { j } - x e ^ { y } \cos z \mathbf { k }$ is

The surface integral $\int \int$ S $x d S$ where S is $z = x ^ { 2 }$ in the first octant bounded by $x = 1 , y = 2$ and the coordinate planes, is

Let $f ( x , y , z ) = \ln \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \right)$ . Its gradient vector field is

The line integral $\int$ C $( x + y ) d x + ( y + z ) d y + ( x + z ) d z$ , where C is the line segment from (0,0,0) to (1,2,4), is

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

Let $f ( x , y ) = \sin x + x y + \cos y .$ Its gradient vector field is

Let $\mathbf { F } ( x , y , z ) = 4 x \mathbf { i } - 3 y \mathbf { j } + 5 \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = z$ such that $x ^ { 2 } + y ^ { 2 } \leq 4$ Then the flux of $\text { F }$ through S is

Using Green's Theorem, the line integral $\int$ C $4 y d x + 3 x d y$ where C is the quadrilateral with vertices at (0, 0), (1, 0), (1, 1), and (0, 1), is

If $\mathbf { F } ( x , y ) = ( 6 x - 5 y ) \mathbf { i } - \left( 5 x - 6 y ^ { 2 } \right) \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ is

The surface integral $\int \int$ S $( x + y ) d S$ where S is $4 x + 3 y + 6 z = 12$ in the first octant, is

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