Question 1

Applying Green's Theorem, the area of the region bounded by $x = \cos ^ { 3 } t$ and $y = \sin ^ { 3 } t$ with $0 \leq t \leq 2 \pi$ is

Accepted Answer

A)  $\frac { 3 \pi } { 8 }$ 
B)  $\frac { 5 \pi } { 8 }$ 
C)  $\frac { 7 \pi } { 8 }$ 
D)  $\frac { 9 \pi } { 8 }$ 
E)  $\frac { 11 \pi } { 8 }$ 
A)  $\frac { 3 \pi } { 8 }$ 
B)  $\frac { 5 \pi } { 8 }$ 
C)  $\frac { 7 \pi } { 8 }$ 
D)  $\frac { 9 \pi } { 8 }$ 
E)  $\frac { 11 \pi } { 8 }$

Question 2

The surface integral $\int \int$S $x d S$ where S is $x + y + z = 1$ in the first octant, is

Accepted Answer

A)  $\frac { \sqrt { 3 } } { 20 }$ 
B)  $- \frac { \sqrt { 3 } } { 21 }$ 
C)  $\frac { \sqrt { 3 } } { 21 }$ 
D)  $\frac { \sqrt { 3 } } { 6 }$ 
E)  $- \frac { \sqrt { 3 } } { 24 }$ 
A)  $\frac { \sqrt { 3 } } { 20 }$ 
B)  $- \frac { \sqrt { 3 } } { 21 }$ 
C)  $\frac { \sqrt { 3 } } { 21 }$ 
D)  $\frac { \sqrt { 3 } } { 6 }$ 
E)  $- \frac { \sqrt { 3 } } { 24 }$

Question 3

The line integral $\int $C $\left( x ^ { 2 } + x y \right) d x + \left( y ^ { 2 } - x y \right) d y$ , where C is the curve $2 y = x ^ { 2 }$ from (0,0) to (2,2), is

Accepted Answer

A)  $\frac { 76 } { 15 }$ 
B)  $\frac { 68 } { 15 }$ 
C)  $\frac { 64 } { 15 }$ 
D)  $\frac { 62 } { 15 }$ 
E)  $\frac { 59 } { 15 }$ 
A)  $\frac { 76 } { 15 }$ 
B)  $\frac { 68 } { 15 }$ 
C)  $\frac { 64 } { 15 }$ 
D)  $\frac { 62 } { 15 }$ 
E)  $\frac { 59 } { 15 }$

Question 4

The area of the surface $\mathbf { r } ( u , v ) = \cos u \mathbf { i } + \sin u \mathbf { j } + v \mathbf { k }$ , where $0 \leq u \leq 2 \pi$ $0 \leq v \leq 3$ is

Accepted Answer

A) 8$\pi$ 
B) 6$\pi$ 
C) 4$\pi$ 
D) 2$\pi$ 
E) $\pi$ 
A) 8$\pi$ 
B) 6$\pi$ 
C) 4$\pi$ 
D) 2$\pi$ 
E) $\pi$

Question 5

Let $\mathbf { F } ( x , y ) = y \sin x \mathbf { i } - \cos x \mathbf { j }$ where C is the line segment from $\left( \frac { \pi } { 2 } , 0 \right)$ to $( \pi , 1 )$ . Then $\int $C $\mathbf { F } \bullet d \mathbf { r }$ is

Accepted Answer

A) 1 
B) 2 
C)  $\frac { 5 } { 2 }$ 
D) 3 
E)  $\frac { 7 } { 2 }$ 
A) 1 
B) 2 
C)  $\frac { 5 } { 2 }$ 
D) 3 
E)  $\frac { 7 } { 2 }$

Question 6

Let $\mathbf { F } ( x , y , z ) = 2 x \mathbf { i } - 2 x \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is the region bounded by and $x = 0 , x = 3$ . Using the Divergence Theorem, $\iint S$ $y = 0 , y = 3$ is

Accepted Answer

A) 54 
B) 63 
C) 72 
D) 81 
E) 108 
A) 54 
B) 63 
C) 72 
D) 81 
E) 108

Question 7

A parameterization of $z = 5 x ^ { 2 }$ is

Accepted Answer

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

Question 8

Applying Green's Theorem, the area of the region bounded by $x = t - \sin t$ and $y = 1 - \cos t$ with $0 \leq t \leq 2 \pi$ is

Accepted Answer

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

Question 9

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

Accepted Answer

A)  $- x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
B)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } - \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } - \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
C)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } - \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
D)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
E)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } - \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
A)  $- x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
B)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } - \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } - \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
C)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } - \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
D)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } + \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$ 
E)  $x e ^ { y } ( \sin z - \cos z ) \mathbf { i } - \left( e ^ { y } \cos z + 2 z \right) \mathbf { j } + \left( e ^ { y } \cos z - 2 y \right) \mathbf { k }$

Question 10

Let $f ( x , y , z ) = \ln ( x y z )$ . Its gradient vector field is

Accepted Answer

A)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$ 
B)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$ 
C)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$ 
D)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$ 
E)  $\nabla f ( x , y ) = - \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$ 
A)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$ 
B)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$ 
C)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$ 
D)  $\nabla f ( x , y ) = \frac { 1 } { x } \mathbf { i } - \frac { 1 } { y } \mathbf { j } - \frac { 1 } { z } \mathbf { k }$ 
E)  $\nabla f ( x , y ) = - \frac { 1 } { x } \mathbf { i } + \frac { 1 } { y } \mathbf { j } + \frac { 1 } { z } \mathbf { k }$

Question 11

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { x y } \mathbf { i } + \frac { 5 } { x y } \mathbf { j }$ is

Accepted Answer

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

Question 12

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

Accepted Answer

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

Question 13

Using Green's Theorem, the line integral $\int $C $e ^ { x + y } d x + e ^ { x + y } d y$ where C is $x ^ { 2 } + y ^ { 2 } = 4$ is

Accepted Answer

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

Question 14

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

Accepted Answer

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

Question 15

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 (1, 0) to (0, 1) is

Accepted Answer

A)  $\frac { 25 } { 12 }$ 
B)  $\frac { 23 } { 12 }$ 
C)  $\frac { 19 } { 12 }$ 
D)  $\frac { 17 } { 12 }$ 
E)  $\frac { 13 } { 12 }$ 
A)  $\frac { 25 } { 12 }$ 
B)  $\frac { 23 } { 12 }$ 
C)  $\frac { 19 } { 12 }$ 
D)  $\frac { 17 } { 12 }$ 
E)  $\frac { 13 } { 12 }$

Question 16

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

Accepted Answer

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

Question 17

Let $\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = 4$ between $z = 0$ and $z = 3$ Then the flux of $\text { F }$ through S is

Accepted Answer

A) 18? 
B) 20? 
C) 24? 
D) 26? 
E) 28? 
A) 18? 
B) 20? 
C) 24? 
D) 26? 
E) 28?

Question 18

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

Accepted Answer

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

Question 19

Using Green's Theorem, the line integral $\int $C $\left( \sin ^ { 4 } x + e ^ { 2 x } \right) d x + \left( \cos ^ { 3 } y - e ^ { y } \right) d y$ where C is $x ^ { 2 } + y ^ { 2 } = 16$ is

Accepted Answer

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

Question 20

Using Green's Theorem, the line integral $\int $C $\left( - x ^ { 2 } + x \right) d y$ where C is the line segment from (0, 0) to (2, 1), and $x = 2 y ^ { 2 }$ from (2, 1) to (0, 0), is

Accepted Answer

A)  $- \frac { 1 } { 5 }$ 
B)  $\frac { 2 } { 5 }$ 
C)  $\frac { 3 } { 5 }$ 
D)  $\frac { 4 } { 5 }$ 
E)  $\frac { 6 } { 5 }$ 
A)  $- \frac { 1 } { 5 }$ 
B)  $\frac { 2 } { 5 }$ 
C)  $\frac { 3 } { 5 }$ 
D)  $\frac { 4 } { 5 }$ 
E)  $\frac { 6 } { 5 }$

Applying Green's Theorem, the area of the region bounded by $x = \cos ^ { 3 } t$ and $y = \sin ^ { 3 } t$ with $0 \leq t \leq 2 \pi$ is

The surface integral $\int \int$ S $x d S$ where S is $x + y + z = 1$ in the first octant, is

The line integral $\int$ C $\left( x ^ { 2 } + x y \right) d x + \left( y ^ { 2 } - x y \right) d y$ , where C is the curve $2 y = x ^ { 2 }$ from (0,0) to (2,2), is

The area of the surface $\mathbf { r } ( u , v ) = \cos u \mathbf { i } + \sin u \mathbf { j } + v \mathbf { k }$ , where $0 \leq u \leq 2 \pi$ $0 \leq v \leq 3$ is

Let $\mathbf { F } ( x , y ) = y \sin x \mathbf { i } - \cos x \mathbf { j }$ where C is the line segment from $\left( \frac { \pi } { 2 } , 0 \right)$ to $( \pi , 1 )$ . Then $\int$ C $\mathbf { F } \bullet d \mathbf { r }$ is

Let $\mathbf { F } ( x , y , z ) = 2 x \mathbf { i } - 2 x \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is the region bounded by and $x = 0 , x = 3$ . Using the Divergence Theorem, $\iint S$ $y = 0 , y = 3$ is

A parameterization of $z = 5 x ^ { 2 }$ is

Applying Green's Theorem, the area of the region bounded by $x = t - \sin t$ and $y = 1 - \cos t$ with $0 \leq t \leq 2 \pi$ is

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

Let $f ( x , y , z ) = \ln ( x y z )$ . Its gradient vector field is

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { x y } \mathbf { i } + \frac { 5 } { x y } \mathbf { j }$ is

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

Using Green's Theorem, the line integral $\int$ C $e ^ { x + y } d x + e ^ { x + y } d y$ where C is $x ^ { 2 } + y ^ { 2 } = 4$ is

Let $f ( x , y ) = \ln \left( \frac { y } { x } \right) + e ^ { \frac { y } { x } }$ . Its gradient vector field 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 (1, 0) to (0, 1) is

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

Let $\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = 4$ between $z = 0$ and $z = 3$ Then the flux of $\text { F }$ through S is

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

Using Green's Theorem, the line integral $\int$ C $\left( \sin ^ { 4 } x + e ^ { 2 x } \right) d x + \left( \cos ^ { 3 } y - e ^ { y } \right) d y$ where C is $x ^ { 2 } + y ^ { 2 } = 16$ is

Using Green's Theorem, the line integral $\int$ C $\left( - x ^ { 2 } + x \right) d y$ where C is the line segment from (0, 0) to (2, 1), and $x = 2 y ^ { 2 }$ from (2, 1) to (0, 0), is

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Exam 16: Vector Calculus

Applying Green's Theorem, the area of the region bounded by x=cos⁡3tx = \cos ^ { 3 } tx=cos3t and y=sin⁡3ty = \sin ^ { 3 } ty=sin3t with 0≤t≤2π0 \leq t \leq 2 \pi0≤t≤2π is

The surface integral ∫∫\int \int∫∫ S xdSx d SxdS where S is x+y+z=1x + y + z = 1x+y+z=1 in the first octant, is

The line integral ∫\int ∫ C (x2+xy)dx+(y2−xy)dy\left( x ^ { 2 } + x y \right) d x + \left( y ^ { 2 } - x y \right) d y(x2+xy)dx+(y2−xy)dy , where C is the curve 2y=x22 y = x ^ { 2 }2y=x2 from (0,0) to (2,2), is

The area of the surface r(u,v)=cos⁡ui+sin⁡uj+vk\mathbf { r } ( u , v ) = \cos u \mathbf { i } + \sin u \mathbf { j } + v \mathbf { k }r(u,v)=cosui+sinuj+vk , where 0≤u≤2π0 \leq u \leq 2 \pi0≤u≤2π 0≤v≤30 \leq v \leq 30≤v≤3 is

Let F(x,y,z)=2xi−2xj+z2k\mathbf { F } ( x , y , z ) = 2 x \mathbf { i } - 2 x \mathbf { j } + z ^ { 2 } \mathbf { k }F(x,y,z)=2xi−2xj+z2k and S is the region bounded by and x=0,x=3x = 0 , x = 3x=0,x=3 . Using the Divergence Theorem, ∬S\iint S∬S y=0,y=3y = 0 , y = 3y=0,y=3 is

A parameterization of z=5x2z = 5 x ^ { 2 }z=5x2 is

Applying Green's Theorem, the area of the region bounded by x=t−sin⁡tx = t - \sin tx=t−sint and y=1−cos⁡ty = 1 - \cos ty=1−cost with 0≤t≤2π0 \leq t \leq 2 \pi0≤t≤2π is

Let F(x,y,z)=(y2+z2)i+xeycos⁡zj−xeycos⁡zk\mathbf { F } ( x , y , z ) = \left( y ^ { 2 } + z ^ { 2 } \right) \mathbf { i } + x e ^ { y } \cos z \mathbf { j } - x e ^ { y } \cos z \mathbf { k }F(x,y,z)=(y2+z2)i+xeycoszj−xeycoszk . Then curl F is

Let f(x,y,z)=ln⁡(xyz)f ( x , y , z ) = \ln ( x y z )f(x,y,z)=ln(xyz) . Its gradient vector field is

The domain of the vector field F(x,y)=xyi+5xyj\mathbf { F } ( x , y ) = \sqrt { x y } \mathbf { i } + \frac { 5 } { x y } \mathbf { j }F(x,y)=xy​i+xy5​j is

Using Green's Theorem, the line integral ∫\int ∫ C ex+ydx+ex+ydye ^ { x + y } d x + e ^ { x + y } d yex+ydx+ex+ydy where C is x2+y2=4x ^ { 2 } + y ^ { 2 } = 4x2+y2=4 is

Let f(x,y)=ln⁡(yx)+eyxf ( x , y ) = \ln \left( \frac { y } { x } \right) + e ^ { \frac { y } { x } }f(x,y)=ln(xy​)+exy​ . Its gradient vector field is

The work done by the force F(x,y)=−x2yi+2yj\mathbf { F } ( x , y ) = - x ^ { 2 } y \mathbf { i } + 2 y \mathbf { j }F(x,y)=−x2yi+2yj moving along the line segment from (1, 0) to (0, 1) is

Let F(x,y,z)=x2i+y2j+z2k\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }F(x,y,z)=x2i+y2j+z2k . Then curl F is

Let F(x,y,z)=xi+yj+2zk\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + 2 z \mathbf { k }F(x,y,z)=xi+yj+2zk and S is x2+y2=4x ^ { 2 } + y ^ { 2 } = 4x2+y2=4 between z=0z = 0z=0 and z=3z = 3z=3 Then the flux of F \text { F } F through S is

Let f(x,y)=2x3+xy2+xz2f ( x , y ) = 2 x ^ { 3 } + x y ^ { 2 } + x z ^ { 2 }f(x,y)=2x3+xy2+xz2 . Its gradient vector field is

Using Green's Theorem, the line integral ∫\int ∫ C (sin⁡4x+e2x)dx+(cos⁡3y−ey)dy\left( \sin ^ { 4 } x + e ^ { 2 x } \right) d x + \left( \cos ^ { 3 } y - e ^ { y } \right) d y(sin4x+e2x)dx+(cos3y−ey)dy where C is x2+y2=16x ^ { 2 } + y ^ { 2 } = 16x2+y2=16 is

Using Green's Theorem, the line integral ∫\int ∫ C (−x2+x)dy\left( - x ^ { 2 } + x \right) d y(−x2+x)dy where C is the line segment from (0, 0) to (2, 1), and x=2y2x = 2 y ^ { 2 }x=2y2 from (2, 1) to (0, 0), is

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Applying Green's Theorem, the area of the region bounded by $x = \cos ^ { 3 } t$ and $y = \sin ^ { 3 } t$ with $0 \leq t \leq 2 \pi$ is

The surface integral $\int \int$ S $x d S$ where S is $x + y + z = 1$ in the first octant, is

The line integral $\int$ C $\left( x ^ { 2 } + x y \right) d x + \left( y ^ { 2 } - x y \right) d y$ , where C is the curve $2 y = x ^ { 2 }$ from (0,0) to (2,2), is

The area of the surface $\mathbf { r } ( u , v ) = \cos u \mathbf { i } + \sin u \mathbf { j } + v \mathbf { k }$ , where $0 \leq u \leq 2 \pi$ $0 \leq v \leq 3$ is

Let $\mathbf { F } ( x , y , z ) = 2 x \mathbf { i } - 2 x \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is the region bounded by and $x = 0 , x = 3$ . Using the Divergence Theorem, $\iint S$ $y = 0 , y = 3$ is

A parameterization of $z = 5 x ^ { 2 }$ is

Applying Green's Theorem, the area of the region bounded by $x = t - \sin t$ and $y = 1 - \cos t$ with $0 \leq t \leq 2 \pi$ is

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

Let $f ( x , y , z ) = \ln ( x y z )$ . Its gradient vector field is

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { x y } \mathbf { i } + \frac { 5 } { x y } \mathbf { j }$ is

Using Green's Theorem, the line integral $\int$ C $e ^ { x + y } d x + e ^ { x + y } d y$ where C is $x ^ { 2 } + y ^ { 2 } = 4$ is

Let $f ( x , y ) = \ln \left( \frac { y } { x } \right) + e ^ { \frac { y } { x } }$ . Its gradient vector field 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 (1, 0) to (0, 1) is

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

Let $\mathbf { F } ( x , y , z ) = x \mathbf { i } + y \mathbf { j } + 2 z \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = 4$ between $z = 0$ and $z = 3$ Then the flux of $\text { F }$ through S is

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

Using Green's Theorem, the line integral $\int$ C $\left( \sin ^ { 4 } x + e ^ { 2 x } \right) d x + \left( \cos ^ { 3 } y - e ^ { y } \right) d y$ where C is $x ^ { 2 } + y ^ { 2 } = 16$ is

Using Green's Theorem, the line integral $\int$ C $\left( - x ^ { 2 } + x \right) d y$ where C is the line segment from (0, 0) to (2, 1), and $x = 2 y ^ { 2 }$ from (2, 1) to (0, 0), is