Exam 16: Vector Calculus

The line integral $\int$ C $\left( y - x ^ { 2 } \right) d s$ , where C is the curve $x = t , y = 2 t , 0 \leq t \leq 1 \text {, }$ is

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

If I = $\int$ C $( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y$ is independent of the path where C is a curve from $\left( 0 , \frac { \pi } { 4 } \right)$ to $\left( \frac { \pi } { 4 } , - \frac { \pi } { 4 } \right)$ then I is

Let $\mathbf { F } ( x , y , z ) = - 3 y \mathbf { i } + 3 x \mathbf { j } + 2 \mathbf { k }$ and C be the boundary of z = 1 inside $x ^ { 2 } + y ^ { 2 } = 9$ . Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

The line integral $\int$ C $( x y - z ) d x + e ^ { x } d y + y d z$ , where C is the line segment from (1,0,0) to (3,4,8), is

If I = $\int$ C $( x + y ) d x + ( 2 y + x ) d y$ is independent of the path where C is a curve from (0, 2) to (1, 3), then I is

The surface integral $\int \int$ S $x y z d S \text {, }$ where S is $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ between $z = 1$ and $z = 2$ is

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } y \mathbf { i } + y ^ { 2 } \mathbf { j } + x z \mathbf { k }$ and S is in the first octant bounded by and the coordinate planes. Using the Divergence Theorem, $\iint S$ $x = 1 ,$ is

The surface integral $\int \int$ S $x y z d S ,$ where S is $x ^ { 2 } + z ^ { 2 } = 4$ between $y = 1$ and $y = 3 \text {, }$ is

The rectangular equation for the parametric surfaceand $x ( u , v ) = 3 \cos u$ is

Using Green's Theorem, the line integral $\int$ C $y ^ { 2 } d x + x ^ { 2 } d y$ where C is the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1), 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 (4, 0) to (0, 4) is

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

If I = $\int$ C $\left( 2 y e ^ { 2 x } - x ^ { 2 } \right) d x + e ^ { 2 x } d y$ is independent of the path where C is a curve from (0, 1) to (1, 2), then I is

The work done by the force $\mathbf { F } ( x , y ) = ( 2 x + y ) \mathbf { i } + ( x - 2 y ) \mathbf { j }$ moving along $\mathbf { r } ( t ) = 3 \cos t \mathbf { i } + 3 \sin t \mathbf { j }$ with $0 \leq t \leq 2 \pi$ 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 (2, 2) and then the line segment from (2, 2) to (0, 2) is

A parameterization of $x + y + z = 6$ is

The work done by the force $\mathbf { F } ( x , y ) = 2 x y \mathbf { i } + \left( x ^ { 2 } + y ^ { 2 } \right) \mathbf { j }$ moving along $y ^ { 2 } = x$ from (0, 0) to (1, 1) is

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is the boundary of the region enclosed on the side by $x ^ { 2 } + y ^ { 2 } = 9$ , below by $z = 0$ and above $z = 4$ Using the Divergence Theorem, $\iint S$ $\mathbf { F } \bullet \mathbf { n } d S$ is

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