Exam 16: Vector Calculus

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

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

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

Let $\mathbf { F } ( x , y , z ) = 2 x \mathbf { i } - 2 x \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } = 4,0 \leq z \leq 3$ .Using the Divergence Theorem, $\iint S$ $\mathbf { F } \cdot \mathbf { n } d S$ 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 (1, 1) and then the line segment from (1, 1) to (0, 1) is

The rectangular equation for the parametric surface and $x ( u , v ) = \sin u \cos v$ 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 = x from (0, 0) to (1, 1) is

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

The work done by the force $\mathbf { F } ( x , y ) = ( 2 x + 3 y ) \mathbf { i } + x y \mathbf { j }$ moving along $\mathbf { r } ( t ) = 4 \sin t \mathbf { i } - \cos t \mathbf { j }$ with $0 \leq t \leq \frac { \pi } { 2 }$ is

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

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

Let $\mathbf { F } ( x , y , z ) = 3 x ^ { 2 } y \mathbf { i } - 2 x y ^ { 3 } \mathbf { j }$ Then curl F is

The line integral $\int$ C $x y d s ,$ where C is the curve $x = 4 t , y = t ^ { 2 } , 0 \leq t \leq 2$ is

If $\mathbf { F } ( x , y ) = \left( \frac { 1 } { x ^ { 2 } } + \frac { 1 } { y ^ { 2 } } \right) \mathbf { i } + \left( \frac { 1 - 2 x } { y ^ { 3 } } \right) \mathbf { j }$ 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

The outer unit normal vector to $z = 9 - x ^ { 2 } - y ^ { 2 }$ is

Let $f ( x , y ) = \tan ^ { - 1 } \left( \frac { y } { x } \right)$ . Its gradient vector field is

The area of the surface $\mathbf { r } ( u , v ) = 4 u \cos v \mathbf { i } + 4 u \sin v \mathbf { j } + u ^ { 2 } \mathbf { k }$ , where is

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

The surface integral $\int \int$ S $\sqrt { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } } d S$ where S is $z ^ { 2 } = x ^ { 2 } + y ^ { 2 }$ between $z = 0$ and $z = 2$ is