Exam 16: Vector Calculus

Let $\mathbf { F } ( x , y , z ) = 4 y \mathbf { i } - 3 z \mathbf { j } + x \mathbf { k }$ and C be the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1). Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

Let $\mathbf { F } ( x , y , z ) = x ^ { 2 } \mathbf { i } + y ^ { 2 } \mathbf { j } + z ^ { 2 } \mathbf { k }$ and S is $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ . Using the Divergence Theorem, $\iint S$ $\mathbf { F } \bullet \mathbf { n } d S$ is

Applying Green's Theorem, the area of the region bounded by $y = 2 x ^ { 2 }$ and $y = 8 x$ is

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

If I = $\int$ C $\frac { 1 } { y } d x - \frac { x } { y ^ { 2 } } d y$ is independent of the path where C is a curve from (5, -1) to (9, -3) then I is

If $\mathbf { F } ( x , y ) = \left( \frac { 2 x - 1 } { y } \right) \mathbf { i } + \left( \frac { x - x ^ { 2 } } { y ^ { 2 } } \right) \mathbf { j }$ is a conservative field, then its potential function $f ( x , y )$ 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 (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 (3, 0) to (3, 3) and then the line segment from (3, 3) to (0, 3) is

The work done by the force $\mathbf { F } ( x , y ) = ( y - x ) \mathbf { i } + x ^ { 2 } y \mathbf { j }$ moving along the line segment from (1, 1) to (2, 2) and then the line segment from (2, 2) to (2, 4) is

Let $\mathbf { F } ( x , y , z ) = \frac { x \mathbf { i } + y \mathbf { j } + z \mathbf { k } } { x ^ { 2 } + y ^ { 2 } + z ^ { 2 } }$ and S is the region bounded between $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 1$ and $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ . Using the Divergence Theorem, $\iint S$ $\mathbf { F } \cdot \mathbf { n } d S$ is

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

A parameterization of $x ^ { 2 } + 4 y ^ { 2 } = 16$ is

A parameterization of $x ^ { 2 } + y ^ { 2 } = 9$ is

The domain of the vector field $\mathbf { F } ( x , y ) = \frac { x y } { x ^ { 2 } + y ^ { 2 } } \mathbf { i } + \frac { 5 } { x ^ { 2 } - y ^ { 2 } } \mathbf { j }$ is

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

Let $\mathbf { F } ( x , y , z ) = z ^ { 2 } \mathbf { i } + x ^ { 2 } \mathbf { j } + y ^ { 2 } \mathbf { k }$ and C be the boundary of the surface $4 - x ^ { 2 } - y ^ { 2 } = z$ above z = 0. Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

The work done by the force $\mathbf { F } ( x , y ) = 3 y \mathbf { i } + 4 x \mathbf { j }$ moving along $\mathbf { r } ( t ) = 2 t ^ { 2 } \mathbf { i } - t \mathbf { j }$ with $0 \leq t \leq 1$ is

Let $\mathbf { F } ( x , y , z ) = ( y - x ) \mathbf { i } + ( x - z ) \mathbf { j } + ( x - y ) \mathbf { k }$ and C be the triangle with vertices (2, 0, 0), (0, 2, 0), and (0, 0, 2). Using Stokes' Theorem, $\int _ { C } \mathbf { F } \cdot \mathbf { T } d s$ is

Let $\mathbf { F } ( x , y , z ) = ( x + y ) \mathbf { i } + y \mathbf { j } + z \mathbf { k }$ and S is $z = 1 - x ^ { 2 } - y ^ { 2 }$ between $z = 1 - x ^ { 2 } - y ^ { 2 }$ and $z = 1$ Then the flux of $\text { F }$ through S is

The domain of the vector field $\mathbf { F } ( x , y ) = \sqrt { 1 - x ^ { 2 } } \mathbf { i } + \ln y \mathbf { j }$ is