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For every problem there is always, at least, a solution which seems quite plausible. It is simple and clean, direct, neat and nice, and yet very wrong, #Anawim, justtothepoint.com
A vector field is an assignment of a vector $\vec{F}$ to each point (x, y) in a space, i.e., $\vec{F} = M\vec{i}+N\vec{j}$ where M and N are functions of x and y.
A vector field on a plane can be visualized as a collection of arrows, each attached to a point on the plane. These arrows represent vectors with specific magnitudes and directions.
Work is defined as the energy transferred when a force acts on an object and displaces it along a path. In the context of vector fields, we calculate the work done by a force field along a curve or trajectory C using a line integral. The work done by a force field $\vec{F}$ along a curve C is: W = $\int_{C} \vec{F}·d\vec{r} = \int_{C} Mdx + Ndy = \int_{C} \vec{F}·\hat{\mathbf{T}}ds$, where $\hat{\mathbf{T}}$ is the unit tangent vector.
A vector field is conservative if there exist a scalar function such that $\vec{F}$ = ∇f (the vector field is its gradient). This scalar function is known or referred to as the potential function associated with the vector field.
Theorem. Fundamental theorem of calculus for line integral. If $\vec{F}$ is a conservative vector field in a simply connected region of space (i.e., a region with no holes), and if f is a scalar potential function for $\vec{F}$ in that region, then $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P0 and P1 are the initial and final points of the curve C, respectively.
The line integral of the vector field $\vec{F}$ along the curve C is defined as: $\int_{C} \vec{F}d\vec{r}$ where $d\vec{r}$ is an infinitesimal vector tangent to the curve, given by: $d\vec{r} = ⟨dx, dy, dz⟩$.
$\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
The curl of a vector field measures the tendency of the field to induce rotation or “twisting” at a point. It is going to measure how much a vector field fail to be conservative.
Definition. Given a vector field $\vec{F} = P\hat{\mathbf{i}}+ Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$ where P, Q, and R are the components of the field along the x-, y-, and z-directions, respectively, the curl of $\vec{F}$ is defined as $curl(\vec{F})=(R_y-Q_z)\hat{\mathbf{i}} + (P_z-R_x)\hat{\mathbf{j}}+(Q_x-P_y)\hat{\mathbf{k}}$
By construction, if the vector field $\vec{F}$ is defined in a simply connected region (a region without holes or topological obstacles), then $\vec{F}$ is conservative (meaning it can be expressed as the gradient of a scalar function, $\vec{F}= ∇f$ where ∇ is the gradient operator) if and only if its curl equals the zero vector, curl($\vec{F}$) = 0.
This means that there is no rotational component in the field, and the field can be derived from a scalar potential. If the curl is not zero, the field has some rotational aspect and cannot be conservative.
Recall our previous notation, the operator ∇ = $⟨\frac{∂}{∂x},\frac{∂}{∂y},\frac{∂}{∂z}⟩$ and using this notation ∇f = $⟨\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}⟩$
Similarly, ∇·⟨P, Q, R⟩ = $\frac{∂P}{∂x} + \frac{∂Q}{∂y} + \frac{∂R}{∂z} = div \vec{F}$
$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ P & Q & R\end{smallmatrix}| = (\frac{∂R}{∂y}-\frac{∂Q}{∂z})\hat{\mathbf{i}}-(\frac{∂R}{∂x}-\frac{∂P}{∂z})\hat{\mathbf{j}} + (\frac{∂Q}{∂x}-\frac{∂P}{∂y})\hat{\mathbf{k}}$
Geometrically, the curl measures the rotation, twist, or circulation of the field around a point.
For example, consider a velocity field $\vec{v}$ representing the flow of a fluid that induces a rotation around the z-axis at an angular velocity w; the velocity field is given by $\vec{v} = ⟨-wy, wx, 0⟩$. If the curl is zero, the field is conservative, meaning it has no rotational component and can be derived from a scalar potential function.
Let’s compute the curl of $\vec{v}$: $\vec{v} = \langle -wy, wx, 0 \rangle$
$ \nabla \times \vec{v} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) \hat{\mathbf{i}} - \left( \frac{\partial R}{\partial x} - \frac{\partial P}{\partial z} \right) \hat{\mathbf{j}} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \hat{\mathbf{k}}$
Compute each term: $\frac{\partial R}{\partial y} = 0, \frac{\partial Q}{\partial z} = 0, \frac{\partial R}{\partial x} = 0, \frac{\partial P}{\partial z} = 0, \frac{\partial Q}{\partial x} = w, \quad \frac{\partial P}{\partial y} = -w.$
Substitute these into the curl formula: $\nabla \times \vec{v} = (0 - 0) \hat{\mathbf{i}} - (0 - 0) \hat{\mathbf{j}} + (w - (-w)) \hat{\mathbf{k}} = 2w \hat{\mathbf{k}} $
It indicates that the curl captures the rotational aspect of the velocity field with the magnitude 2w representing the strength of this rotation and the direction $\hat{\mathbf{k}}$ (the unit vector along the z-axis) along the z-axis indicating the axis of rotation.
$curl \vec{F} = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ a & b & c\end{smallmatrix}| = (\frac{∂c}{∂y}-\frac{∂b}{∂z})\hat{\mathbf{i}}-(\frac{∂c}{∂x}-\frac{∂a}{∂z})\hat{\mathbf{j}} + (\frac{∂b}{∂x}-\frac{∂a}{∂y})\hat{\mathbf{k}} = ⟨0, 0, 0⟩$
In this case, the vector field only depends on the x-coordinate. The field points in the x-direction, and its magnitude increases as x increases.
$curl \vec{F} = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ x & 0 & 0\end{smallmatrix}| = (\frac{∂0}{∂y}-\frac{∂0}{∂z})\hat{\mathbf{i}}-(\frac{∂0}{∂x}-\frac{∂x}{∂z})\hat{\mathbf{j}} + (\frac{∂0}{∂x}-\frac{∂x}{∂y})\hat{\mathbf{k}} = ⟨0, 0, 0⟩$. This field has no rotational component, but it does have nonzero divergence:
$div \vec{F} = ∇·\vec{F} = \frac{∂x}{∂x} + \frac{∂0}{∂y} + \frac{∂0}{∂z} = 1 + 0 + 0 = 1.$ Thus, this field spreads out as x increases, but it does not rotate.
This field rotates around the z-axis. We can compute its curl to confirm the rotational component:
$curl \vec{F} = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ -y & x & 0\end{smallmatrix}| = (\frac{∂0}{∂y}-\frac{∂x}{∂z})\hat{\mathbf{i}}-(\frac{∂0}{∂x}-\frac{∂(-y)}{∂z})\hat{\mathbf{j}} + (\frac{∂x}{∂x}-\frac{∂(-y)}{∂y})\hat{\mathbf{k}} = ⟨0-0, 0-0, 1-(-1)⟩ = ⟨0, 0, 2⟩.$ This field has a rotational component around the z-axis, with a strength of 2. The positive value indicates a counterclockwise rotation in the xy-plane.
This field has components that are more complex and depend on products of variables. Let’s compute its curl:
$curl \vec{F} = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ cos(xy) & sin(xz) & xz^2\end{smallmatrix}| = (\frac{∂(xz^2)}{∂y}-\frac{∂sin(xz)}{∂z})\hat{\mathbf{i}}-(\frac{∂cos(xy)}{∂z}-\frac{∂(xz^2)}{∂x})\hat{\mathbf{j}} + (\frac{∂sin(xz)}{∂x}-\frac{∂cos(xy)}{∂y})\hat{\mathbf{k}} = ⟨0-xcos(xz), 0-z^2, zcos(xz)+xsin(xy)⟩ = ⟨-xcos(xz),-z^2, zcos(xz)+xsin(xy)⟩$
This result indicates that the field has a nontrivial rotational component that varies in both magnitude and direction depending on x, y, and z.
Theorem. Curl of the Gradient of a Function. If f is a scalar function of three variables that has continuous second order partial derivatives, then $curl(∇f)=0.$ In words, the gradient of a scalar function always produces a vector field with no rotational component.
Prof:
Given a scalar function f with continuous second-order partial derivatives, let’s compute the curl of the gradient of f.
The gradient of f is given by: ∇f = $⟨\frac{∂}{∂x}, \frac{∂}{∂y}, \frac{∂}{∂z}⟩ = ⟨f_x, f_y, f_z⟩$
Now, to compute the curl of the gradient ∇f: $curl(∇f) = ∇ x ∇f = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ f_x & f_y & f_z\end{smallmatrix}| = ⟨f_{zy}-f_{yz}, -(f_{zx}-f_{xz}), f_{yx}-f_{yx}⟩ = ⟨f_{zy}-f_{yz}, f_{xz}-f_{zx}, f_{yx}-f_{yx}⟩$
Since f has continuous second order partial derivatives, by Clairaut’s theorem or Schwarz’s theorem, the mixed partial derivatives are equal (symmetry of second derivate) $f_{zy} = f_{yz}, f_{xz}= f_{zx}, f_{yx} = f_{yx}$
Therefore, $curl(∇f) = ⟨f_{zy}-f_{yz}, f_{xz}-f_{zx}, f_{yx}-f_{yx}⟩ = ⟨0, 0, 0⟩$∎
A vector field $\vec{F}$ is said to be conservative if there exists a scalar potential function f such that $\vec{F} = ∇f.$ In other words, a conservative vector field is one that can be expressed as the gradient of some scalar function.
From the theorem, we know the curl of the gradient of any scalar function is zero: curl(∇f) = 0
if $\vec{F}$ is conservative then curl($\vec{F}$) = curl(∇f) =[Th. Curl of the Gradient of a Function] 0. In words, the curl of a conservative vector field is $\vec{0}$. This is a key characteristic of conservative fields: they have no rotational component, which is why they can be derived from a potential function. The converse is also true under certain conditions.
For a vector field $\vec{F}=⟨P, Q, R⟩$ to be conservative, the following conditions must hold:
A domain D is open if, for every point in D, there is a small neighborhood (an ϵ-neighborhood) around that point that is entirely contained within D. This implies that the domain does not include its boundary.
A domain is simply connected if it has no “holes” or topological obstacles. This means that any closed loop within the domain can be continuously contracted to a single point without leaving the domain.
These conditions ensure that the vector field behaves well enough to have a potential function.
The concept of a conservative field is closely related to potential energy in physics. For example, in a gravitational or electrostatic field, the work done by the field when moving a particle from one point to another depends only on the starting and ending points, not on the path taken. This is because such fields are conservative, meaning they can be described by a potential function.
The conditions for a vector field to be conservative are: the curl must be zero, the field's partial derivatives must be continuous, and the domain must be open and simply connected.
Curl Condition. The first condition for a vector field to be conservative is that its curl must be zero: $curl(\vec{F})=\vec{0}$. $curl \vec{F} = ∇ x \vec{F} = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ z & 2yz & x+y^2\end{smallmatrix}\Biggr \vert = ⟨2y-2y, 1-1, 0-0⟩ = ⟨0, 0, 0⟩$
The second condition is that the vector field components must have continuous first-order partial derivatives. The components of $\vec{F}$ are: P = z, Q = 2yz, and R = x + y2. All of these are polynomials, and polynomials have continuous partial derivatives everywhere. Thus, the second condition is also satisfied.
The domain D of $\vec{F}$ must be open and simply connected. The domain D is ℝ3 is indeed open and simply connected. There are no holes or obstacles in ℝ3, meaning that any loop in the domain can be contracted to a point.
Since all three conditions are satisfied (curl is zero, continuous partial derivatives, and an open and simply connected domain), the vector field $\vec{F}$ is conservative.
Since $\vec{F}$ is conservative, there exists a scalar potential function f such that $\vec{F} = ∇f$. We need to find f such that: fx = z, fy = 2yz, fz = x + y2.
fx = P = z ⇒[Integrate fx with respect to x, $\int \frac{∂f}{∂x}dx = \int zdx$] f = xz + g(y, z), where g(y, z) is an arbitrary function of y and z (since we are treating y and z as constants when integrating with respect to x).
Differentiate f with respect to y. $\frac{∂f}{∂y} = \frac{∂}{∂y}(xz + g(y, z)) = \frac{∂g}{∂y}$. Set this equal to fy = Q = 2yz, $\frac{∂g}{∂y}=2yz$ and integrate with respect to y, g(y, z) = y2z + h(z) where h(z) is an unknown function of z.
Differentiate f with respect to z. $\frac{∂f}{∂z} = \frac{∂}{∂z}(xz + y^2z + h(z)) = x + y^2 + h’(z)$. Set this equal to fz = R = x + y2, $x + y^2 + h’(z) = x + y^2 ⇒h’(z) = 0 ⇒ h(z) = c.$
We can ignore the constant C when computing the potential function since it does not affect the gradient: f(x, y, z) = xz + y2z.
The flux of a vector field through a surface is a measure of how much of the field “flows” or “passes” through the surface. In this problem, instead of directly calculating the flux through the surface, we can use the Divergence Theorem, which relates the flux through a closed surface to the divergence of the vector field over the volume enclosed by the surface.
The Divergence Theorem states that the flux of a vector field $\vec{F}$ through a closed surface S is equal to the integral of the divergence of $\vec{F}$ over the volume D enclosed by the surface:
$\oint_S \vec{F} \cdot d\vec{S} = \int \int \int_{D} div \vec{F} dV =$[Notation] $ \int \int \int_{D} ∇·\vec{F} dV$
Calculate the Divergence of $\vec{F}$: $∇⋅\vec{F} = P_x +Q_y + R_z =[\text{Given: P = x⁴y, Q = -2x³y², R = z²}] = \frac{∂}{∂x}(x^4y)+\frac{∂}{∂y}(-2x^3y^2)+\frac{∂}{∂z}(z^2) = 4x^3y -4x^3y + 2z = 2z.$
Set Up the Volume Integral
The volume integral of the divergence over the region D is given by: $\int \int \int_{D} (∇·\vec{F})dV = \int \int \int_{D} 2zdV$
To evaluate this, we switch to cylindrical coordinates because the region is symmetric about the z-axis. In cylindrical coordinates: x = rcos(θ), y = rsin(θ), z = z, and dV = rdrdθdz.
The limits of integration are: 0 ≤ θ ≤ 2π, 0 ≤ r ≤ R (the radius of the cylinder x2+y2 = R2), and z from 0 to h (the solid cylinder is bounded by the planes z = 0 and z = h).
Flux = $\int_{0}^{2π}\int_{0}^{R}\int_{0}^{h} 2z·rdzdrdθ$
Inner integral with respect to z: $\int_{0}^{h} 2z·rdz = rz^2\bigg|_{0}^{h} = rh^2$.
Middle integral with respect to r: $\int_{0}^{R} rh^2 dr = \frac{1}{2}h^2r^2\bigg|_{0}^{R} = \frac{1}{2}h^2R^2$.
Outer integral with respect to θ: $\int \int \int_{D} (∇·\vec{F})dV = \int \int \int_{D} 2zdV = \int_{0}^{2π} \frac{1}{2}h^2R^2dθ = \frac{1}{2}h^2R^2θ\bigg|_{0}^{2π} = \frac{1}{2}h^2R^22π = πh^2R^2$
The flux of the vector field $\vec{F} = ⟨x^4y, -2x^3y^2, z^2⟩$ through the surface of the solid bounded by the planes z = 0, z = h and the cylinder x2 + y2 = R2 is $πh^2R^2$. This result provides a measure of how much of the vector field flows through the cylindrical surface.
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