To raise new questions, new possibilities, to regard old problems from a new angle, requires creative imagination and marks real advance in science, Albert Einstein.
I have yet to see any problem, however complicated, which, when looked at in the right way did not become still more complicated, Paul Anderson.
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) with a potential function f, then the line integral of the vector field along a curve C $\int_{C} \vec{F}·d\vec{r} = \int_{C} ∇f·d\vec{r} = f(P_1)-f(P_0)$ where P_{0} and P_{1} 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: $d\vec{r} = ⟨dx, dy, dz⟩$.
$\int_{C} \vec{F}d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$
Let our vector field $\vec{F} = P\hat{\mathbf{i}}+ Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$, we define the curl of $\vec{F}$ as $curl(\vec{F})=(R_y-Q_z)\hat{\mathbf{i}} + (P_z-R_x)\hat{\mathbf{j}}+(Q_x-P_y)\hat{\mathbf{k}}$
$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ P & Q & R\end{smallmatrix}| = (\frac{∂R}{∂y}-\frac{∂Q}{∂z})\hat{\mathbf{i}}-(\frac{∂R}{∂x}-\frac{∂P}{∂z})\hat{\mathbf{j}} + (\frac{∂Q}{∂x}-\frac{∂P}{∂i})\hat{\mathbf{k}}$
Theorem. The curl of a conservative field is $\vec{0}$. Conversely, if $curl(\vec{F})=\vec{0}$, the components P, Q, and R have continuous first-order partial derivatives, and its domain is open and simply-connected, the vector field is conservative.
Given a vector field $\vec{F}$ and a surface S with boundary curve C, Stokes’ Theorem states: $\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})·\hat{\mathbf{n}}dS$
According to Stoke’s theorem, the circulation of a vector field $\vec{F}$ around a closed curve C is equal to the flux of the curl of $\vec{F}$ through any surface S whose boundary is C. Mathematically,
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$
We need to find the flux of the curl of $\vec{F}$ over a surface S with C as its boundary. We can choose any surface whose boundary is the curve C. The simplest choice is the triangular plane defined by the points: A(1, 0, 0), B(0, 1, 0), and C(0, 0, 2).
To compute the surface integral, we need the normal vector to the plane containing the triangle. The normal vector is given by the cross product of two vectors that lie on the plane. We can choose $\vec{AB}$ and $\vec{AC}$ as those two vectors:
$\vec{AB} = B -A = (-1, 1, 0), \vec{AC} = C -A = (-1, 0, 2)$, the normal vector is given by the cross product of these two vectors, $\hat{\mathbf{n}} = \vec{AB} x \vec{AC} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ -1 & 1 & 0\\ -1 & 0 & 2\end{smallmatrix}| = ⟨2,-(-2), 1⟩ = ⟨2, 2, 1⟩$.
The general equation of a plane is given by: ax +by +cz = d where ⟨a, b, c⟩ is the normal vector. Using A = (1, 0, 0) to calculate d, 2·1+2·0+1·0=d ⇒ d = 2. Therefore, the equation of the plane is 2x +2y +z = 2. The normal vector is ⟨2, 2, 1⟩, since the plane is orientated upwards the surface that lies on this plane and is bounded by C is a compatible surface because induces a positive (counter-clockwise) direction on C.
Parametrize the surface S. Next, we parametrize the surface S using coordinates x and y. Since we are working on a plane, we can express the z-coordinate as a function of x and y. The equation of the plane passing through the points A, B, and C can be written as: x + 2y +z = 2 ⇒[Solving for z:]z = 2 −2x −2y.
Using x and y as parameters: $\vec{r}(x, y)=⟨x, y, 2-2x-2y⟩$
Calculating the Curl of $\vec{F}$. $curl(\vec{F}) = ∇ x \vec{F} = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ z^2 & y^2 & xy\end{smallmatrix}\Biggr \vert = (\frac{∂(xy)}{∂y}-\frac{∂(y^2)}{∂z})\hat{\mathbf{i}}-(\frac{∂(xy)}{∂x}-\frac{∂z^2}{∂z})\hat{\mathbf{j}} + (\frac{∂y^2}{∂x}-\frac{∂z^2}{∂y})\hat{\mathbf{k}} = ⟨x, -(y+2z), 0-0⟩ = ⟨x, 2z-y, 0⟩ =$[Substitute z=2−2x−2y into the Curl] = ⟨x, 2(2-2x-2y)-y, 0⟩ = ⟨x, 4 -4x -5y, 0⟩
Evaluate the Surface Integral:
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨x, 4 -4x -5y, 0⟩·⟨2, 2, 1⟩dS = \int \int_{S} (2x+8-8x-10y+0)dS = \int \int_{S} (-6x-10y+8)dS$
Convert to Double Integral Over the xy-Region:(Figure B). The shadow or the projection of the surface S onto the xy-plane is the triangle with vertices (1, 0), (0, 1) and (0, 0). The equation of the plane is 2x + 2y + 2z = 0, set z = 0 ⇒x + y = 0. The region can be described by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 -x
$\int_{0}^{1}\int_{0}^{-x+1} (-6x-10y+8)dydx$
$\int_{0}^{-x+1} (-6x-10y+8)dy = -6xy-5y^2+8y\bigg|_{0}^{1-x} = -6x(1-x)-5(1-x)^2+8(1-x) = -6x+6x^2-5(1+x^2-2x)+8-8x = -6x+6x^2-5-5x^2+10x+8-8x = x^2-4x+3$
$\int_{0}^{1} (x^2-4x+3)dx = \frac{x^3}{3}-2x^2+3x\bigg|_{0}^{1}$
$\frac{1}{3}-2+3 = \frac{1}{3}+1 = \frac{4}{3}$.
Thus, the value of the line integral is: $\frac{4}{3}$.
According to Stoke’s theorem, the circulation of a vector field $\vec{F}$ around a closed curve C is equal to the flux of the curl of $\vec{F}$ through any surface S whose boundary is C. Mathematically,
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$
According to Stokes’ theorem, this surface integral can be related to a line integral around the boundary curve C of the surface S.
Describe the Surface and Boundary. The surface S is defined by the equation z = x^{2} + y^{2}, which represents a paraboloid. The surface is bounded by the vertical cylinder x^{2}+y^{2}=1, which is a circle of radius 1 in the xy-plane, and a cylinder in 3D.
The curve C is the intersection of the paraboloid and the cylinder. Substituting x^{2} +y^{2} = 1 into the equation of the surface z = x^{2} + y^{2}, we find that at the boundary, z = 1. Therefore, the boundary curve C is a horizontal circle at z = 1 with radius 1 in the xy-plane.
Parametrize the curve C. Since the boundary C is a circle of radius 1 in the plane z = 1, we can parametrize the curve C using the standard parametrization for a circle. Let the parameter t represent the angle around the circle: x = cos(t), y = sin(t), z = 1, 0 ≤ t ≤ 2π.
Next, let’s calculate the differentials, dx = -sin(t)dt, dy = cos(t)dt, dz = 0 (since z = 1 is constant).
Set up the line integral.
$\oint_C \vec{F} \cdot d\vec{r} =$[Substituting the parameterization into $\vec{F}$ we get:]$ \oint_C y^2zdx +xzdy +x^2y^2dz = \int_{0}^{2π} (sin^2(t))1(-sin(t))dt+cos(t)1(cos(t)dt)+0 = \int_{0}^{2π} (-sin^3(t)+cos^2(t))dt = π$
Evaluate the line integral
To evaluate this, we split it into two parts:
$\int_{0}^{2π} -sin^3(t)dt = 0$. The integral of -sin^{3}(t) over one full period [0, 2π] is zero because sin^{3}(t) is an odd function and its integral over a symmetric interval around zero cancels out.
$\int_{0}^{2π} cos^2(t)dt$ = [We are going to use the well-known identity, $cos^2(t)=\frac{1+cos(2t)}{2}$]=$\int_{0}^{2π} \frac{1+cos(2t)}{2}dt = \frac{1}{2}t + sin(2t)\bigg|_{0}^{2π} =$
= $\frac{1}{2}(2π-0)+(0-0)=\frac{1}{2}2π=π.$
We didn’t need to calculate the second integral because the integral of cos(2t) over a full period is zero, too.
Thus, the value of the surface integral, according to Stokes’ theorem, is π.
We can use Stokes’ theorem to convert this line integral into a surface integral. According to Stoke’s theorem, the circulation of a vector field $\vec{F}$ around a closed curve C is equal to the flux of the curl of $\vec{F}$ through any surface S whose boundary is C. Mathematically,
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS$
Describing the Surfaces and the Boundary Curve C.
Understanding that z = y^{2}-x^{2} is a hyperbolic paraboloid often referred to as a saddle surface because it has both upward- and downward-curving sections, resembling a saddle. This surface opens up in the y-direction and down in the x-direction. x^{2} + y^{2} = 1 is a vertical cylinder with radius 1, centered on the z-axis, and a circle of radius 1 in the xy-plane. The intersection of these two surfaces gives a closed curve C that lies on both surfaces, the cylinder and the saddle surface. Viewed from above (in the xy-plane), this curve appears as the circle x^{2} + y^{2} = 1, and the orientation is counterclockwise.
Since positive orientation is upwards, we are standing on the saddle surface (a kind of hills with valleys and peaks) z = y^{2}-x^{2}, the normal vector can be derived from the surface equation. The gradient of z (which helps find the direction of the steepest slope on our surface by showing how z changes as x and y changes) gives the normal vector to the surface: $\hat{\mathbf{n}}=∇z = ⟨-z_x, -z_y, 1⟩ = ⟨2x, -2y, 1⟩$.
If our surface is visualized as a hill, standing at any point on the hill, the normal vector is like an arrow pointing straight up from where you're standing. This vector is not necessarily unit length, but it’s suitable for our integral as is.
$curl(\vec{F}) = ∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\ x^2y & \frac{1}{3}x^3 & xy\end{smallmatrix}| = (\frac{∂(xy)}{∂y}-\frac{∂(\frac{1}{3}x^3)}{∂z})\hat{\mathbf{i}}-(\frac{∂(xy)}{∂x}-\frac{∂x^2y}{∂z})\hat{\mathbf{j}} + (\frac{∂\frac{1}{3}x^3}{∂x}-\frac{∂x^2y}{∂y})\hat{\mathbf{k}} = ⟨x, -y, x^2-x^2⟩= ⟨x, -y, 0⟩$
Set up and Evaluate the Surface Integral
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇x\vec{F})\hat{\mathbf{n}}dS = \int \int_{S} ⟨x, -y, 0⟩·⟨2x, -2y, 1⟩dS = \int \int_{D} (2x^2+2y^2)dA =$ where D is the unit circle·
=[We switch to polar coordinates to evaluate the integral, where: x = rcos(θ), y=rsin(θ), $x^2+y^2=r^2, dA = rdrdθ$] $2\int_{0}^{2π} \int_{0}^{1} r^2·rdrdθ =$
Evaluate the inner integral: $\int_{0}^{1} r^3dr = \frac{r^4}{4}\bigg|_{0}^{1} = \frac{1}{4}$
Now, the outer integral: $2\frac{1}{4} \int_{0}^{2π}dθ = \frac{1}{2}θ\bigg|_{0}^{2π} = \frac{2π}{2}=π.$
$\int \int_{S} \vec{F}·\hat{\mathbf{n}}dS = [\text{The Divergence Theorem relates the flux of a vector field through a closed surface S to the volume integral of the divergence of the vector field over the region D enclosed by S.}] \int\int \int_{D} ∇·\vec{F}dV =$
Compute the divergence of $\vec{F}$: $∇·\vec{F} =\frac{∂}{∂x}(x^3)+\frac{∂}{∂y}(y^3)+\frac{∂}{∂z}(z^2)$ where $\frac{∂}{∂x}(x^3)=3x^2 \frac{∂}{∂y}(y^3)=3y^2, \frac{∂}{∂z}(z^2)=2z$. Thus, the divergence of $\vec{F}$: $∇·\vec{F} = 3x^2+3y^2+2z.$
$= \int\int \int_{D} (3x^2+3y^2+2z)dV$
Let’s use cylindrical coordinates: x = rcosθ, y = rsinθ, r^{2} = x^{2} + y^{2},dV=rdrdθdz. Next, we can set up the triple integral for the region D, which is a cylindrical volume with 0 ≤ r ≤ 2, 0 ≤ θ ≤2π, and 0 ≤z ≤ 1:
$ = \int_{0}^{2π} \int_{0}^{2}\int_{0}^{1} (3r^2+2z)rdzdrdθ = \int_{0}^{2π} \int_{0}^{2}\int_{0}^{1} 3r^3z+rz^2\bigg|_{0}^{1}drdθ = $
$\int_{0}^{2π} \int_{0}^{2} (3r^3+r)drdθ = \int_{0}^{2π} \frac{3r^4}{4}+\frac{r^2}{2}\bigg|_{0}^{2}dθ =$
$\int_{0}^{2π} \frac{3·16}{4}+\frac{4}{2}-(0+0)dθ = 14θ\bigg|_{0}^{2π} = 28π$
We will solve this using Stokes’ Theorem, which relates a surface integral of the curl of a vector field to a line integral over the boundary of the surface. Stokes’ Theorem is given by:
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇ x \vec{F})\hat{\mathbf{n}}dS$
Thus, instead of directly computing the surface integral, we can instead evaluate the line integral around the boundary C of the cone.
Parametrize C. The boundary of the cone surface $z = 1 - \sqrt{x^2+y^2}$ is a circle of radius 1 in the xy-plane, oriented counter-clockwise. This is because at the point where z = 0, the equation of the cone becomes $\sqrt{x^2+y^2} = 1$, which describes a circle of radius 1 centered at the origin.
We parametrize the curve C as: x = cos(t), y = sin(t), z = 0, where the parameter t ranges from 0 to 2π as we move counterclockwise around the boundary. Thus, the position vector $\vec{r}(t)$ is: $\vec{r}(t)=⟨cos(t), sin(t), 0⟩$
Taking the differentials: $d\vec{r}(t)=⟨-sin(t), cos(t), 0⟩dt$.
Evaluating $\vec{F}$ on the boundary C.We now substitute the parametrization of C into the given vector field $\vec{F} = (x-y)\hat{\mathbf{i}} - 2xz\hat{\mathbf{j}} - x^2\hat{\mathbf{k}}$, so: $\vec{F}(\vec{r}(t)) = ⟨cos(t)-sin(t), 0, -cos^2(t)⟩$
Calculating the line integral. The line integral we need to evaluate is: $\oint_C \vec{F} \cdot d\vec{r} = \oint_C ⟨cos(t)-sin(t), 0, -cos^2(t)⟩·⟨-sin(t), cos(t), 0⟩dt = \int_{0}^{2π} (-sin(t)cos(t)+sin^2(t))dt = \int_{0}^{2π} (-sin(t)cos(t)+\frac{1}{2}(1-cos(2t)))dt = \int_{0}^{2π} (-sin(t)cos(t))dt + -\int_{0}^{2π}\frac{1}{2}cos(2t)dt+\frac{1}{2}\int_{0}^{2π}dt =$[The integral of sin(t)cos(t) over a full period [0, 2π] is zero (this follows from the fact that sin(2t) is periodic and symmetric about the origin). The integral of cos(2t) over [0, 2π] is zero (because cos(2t) completes two full periods in this interval).] $\frac{1}{2}\int_{0}^{2π}dt = \frac{1}{2}t\bigg|_{0}^{2π} = π$
We will solve this using Stokes’ Theorem, which relates a surface integral of the curl of a vector field to a line integral over the boundary of the surface. Stokes’ Theorem is given by:
$\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇ x \vec{F})\hat{\mathbf{n}}dS$
Thus, instead of directly computing the surface integral, we can instead evaluate the line integral around the boundary C of the ellipsoid.
Parametrize C. The boundary of the ellipsoid surface $2x^2+2y^2+z^2=8$ is the circle that forms where the ellipsoid intersects the xy-plane (i.e., where z = 0). This is because at the point where z = 0, the equation of the ellipsoid becomes $2x^2+2y^2 = 8↭ x^2 + y^2 = 4$, which describes a circle with radius 2 centered at the origin.
We parametrize the curve C as: x = 2cos(t), y = 2sin(t), z = 0, where the parameter t ranges from 0 to 2π as we move counterclockwise around the boundary. Thus, the position vector $\vec{r}(t)$ is: $\vec{r}(t)=⟨2cos(t), 2sin(t), 0⟩$
Taking the differentials: $d\vec{r}(t)=⟨-2sin(t), 2cos(t), 0⟩dt$.
Evaluating $\vec{F}$ on the boundary C.We now substitute the parametrization of C into the given vector field $\vec{F} = (3x-y)\hat{\mathbf{i}} +(x+3y)\hat{\mathbf{j}} +(1+x^2+y^2+z^2)\hat{\mathbf{k}}$, so: $\vec{F}(\vec{r}(t)) = ⟨3·2cos(t)-2sin(t), 2cos(t)+3·2sin(t), 1+4(cos^2(t)+sin^2(t))+0⟩ = ⟨3·2cos(t)-2sin(t), 2cos(t)+3·2sin(t), 5⟩$
Calculating the line integral. The line integral we need to evaluate is: $\oint_C \vec{F} \cdot d\vec{r} = \int_{0}^{2π} ⟨3·2cos(t)-2sin(t), 2cos(t)+3·2sin(t), 5⟩·⟨-2sin(t), 2cos(t), 0⟩dt = \int_{0}^{2π} (-12sin(t)cos(t)+4sin^2(t)+4cos^2(t)+12sin(t)cos(t)+0)dt =[\text{Notice that the terms −12cos(t)sin(t) and 12cos(t)sin(t) cancel out}] \int_{0}^{2π} 4(sin^2(t)+cos^2(t))dt = \int_{0}^{2π} 4dt = 4t\bigg|_{0}^{2π} = 8π.$
Stokes’ theorem, depicted in Figure i, is one of the most powerful and elegant theorems in calculus. It asserts that the line integral of a vector field $\vec{F}$ around a closed curve C is equal to the surface integral of the flux of the curl of the vector field over any surface S that has C as its boundary.
Mathematically, this property is expressed as: $\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S} (∇ x \vec{F})\hat{\mathbf{n}}dS$
The remarkable feature of Stokes’ theorem is its surface independence property. This means that the value of the line integral around C is independent of the surface S chosen, as long as S has C as its boundary.
Consider two different surfaces S_{1} and S_{2} that share the same common boundary curve C. According to Stokes’ theorem, the line integral of the vector field $\vec{F}$ around the curve C is equal to the surface integral of the curl of the vector field over both S_{1} and S_{w}, which means, $\oint_C \vec{F} \cdot d\vec{r} = \int \int_{S_1} (∇ x \vec{F})\hat{\mathbf{n}}dS = \int \int_{S_2} (∇ x \vec{F})\hat{\mathbf{n}}dS$ (Figure i).
This is a powerful statement because it means that the line integral depends only on the behavior of the vector field around the boundary curve and not on the specific shape or position of the surface.
To understand why the choice of surface does not affect the result, let’s explore the difference between the integrals over two surfaces, S_{1} and S_{2}, that share the same boundary C. Suppose we subtract the surface integrals over S_{2} from those over S_{1}:
$\int \int_{S_1} (∇ x \vec{F})\hat{\mathbf{n}}dS - \int \int_{S_2} (∇ x \vec{F})\hat{\mathbf{n}}dS = \int \int_{S=S_1-S_2} (∇ x \vec{F})\hat{\mathbf{n}}dS$
The combined surface S = S_{1} -S_{2} is a closed surface that encloses a volume D. According to the divergence theorem, the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence of the vector field inside the enclosed region:
$\int \int_{S} \vec{F}\hat{\mathbf{n}}dS = \int \int \int_{D} div(\vec{F})dV$
In this case, the vector field is $∇ x \vec{F}$:
$\int \int \int_{D} div(∇ x \vec{F})dV$ [We now aim to demonstrate that $div(∇ x \vec{F}) = 0$ always] = 0.
Consider $\vec{F} = ⟨P, Q, R⟩$. The curl of $\vec{F}$ is given by $∇ x \vec{F} = |\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{∂}{∂x} & \frac{∂}{∂y} & \frac{∂}{∂z}\\P & Q & R\end{smallmatrix}| = ⟨R_y-Q_z, P_z-R_x, Q_x-P_y⟩$
Next, let’s compute the divergence of $∇ x \vec{F}$:
$div(∇ x \vec{F}) =$ [Recall $div(\vec{F}) = P_x+Q_y+R_z$] = (R_{y}-Q_{z})_{x} + (P_{z}-R_{x})_{y}+ (Q_{x}-P_{y})_{z} = R_{yx} -Q_{zx} + P_{zy} -R_{xy} + Q_{xz} -P_{yz} =[After rearranging terms, we notice that the cross partial derivatives cancel out due to the equality of mixed partial derivatives (by Clairaut’s theorem): P_{zy} = P_{yz}, R_{yx} = R_{xy}, Q_{zx} = Q_{xz}] 0
Therefore, $\int \int \int_{D} div(∇ x \vec{F})dV = 0$, and this confirms that: $\int \int_{S_1} (∇ x \vec{F})\hat{\mathbf{n}}dS = \int \int_{S_2} (∇ x \vec{F})\hat{\mathbf{n}}dS$. Hence, the surface independence property of Stokes’ theorem is established, demonstrating that the line integral around C is indeed independent of the surface chosen, as long as it has C as its boundary.
Definition of Curl. In multivariable calculus, the curl of a vector field $\vec{F}$ is a mathematical tool that measures the rotation or the twisting of the field at a particular point.
Let’s start with the formal definition. Given a vector field $\vec{F} = ⟨P, Q, R⟩$, the curl of $\vec{F}$ is defined as $curl(\vec{F}) = ∇ × \vec{F} = (\frac{∂R}{∂y}-\frac{∂Q}{∂z})\hat{\mathbf{i}}+(\frac{∂P}{∂z}-\frac{∂R}{∂x})\hat{\mathbf{j}}+(\frac{∂Q}{∂x}-\frac{∂P}{∂y})\hat{\mathbf{k}}$ where $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$ are the unit vectors in the x-, y-, and z- directions, respectively.
In two dimensions, given a vector field $\vec{F} = ⟨M, N⟩$, the curl simplifies to $curl(\vec{F}) = ∇ x \vec{F} = \frac{∂N}{∂x}-\frac{∂M}{∂y} = N_x - M_y$.
It measures how far the vector field is from being conservative and how much the vector field $\vec{F}$ “rotates” around a point (it quantifies the “circulation density” at each point in space). If the curl is zero, the field is said to be irrotational, meaning it has no tendency to rotate around any point.
To grasp the physical intuition behind the curl, imagine that the vector field $\vec{F}$ represents the velocity field of a fluid. In this case, the curl of this velocity field represents the local rotation or twisting motion of the fluid at each point. In other words, it describes how the fluid is swirling or spinning at each point.
Let’s consider a few concrete examples to see how the curl behaves in different situations:
As we have previously stated, the curl is also tied to important topological properties of vector fields. In particular:
If the curl of a vector field $\vec{F}$ is zero everywhere in a simply connected region, the field is said to be irrotational and conservative. This means the vector field can be written as the gradient of some scalar potential function f, i.e., $\vec{F} = ∇f$
However, in regions with “holes” or complex topology, a vector field can have zero curl but still fail to be conservative. This is because the topology of the region plays a role in determining whether a vector field can be expressed as a gradient of a potential function.
In physics and mechanics, torque is the rotational analogue or equivalent of linear force. It measures the ability or effectiveness of a force to cause an object to rotate about an axis or pivot point. Just as a force pushes or pulls an object in a straight line, torque “twists” or rotates an object.
Mathematically, torque is defined as the cross product between two vectors:
The formula for torque is: $\vec{τ} = \vec{r}x\vec{F}$ (Figure ii) where:
The magnitude of the torque is given by τ = rFsin(θ) where θ is the angle between $\vec{r}$ and $\vec{F}$. Torque measures how effectively the force causes rotation; if the force is applied perpendicular to the position vector, it produces the maximum torque. If the force is applied parallel to the position vector, the torque is zero, as no rotational effect occurs.
Newton’s Second Law and Rotational Motion. In linear motion, Newton’s second law states: Force = Mass x Acceleration ↭ $\frac{Force}{Mass} = Acceleration = \frac{d}{dt}(Velocity)$.
The moment of inertia I of a rigid body is a measure of how much resistance it offers to changes in its rotational motion about a specific axis. It depends on both the mass of the body and the distribution of that mass relative to the axis of rotation.
In rotational dynamics, Newton’s second law is expressed as: Torque = Moment of inertia × Angular acceleration or $\vec{τ} = I\vec{α}$ where $\vec{τ}$ is the torque applied to the body, I is the moment of inertia of the body, and α is the angular acceleration, which is the rate of change of angular velocity. It is is analogous to the linear version of Newton’s second law: Force = Mass x Acceleration ↭ $\frac{Force}{Mass} = Acceleration = \frac{d}{dt}(Velocity)$.
For rotational motion, the equivalent law involves torque and angular acceleration: $\frac{Torque}{\text{Moment of inertia}} = \text{Angular acceleration} = \frac{d}{dt}(\text{Angular velocity}).$
The curl of a vector field $\vec{F}$ provides valuable insight into the rotational behavior of the field. In fluid mechanics, the curl of the velocity field gives the angular velocity of the flow. Similarly, the curl of an acceleration field gives the angular acceleration. Therefore, the curl of a force field measures the torque per unit moment of inertia. It indicates how much torque the force field exerts on a small solid placed within it, essentially describing how much rotational motion (spinning) a small object placed in the force field would experience.
In summary:
A force field that is conservative (one that can be derived from a scalar potential function) has a zero curl, $curl(\vec{F}) = ∇ × \vec{F} = 0$. This means such a field does not induce rotational motion. In other words, if you leave an object in this force field, it might accelerate in some direction, but the force field itself does not cause it to spin.
Consider a specific example: the gravitational field around the Earth. The Earth’s gravitational field is conservative, meaning its curl is zero. If we place an object in this field, the object will experience a force pulling it toward the Earth, but this force will not cause the object to rotate.
The Earth’s own rotation is due to its initial angular momentum and is not caused by the gravitational force. In other words, the curl of the gravitational force field is zero, indicating no inherent rotational effect within the field itself.
This relationship between curl and torque helps bridge the gap between abstract calculus concepts and their physical interpretations, deepening our understanding of both linear and rotational dynamics.