Line integrals in Space

No pressure, no diamonds, Thomas Carlyle.

Vector Fields and Line Integrals in Space

A vector field is an assignment of a vector $\vec{F}$ to each point $(x, y)$ in the space. In two dimensions, this is written or expressed as $\vec{F}(x,y) = P(x,y)\hat{\mathbf{i}} + Q(x,y)\hat{\mathbf{j}}$, where $P$ and $Q$ are scalar functions of $x$ and $y$, and $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ are the standard unit vectors.

A vector field on a plane can be visualized as a continuous collection of arrows, each attached to a specific coordinate point on the plane. These arrows represent vectors with specific magnitudes and directions.

Work and Line Integrals

In physics, work is defined as the energy transferred when a force acts upon an object, causing a displacement. If the force is constant and the path is a straight line, work is simply the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$).

However, what if the path is curved and the force changes at every point in space? We must use a line integral. The work done by a force field $\vec{F}$ moving an object along a trajectory $C$ is defined as: $W = \int_{C} \vec{F} \cdot d\vec{r} = \int_{C} (P dx + Q dy)$

Alternatively, we can express this using the unit tangent vector $\hat{\mathbf{T}}$ and arc length $s$: $W = \int_{C} \vec{F} \cdot \hat{\mathbf{T}} ds$

The Intuition: The dot product $\vec{F} \cdot d\vec{r}$ isolates the component of the force vector that points exactly in the direction of the curve at that specific microscopic instant. Forces pushing perpendicular to your motion do zero work!

Conservative Vector Fields

A vector field is conservative if there exist a scalar function $f$ such that $\vec{F} = \nabla f$. This means the vector field is exactly the gradient of some underlying scalar field $f$. This scalar function is known or referred to as the potential function associated with the vector field.

Theorem. The 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 the work done moving from point $P_0$ to point $P_1$ is entirely independent of the path taken. Indeed, it depends only on the endpoints: $\int_{C} \vec{F}\cdot d\vec{r} = \int_{C} \nabla f\cdot d\vec{r} = f(P_1)-f(P_0)$ where $P_0$ and$P_1$ are the initial and final points of the curve $C$, respectively.

Line integrals in 3D Space

A line integral in space extends these concepts to three dimensions, allowing us to integrate a function along a 3D curve. This mathematical tool is the backbone of advanced physics, used to calculate:

• Work done by a gravitational or electric field moving a mass or charge.
• Circulation of a fluid or magnetic field around a closed loop (central to Ampere’s Law and fluid dynamics).
• Electromotive Force (EMF) in Faraday’s Law of Induction.

Consider a vector field $\vec{F}$ in three dimensional space. A vector field can be expressed as: $\vec{F} = P\hat{\mathbf{i}} + Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$ where $P$, $Q$, and $R$ are functions of the coordinates $x, y$, and $z$ respectively. Each of these functions represents the components of the vector field in the $\hat{\mathbf{i}}$ ($x$-axis), $\hat{\mathbf{j}}$ ($y$-axis), and $\hat{\mathbf{k}}$ ($z$-axis) directions, respectively.

The vector field $\vec{F}$ can represent various physical phenomena where a vector quantity is defined at every point in space, such as:

• A force field, where each point in space has a force vector associated with it (e.g., gravitational or electric fields).
• An electromagnetic field. The electric field vector represents the force exerted on a charged particle, while the magnetic field vector describes the force exerted on a moving charged particle.
• A velocity field in fluid dynamics, where the vector at each point represents the velocity of the fluid at that point. This field is crucial for understanding the movement of fluids, such as air or water, in various scenarios.

To perform a line integral, we first need to describe the curve $C$ along which we are integrating. A curve in three dimensions can be parametrized by a parameter $t$ (often representing time), which varies over an interval $[a, b]$. The curve is represented as: $\vec{r}(t)=\langle x(t), y(t), z(t)\rangle $ where $a \leq t \leq b$. The functions $x(t), y(t)$, and $z(t)$ are the coordinates that describe the positions of points on the curve $C$ as the parameter $t$ varies from $a$ to $b$.

The line integral of the vector field $\vec{F}$ along the curve $C$ is defined as: $\int_{C} \vec{F} \cdot d\vec{r}$ where $d\vec{r}$ represents an infinitesimal vector (small step) strictly tangent to the curve at each point. The vector $d\vec{r}$ can be expressed as: $d\vec{r} = \langle dx, dy, dz\rangle $.

In the context of physics, when the vector field represents a force, the line integral of the vector field along a curve $C$ gives the work done by the force as it moves a particle along the curve.

To compute the line integral, we express $d\vec{r}$ in terms of the parameter $t$. Given the parametrization $\vec{r}(t)=\langle x(t), y(t), z(t)\rangle $, the differential vector $d\vec{r}$ can be written as: $d\vec{r} = \frac{d\vec{r}}{dt}dt = \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\rangle dt$.

Now, substituting $d\vec{r}$ into the line integral expression: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}dt$

Given the vector field $\vec{F} = P\hat{\mathbf{i}} + Q\hat{\mathbf{j}}+R\hat{\mathbf{k}}$, we can express it as: $\vec{F}(\vec{r}(t)) = \langle P(x(t), y(t), z(t)), Q(x(t), y(t), z(t)), R(x(t), y(t), z(t))\rangle $

Thus, the dot product $\vec{F} \cdot d\vec{r}$ becomes, $\vec{F} \cdot d\vec{r} = \langle P, Q, R \rangle \cdot \left\langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right\rangle dt = \left( P \frac{dx}{dt} + Q \frac{dy}{dt} + R \frac{dz}{dt} \right) dt$

Substituting this into the line integral gives: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$

Conclusion: Given a vector field $\vec{F} = \langle P(x, y, z), Q(x, y, z), R(x, y, z)\rangle $ and a curve $C$ parametrized by $\vec{r}(t)=\langle x(t), y(t), z(t)\rangle $, the line integral is defined as: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$

Let’s work through exercises to illustrate these concepts.

Solved Exercises

• Work Done in 2D. Calculate the work done by the vector field $\vec{F} = \langle x\sin(y), y\rangle $ along the curve defined by $y = x^2$ from the point $(-1, 1)$ to the point $(2, 4)$.

Work done by a force field along a path is calculated using the line integral of the vector field along the curve. If $\vec{F} =\langle P(x,y),Q(x,y)\rangle $ is the vector field and $C$ is the curve, the work done is given by the line integral: $\int_{C} \vec{F}\cdot d\vec{r}$ where $d\vec{r}$ is the differential displacement vector along the curve $C$. The vector field $\vec{F}\cdot d\vec{r}$ represents the dot product of the force field with the tangent to the curve.

1. Understanding the Curve. $(-1, 1)$ to $(2, 4)$ is not a line segment, they lie on the parabola $y = x^2$. To find the work done along this curve, we can parametrize the curve using a parameter t. A natural choice for this parameterization is: $x = t, y = x^2= t^2$.
   Parameter Range. Since $x = t$ ranges from $-1$ to $2$, the parameter $t$ also varies from $-1$ to $2$, $-1 \leq t \leq 2$.

2. Compute the Differentials. $dx = \frac{d}{dt}(t)dt = dt, dy = \frac{d}{dt}(t^2)dt = 2tdt$

3. Substitute Parametrization into the vector field $\vec{F}$: $P = x\sin(y) = t\sin(t^2), Q = y = t^2$.

4. Evaluate the Line Integral. $\int_{C} \vec{F} \cdot d\vec{r} = \int_{C} [P\frac{dx}{dt}+Q\frac{dy}{dt}]dt$ =[Substitute $P, Q$, and their differentials:] $\int_{-1}^{2} [t\sin(t^2)\cdot 1+t^2\cdot 2t]dt = \int_{-1}^{2} [t\sin(t^2)+2t^3]dt$

5. Integrate each term separately.
   $\int_{-1}^{2} t\sin(t^2)dt=$[To solve this integral, we use a substitution method. Let $u = t^2$. Then, $du = 2tdt, \frac{1}{2}du = tdt$] $\frac{1}{2}\int_{-1}^{2} \sin(u)du = -\frac{1}{2}\cos(u)+C =$[Substitute back $u = t^2$] $\frac{-1}{2}\cos(t^2)+C$
   For the second term: $\int_{-1}^{2} 2t^3dt = 2\frac{t^4}{4} = \frac{t^4}{2}$

6. Combine the results and evaluate the definite integral: $\left[ \frac{-1}{2}\cos(t^2) +\frac{t^4}{2} \right]_{-1}^{2} =$
   $[\frac{-1}{2}\cos(4) +\frac{16}{2}]-[\frac{-1}{2}\cos(1) +\frac{1}{2}] = \frac{-1}{2}\cos(4) + 8 + \frac{1}{2}\cos(1)-\frac{1}{2} = \frac{-1}{2}\cos(4) + \frac{1}{2}\cos(1) +\frac{15}{2}.$

7. Conclusion: The work done by the vector field $\vec{F} = \langle x\sin(y), y\rangle $ along the curve $y = x^2$ from $(-1, 1)$ to $(2, 4)$ is $\int_{C} \vec{F} \cdot d\vec{r} =\frac{-1}{2}\cos(4) + \frac{1}{2}\cos(1) +\frac{15}{2}.$

• Line Integral of a Helix. Evaluate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$ where the vector field is $\vec{F} = \langle y, x, z\rangle $, and the curve $C$ is parametrized by $x = \cos(t), y = \sin(t), z = t^2, 0 \leq t \leq 2\pi$.

Given a vector field $\vec{F} = \langle P(x, y, z), Q(x, y, z), R(x, y, z)\rangle $ and a curve $C$ parametrized by $\vec{r}(t)=\langle x(t), y(t), z(t)\rangle $, the line integral is defined as: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$

1. Compute the Differentials: dx = $\frac{d}{dt}(\cos(t))dt = -\sin(t)dt, dy = \cos(t)dt, dz = 2tdt$

2. Substitute Parametrization into the vector field $\vec{F}$: $P = y = \sin(t), Q = x = \cos(t), R = z = t^2.$

3. Evaluate the Line Integral.
   $\int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{2\pi} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt$ =[Substitute $P, Q, R$, and their differentials:] $\int_{0}^{2\pi}[\sin(t)(-\sin(t))+\cos(t)\cos(t)+t^2(2t)]dt =[\text{Simplifying inside the integral:}] \int_{0}^{2\pi} [-\sin^2(t)+\cos^2(t)+2t^3]dt$

4. Integrate Each Term Separately.
   We can use the identities, $\sin^2(t)=\frac{1-\cos(2t)}{2}, \cos^2(t)=\frac{1+\cos(2t)}{2}$
   $\int_{0}^{2\pi} -\sin^2(t)dt = -\frac{1}{2}\int_{0}^{2\pi}(1-\cos(2t))dt = -\frac{1}{2}(t-\frac{1}{2}\sin(2t))\bigg|_{0}^{2\pi} = -\frac{1}{2}(2\pi-0) = -\pi$
   $\int_{0}^{2\pi} \cos^2(t)dt = \frac{1}{2}\int_{0}^{2\pi}(1+\cos(2t))dt = \frac{1}{2}(t+\frac{1}{2}\sin(2t))\bigg|_{0}^{2\pi} = \frac{1}{2}(2\pi-0) = \pi$
   $\int_{0}^{2\pi} 2t^3dt = \frac{1}{2}t^4\bigg|_{0}^{2\pi} = \frac{(2\pi)^4}{2}-\frac{0^4}{2} = 8\pi^4$

5. Combining everything: $-\pi + \pi + 8\pi^4 = 8\pi^4$

6. Conclusion: $\int_{C} \vec{F} \cdot d\vec{r} = 8\pi^4$

• 3D Curve (Two Approaches). Evaluate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$ where the vector field is $\vec{F} = \langle yz, xz, xy\rangle $, and the curve $C$ is parametrized by $x = t^3, y = t^2, z = t, 0 \leq t \leq 1$.

Method 1: Direct Computation

Given a vector field $\vec{F} = \langle P(x, y, z), Q(x, y, z), R(x, y, z)\rangle $ and a curve $C$ parametrized by $\vec{r}(t)=\langle x(t), y(t), z(t)\rangle $, the line integral is defined as: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} (P\frac{dx}{dt} + Q\frac{dy}{dt} + R\frac{dz}{dt})dt = \int_{a}^{b} (P(x(t), y(t), z(t))\frac{dx}{dt} + Q(x(t), y(t), z(t))\frac{dy}{dt} + R(x(t), y(t), z(t))\frac{dz}{dt})dt$

1. Compute the Differentials, dx = $\frac{d}{dt}(t^3)dt = 3t^2dt, dy = 2tdt, dz = dt$

2. Substitute Parametrization into the vector field $\vec{F}$: $P = yz = t^2\cdot t = t^3, Q = zx = t^2\cdot t = t^4, R = xy = t^3t^2 = t^5$.

3. Evaluate the Line Integral.
   $\int_{C} \vec{F} \cdot d\vec{r} = \int_{0}^{1} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt $ =[Substitute $P, Q, R$, and their differentials:]$ \int_{0}^{1} [t^3\cdot 3t^2+t^4\cdot 2t+ t^5\cdot 1] dt =[\text{Simplifying inside the integral:}] \int_{0}^{1} 6t^5dt = t^6\bigg|_{0}^{1} = 1.$

4. Conclusion. The work done by the vector field $\vec{F} = \langle yz, xz, xy\rangle $, along the curve $C$ parametrized by $x = t^3, y = t^2, z = t, 0 \leq t \leq 1$ is: $\int_{C} \vec{F} \cdot d\vec{r} = 1$.

Method 2: The Elegant Way (Fundamental Theorem)

Notice that $\vec{F} = \langle yz, xz, xy \rangle$ is a conservative vector field. We can find a scalar potential function $f(x,y,z)$ such that $\nabla f = \vec{F}$.

By closer inspection, we observe that the scalar function $f(x,y,z) = xyz$ fulfills this condition: $\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \langle yz, xz, xy \rangle = \vec{F}$

Because the field is conservative, the integral is path-independent! We only need the starting point ($t=0$) and ending point ($t=1$):

• At $t=0$: $P_0 = (0,0,0)$
• At $t=1$: $P_1 = (1,1,1)$

Applying the Fundamental Theorem of Line Integrals: $\int_{C} \vec{F} \cdot d\vec{r} = f(P_1) - f(P_0) = f(1,1,1) - f(0,0,0) = (1)(1)(1) - (0)(0)(0) = 1$

Vector field and piecewise curve

• Consider the vector field $\vec{F} = \langle yz, xz, xy \rangle$. Suppose we want to evaluate the line integral along a piecewise curve $C'$ that does not travel in a straight line, but rather moves in “steps” along the coordinate axes from the origin to the point $(1, 1, 1)$.

The curve $C'$ consists of three distinct segments:

1. $C_1$: From the origin $(0, 0, 0)$ to $(1, 0, 0)$ along the x-axis.
2. $C_2$: From $(1, 0, 0)$ to $(1, 1, 0)$ moving parallel to the y-axis.
3. $C_3$: From $(1, 1, 0)$ to $(1, 1, 1)$ moving parallel to the z-axis.

The goal is to evaluate the line integral $\int_{C'} \vec{F} \cdot d\vec{r}$ along the piecewise curve.

The line integral of the vector field $\vec{F}$ along the curve $C$’ is: $\int_{C'} \vec{F}d\vec{r} = \int_{C'} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt = \int_{C'} [yzdx +xzdy + xydz]dt$

Since C′ is a piecewise curve consisting of three segments, the line integral can be broken down into three separate integrals, one for each segment (we exploit the property that integrals are additive): $\int_{C'} \vec{F}d\vec{r} = \int_{C_1} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt + \int_{C_2} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt + \int_{C_3} [P\frac{dx}{dt}+Q\frac{dy}{dt}+R\frac{dz}{dt}]dt$

• Path $C_1$ (Movement strictly in $x$).
  The curve $C_1$ can be parameterized as $\vec{r}(t) = \langle t, 0, 0 \rangle$ where $0 \leq t \leq 1$.
  Differentials: Because $y$ and $z$ are constant at zero, $dy = 0$ and $dz = 0$. We only have $dx = 1\,dt$.
  Integrand components: Substitute the parameterization back into the vector field: $P = yz = (0)(0) = 0$, $Q = xz = (t)(0) = 0$, $R = xy = (t)(0) = 0$. Integral: $\int_{C_1} (0\,dt + 0 + 0) = 0$
• Path $C_2$ (Movement strictly in $y$).
  The curve $C_2$ is parameterized as $\vec{r}(t) = \langle 1, t, 0 \rangle$ where $0 \leq t \leq 1$.
  Differentials: $x$ is constant at $1$, and $z$ is constant at $0$. Thus, $dx = 0$, $dy = 1\,dt$, $dz = 0$.
  Integrand components: $P = yz = (t)(0) = 0$, $Q = xz = (1)(0) = 0$, $R = xy = (1)(t) = t$.
  Integral: Because $dx$ and $dz$ are zero, only the $Q$ term matters, but $Q=0$ here. $\int_{C_2} (0 + 0\,dt + 0) = 0$
• Path $C_3$ (Movement strictly in $z$).
  The curve $C_3$ is parameterized as $\vec{r}(t) = \langle 1, 1, t \rangle$ where $0 \leq t \leq 1$.
  Differentials: $dx = 0$, $dy = 0$, $dz = 1\,dt$.
  Integrand components: $P = yz = (1)(t) = t$, $Q = xz = (1)(t) = t$, $R = xy = (1)(1) = 1$.
  Integral: Here, only the $R$ term survives because $dx=0$ and $dy=0$.
  $\int_{C_3} \vec{F} \cdot d\vec{r} = \int_{0}^{1} (R \cdot dz) = \int_{0}^{1} (1 \cdot 1\,dt) = \int_{0}^{1} dt = 1$

Total Work Done. Adding up the contributions from each segment: $\int_{C'} \vec{F}d\vec{r} = 0 + 0 + 1 = 1$

The Fundamental Theorem Shortcut

We have calculated the work by evaluating three separate integrals. However, because $\vec{F} = \langle yz, xz, xy \rangle$ is a conservative vector field, it can be expressed as the gradient of a scalar potential function: $\vec{F} = \nabla f = \nabla (xyz) = \langle yz, xz, xy \rangle$ where $f(x, y, z) = xyz$.

For conservative vector fields, the line integral between two points depends only on the endpoints and not on the path taken. This means that for any two curves $C$ (a straight line) and $C'$ (our stepped piecewise curve) connecting the same points, the line integrals will be identical.

Using the Fundamental Theorem of Calculus for Line Integrals between $P_0(0, 0, 0)$ and $P_1(1, 1, 1)$: $\int_{C'} \vec{F} \cdot d\vec{r} = \int_{C'} \nabla f \cdot d\vec{r} = f(P_1) - f(P_0) = f(1, 1, 1) - f(0, 0, 0) = 1 - 0 = 1$

Both methods obviously yield exactly $1$, but the potential function skips the integration entirely!

Criteria for a Vector Field to be Conservative

In calculus, a vector field $\vec{F} = \langle P, Q, R\rangle $ is considered conservative if it can be expressed as the gradient of some scalar potential function $f(x, y, z)$. In other words, there exists a scalar function $f$ such that: $\vec{F} = \nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle $.

If such a function $f$ exists, it means that the vector field $\vec{F}$ is a gradient field and has a few important properties, such as the fact that the line integral of $\vec{F}$ between two points depends only on the endpoints, not on the path taken.

To determine if a vector field $\vec{F}= \langle P, Q, R\rangle $ is conservative, you can check the following criteria:

1. The Gradient Condition: The vector field $\vec{F} = \langle P, Q, R\rangle $ is conservative if there exist a scalar function $f$ such that: $\vec{F} = \nabla f = \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle $. This means that each component of $\vec{F}$ corresponds to a partial derivative of $f$: $P = \frac{\partial f}{\partial x}, Q =\frac{\partial f}{\partial y}, R = \frac{\partial f}{\partial z}$.
2. The Exact differential. For the vector field to be conservative, the expression $Pdx + Qdy + Rdz$ must be an exact differential, meaning there exist a function $f(x, y, z)$ whose total differential $df$ matches our vector field: $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = Pdx + Qdy + Rdz$.
3. Equality of Mixed Partial Derivatives (The Curl Test): If $\vec{F} = \nabla f$, then by Clairaut’s Theorem, the mixed second partial derivatives of $f$ must be equal (e.g., $f_{xy} = f_{yx}$). Translated into the components of our vector field $\vec{F} = \langle P, Q, R \rangle$, this requires three equalities to hold true simultaneously: $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \quad \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}, \quad \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}$

   A much cleaner way to state this is using the Curl operator. A vector field is conservative if and only if it is “irrotational,” meaning it has zero curl at all points: $\nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right)\hat{\mathbf{i}} + \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right)\hat{\mathbf{j}} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\hat{\mathbf{k}} = \vec{0}$.

4. Simply Connected Domain (The Topology Check). In addition to the conditions above, for a vector field to be conservative, it must be defined on a simply connected domain. A simply connected domain is one in which any loop within the domain can be continuously contracted or shrunk to a single point without leaving the domain.
   In 2D Space ($\mathbb{R}^2$): A domain with a single “hole” or missing point (like a donut) is not simply connected. A loop drawn around the hole will get caught and cannot shrink to a point.
   In 3D Space ($\mathbb{R}^3$): A single missing point does not break simple connectivity! You can always slide a 1D lasso off a single 0D missing point in a 3D room. However, an infinitely long missing line (like a pole running through the room) acts as a tunnel, snagging the loop and destroying the simple connectivity.
   In such a domain, the vanishing of the curl of $\vec{F}$, $\nabla × \vec{F} = (\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\hat{\mathbf{i}}+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\hat{\mathbf{j}}+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\hat{\mathbf{k}} = \vec{0}$, i.e., the equality of mixed partial derivatives, guarantees that $\vec{F}$ is conservative.

Practical Application: Testing and Finding the Potential

When given a vector field $\vec{F} = \langle P, Q, R\rangle$, follow these steps to test if it is conservative:

1. Check the Curl Condition. Calculate the curl of $\vec{F}$: $\nabla × \vec{F} = (\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z})\hat{\mathbf{i}}+(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x})\hat{\mathbf{j}}+(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})\hat{\mathbf{k}}$. If the result is not the zero vector ($\vec{0}$), stop immediately. The field is not conservative.
2. Ensure the Domain is Simply Connected: Verify that the domain of the vector field is simply connected. If the domain is not simply connected, even if the curl is zero, $\vec{F}$ might not be conservative due to the possibility of “holes” or “tunnels” in the domain.
3. Find the Potential Function: If the curl is zero and the domain is simply connected, try to find a potential function $f(x, y, z)$ such that: $P = \frac{\partial f}{\partial x}, Q =\frac{\partial f}{\partial y}, R = \frac{\partial f}{\partial z}$ using partial integration. Here is the exact algorithm:
   $\star$ Integrate $P$ with respect to $x$: $f(x,y,z) = \int P\,dx + g(y,z)$
   $\star$ Take the partial derivative of your new $f$ with respect to $y$, and set it equal to $Q$: $\frac{\partial f}{\partial y} = Q$. Use this equation to solve for $\frac{\partial g}{\partial y}$.
   $\star$ Integrate $\frac{\partial g}{\partial y}$ with respect to $y$ to find $g(y,z) = h(y,z) + k(z)$. Update your master equation for $f$.
   $\star$ Finally, take the partial derivative of $f$ with respect to $z$, set it equal to $R$, solve for $k(z)$, and complete the function.

Solved Exercises

Exercise 1: Identifying Conservative Fields

• Is the vector field $\vec{F}=\langle yz, xz, xy\rangle $ conservative?

The vector field $\vec{F} = \langle P, Q, R\rangle $ is conservative if there exist a scalar potential function $f$ such that: $\vec{F} = \nabla f = \langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle $. In other words, the vector field is conservative if it can be written as the gradient of some scalar function $f$.

To prove a field is conservative, we must verify two conditions: the domain must be simply connected, and the curl must be zero (which is equivalent to checking the equality of mixed partial derivatives).

1. The vector field $\vec{F}$ is defined in $\mathbb{R}^3$ (all of three-dimensional space), which is a simply connected domain. In $\mathbb{R}^3$, any closed loop can be contracted to a point, and there are no “holes”, “tunnels”, or “boundaries (missing points)” that could prevent the field from being conservative.
2. Compute the partial derivatives. $P = yz, \frac{\partial P}{\partial y} = z, \frac{\partial P}{\partial z} = y. Q = xz, \frac{\partial Q}{\partial x} = z, \frac{\partial Q}{\partial z} = x. R = xy, \frac{\partial R}{\partial x} = y, \frac{\partial R}{\partial y} = x$.
3. Check the equality of mixed partial derivatives. For the curl to be zero ($\nabla \times \vec{F} = \vec{0}$), the cross-derivatives must be equal: $P_y = Q_x = z; P_z = R_x = y$, and $Q_z = R_y = x$. Since all the mixed partial derivatives are equal, we conclude that the vector field $\vec{F}$ is conservative.
4. Finding the potential function. We must find $f(x, y, z)$ such that $\nabla f = \vec{F}$.
   $\star$ First, integrate $P = yz = \frac{\partial f}{\partial x}$ with respect to $x$. $f(x, y, z) = \int yzdx = xyz +g(y, z)$ where $g(y, z)$ is a function of $y$ and $z$.
   $\star$ Next, differentiate $f(x, y, z) = xyz + g(y, z)$ with respect to $y$ and compare it to $Q(x, y, z) = xz$: $\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y} = Q = xz \Rightarrow \frac{\partial g}{\partial y} = 0$. Therefore, $g(y, z)$ does not depend on $y$, so $g(y, z) = h(z)$, where $h(z)$ is a function of $z$ alone, $g(y, z) = h(z)$.
   $\star$ Finally, differentiate $f = xyz + h(z)$ with respect to $z$ and compare it to $R(x, y, z) = xy$: $\frac{\partial f}{\partial z} = xy + h'(z) = R = xy \Rightarrow h'(z) = 0 \Rightarrow h(z) = C \Rightarrow f = xyz + C.$
5. Conclusion. So, the potential function is: $f(x, y, z) = xyz + C$. Therefore, $\vec{F} = \langle yz,xz,xy\rangle $ is indeed conservative, and its potential function is $f(x, y, z) = xyz$ (ignoring the constant C since it does not affect the gradient).

Exercise 2: Exact Differential Forms

• Given the differential form: $Pdx +Qdy + Rdz = (axy)dx + (x^2 +z^3)dy + (byz^2-4z^3)dz, determine the values of constants $a$ and $b$ such that this form is exact. Then, find the potential function.

An exact differential form means that it can be written as the differential of some scalar potential function $f(x, y, z)$, i.e., $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy + \frac{\partial f}{\partial z}dz = Pdx + Qdy + Rdz$. This implies the vector field $\vec{F}$ is conservative.

$\vec{F} = \langle P, Q, R\rangle $ where $P = axy, Q = x^2 +z^3, R = byz^2-4z^3$.

To check if the differential form is exact, we need to verify the equality of mixed partial derivatives:

1. $P_y = Q_x \implies ax = 2x \implies a = 2$
2. $P_z = R_x \implies 0 = 0$ (This is automatically satisfied)
3. $Q_z = R_y \implies 3z^2 = bz^2 \implies b = 3$

Thus, the values of $a$ and $b$ such that the differential form is exact are: $a = 2$ and $b = 3$. The exact differential form becomes $2xydx + (x^2+z^3)dy +(3yz^2 -4z^3)dz$

Finding the Potential Function (Two Approaches)

Approach A: Using Line Integrals (The Geometric Method)

Because the field is conservative, the line integral is path-independent. We can find the potential at any point $(x, y, z)$ by integrating from the origin $(0,0,0)$ to $(x,y,z)$ along the easiest possible path: a “staircase” along the coordinate axes (something similar to Figure B).

To prevent variable confusion, we will use $t$ as our parameter for integration.

• Segment $C_1$: From $(0, 0, 0)$ to $(x, 0, 0)$. Let $\vec{r}(t) = \langle t, 0, 0 \rangle$ for $0 \leq t \leq x$. Here, $dy = dz = 0$, and $y = z = 0$. $$\int_{C_1} \vec{F} \cdot d\vec{r} = \int_{0}^{x} P(t, 0, 0)\,dt = \int_{0}^{x} 2(t)(0)\,dt = 0$$
• Segment $C_2$: From $(x, 0, 0)$ to $(x, y, 0)$. Let $\vec{r}(t) = \langle x, t, 0 \rangle$ for $0 \leq t \leq y$. Here, $dx = dz = 0$, and the $z$-coordinate is $0$. $$\int_{C_2} \vec{F} \cdot d\vec{r} = \int_{0}^{y} Q(x, t, 0)\,dt = \int_{0}^{y} (x^2 + 0^3)\,dt = \left[ x^2 t \right]_{0}^{y} = x^2 y$$
• Segment $C_3$: From $(x, y, 0)$ to $(x, y, z)$. Let $\vec{r}(t) = \langle x, y, t \rangle$ for $0 \leq t \leq z$. Here, $dx = dy = 0$. $$\int_{C_3} \vec{F} \cdot d\vec{r} = \int_{0}^{z} R(x, y, t)\,dt = \int_{0}^{z} (3yt^2 - 4t^3)\,dt = \left[ yt^3 - t^4 \right]_{0}^{z} = yz^3 - z^4$$

Adding these up, the potential function is the sum of the work done along the three paths: $f(x, y, z) = 0 + x^2 y + yz^3 - z^4 = x^2 y + yz^3 - z^4 + C$

Approach B: Using the Algebraic Gradient Method. We find a function $f(x, y, z)$ such that $\vec{F} = \nabla f$.

1. Integrate $f_x = P = 2xy$ with respect to $x$: $f(x,y,z) = \int 2xy \,dx = x^2 y + g(y, z)$
2. Differentiate with respect to $y$ and equate to $Q$: $f_y = x^2 + \frac{\partial g}{\partial y} = x^2 + z^3 \implies \frac{\partial g}{\partial y} = z^3$
   Integrating with respect to $y$ yields $g(y, z) = yz^3 + h(z)$. Now, $f = x^2 y + yz^3 + h(z)$.
3. Differentiate with respect to $z$ and equate to $R$: $f_z = 3yz^2 + h'(z) = 3yz^2 - 4z^3 \implies h'(z) = -4z^3$
   Integrating with respect to $z$ yields $h(z) = -z^4 + C$.
4. Combining everything, we arrive at the exact same potential function: $f(x,y,z) = x^2 y + yz^3 - z^4 + C$

Exercise 3: Trigonometric Vector Fields

• Is the field $\vec{F} = \langle y\cos(x) + y, \sin(x) + x, 1 \rangle$ conservative? If so, find the scalar potential function and evaluate the line integral $\int_{C} \vec{F} \cdot d\vec{r}$ for $C$ from $(0, 2, 1)$ to $(\pi, 1, 3)$.

Step 1. Check the Conservative Criteria. The vector field $\vec{F}$ is defined in $\mathbb{R}^3$ (all of three-dimensional space), which is a simply connected domain. In $\mathbb{R}^3$, any closed loop can be contracted to a single point, and there are no “holes”, “tunnels”, “obstacles”, or “missing points” that could prevent the field from being conservative.

Compute the partial derivatives. We verify the mixed partial derivatives for the components $P = y\cos(x) + y$, $Q = \sin(x) + x$, and $R = 1$:

$\star$ $P_y = \cos(x) + 1$ and $Q_x = \cos(x) + 1 \implies P_y = Q_x$

$\star$ $P_z = 0$ and $R_x = 0 \implies P_z = R_x$

$\star$ $Q_z = 0$ and $R_y = 0 \implies Q_z = R_y$

All mixed partials match. The field is conservative.

Step 2. Construct the Potential function.

1. Integrate $P$ with respect to $x$: $f(x,y,z) = \int (y\cos(x) + y)\,dx = y\sin(x) + yx + g(y, z)$
2. Differentiate with respect to $y$ and equate to $Q$: $f_y = \sin(x) + x + \frac{\partial g}{\partial y} = \sin(x) + x \implies \frac{\partial g}{\partial y} = 0$
   Because $\frac{\partial g}{\partial y} = 0$, $g$ is a function of $z$ alone: $g(z)$.
3. Differentiate with respect to $z$ and equate to $R$: $f_z = g'(z) = 1 \implies g(z) = z + C$
4. The potential function is: $f(x, y, z) = y\sin(x) + xy + z + C$

Step 3: Evaluate the Line Integral

Because the field is conservative, we do not need to parameterize a path. We invoke the Fundamental Theorem of Calculus for Line Integrals, which states that the work done is simply the difference in the potential function between the final and initial points: $\int_{C} \vec{F} \cdot d\vec{r} = f(\pi, 1, 3) - f(0, 2, 1)$

Evaluate the potential at the final point $(\pi, 1, 3)$: $f(\pi, 1, 3) = 1\sin(\pi) + (1)(\pi) + 3 + C = 0 + \pi + 3 + C = \pi + 3 + C$

Evaluate the potential at the initial point $(0, 2, 1)$: $f(0, 2, 1) = 2\sin(0) + (0)(2) + 1 + C = 0 + 0 + 1 + C = 1 + C$

Subtract the initial state from the final state: $\int_{C} \vec{F} \cdot d\vec{r} = (\pi + 3 + C) - (1 + C) = \pi + 2$

The line integral evaluates precisely to $\pi + 2$.

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