Lecture 24: Section 16.3 — The Fundamental Theorem of Line Integrals

In Section 16.2, we learned to compute vector line integrals, $\int_C \vec{F} \cdot d\vec{r}$. This was often a long process of parametrizing a curve.

In Section 16.4, we learned **Green's Theorem**, a powerful tool that connects a line integral over a closed loop to a double integral over the region inside: $$ \oint_C P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA $$

Now in Section 16.3, we will investigate a "special case" that is deeply connected to Green's Theorem. We will look for a shortcut for line integrals, similar to the Fundamental Theorem of Calculus from Calc I.

The big question is: Can we evaluate $\int_C \vec{F} \cdot d\vec{r}$ just by using the start and end points of the path $C$?

The answer is yes, but only for a very special class of vector fields.

Topic 1: Conservative Vector Fields and Potential Functions

The shortcut only works if the vector field $\vec{F}$ is the gradient of some scalar function $f$.

Definitions: Conservative Fields and Potential Functions

A vector field $\vec{F}$ is called a conservative vector field if it is the gradient of some scalar function $f$. That is, $\vec{F} = \nabla f$.

The scalar function $f$ is called the potential function for $\vec{F}$.

(Note: This name comes from physics, where the work done by a conservative force field, like gravity or an electric field, can be described by a change in potential energy, $f$.)

Example 1A: Show that $\vec{F}(x,y) = \langle 2x, 2y \rangle$ is conservative.

Solution (Step-by-step):

To show $\vec{F}$ is conservative, we must find a potential function $f(x,y)$ such that $\nabla f = \vec{F}$.

$\nabla f = \langle f_x, f_y \rangle = \langle 2x, 2y \rangle$

Set up equations:
(1) $f_x = 2x$
(2) $f_y = 2y$
Integrate (1) with respect to $x$:
If $f_x = 2x$, then $f(x,y) = \int 2x \,dx = x^2 + g(y)$.
Important: The "constant" of integration is not just $K$. Since we integrated with respect to $x$, any function that *only* involves $y$ (like $y^2$ or $\sin y$) would have an $x$-derivative of 0. So, our constant is an unknown function $g(y)$.
Differentiate our $f$ with respect to $y$:
Using our result from step 2, $f(x,y) = x^2 + g(y)$, we find $f_y$:
$f_y = \frac{\partial}{\partial y}(x^2 + g(y)) = 0 + g'(y) = g'(y)$.
Use equation (2) to solve for $g(y)$:
We now have two expressions for $f_y$. From (2), $f_y = 2y$. From step 3, $f_y = g'(y)$.
Set them equal: $g'(y) = 2y$.
Integrate with respect to $y$: $g(y) = \int 2y \,dy = y^2 + K$. (Here, $K$ is just a normal constant).
Combine and state the potential function:
Substitute $g(y)$ back into our expression from step 2:
$f(x,y) = x^2 + g(y) = x^2 + y^2 + K$.

Since we found a potential function (we can just pick $K=0$, $f(x,y) = x^2+y^2$), the vector field $\vec{F} = \langle 2x, 2y \rangle$ is conservative.

Example 1B: Show that $\vec{F}(x,y) = \langle y, -x \rangle$ is not conservative.

Solution (Step-by-step):

Let's *try* to find a potential function $f(x,y)$ such that $\nabla f = \langle y, -x \rangle$.

Set up equations:
(1) $f_x = y$
(2) $f_y = -x$
Integrate (1) with respect to $x$:
$f(x,y) = \int y \,dx = xy + g(y)$.
Differentiate our $f$ with respect to $y$:
$f_y = \frac{\partial}{\partial y}(xy + g(y)) = x + g'(y)$.
Use equation (2) to solve for $g(y)$:
Set our two expressions for $f_y$ equal:
$x + g'(y) = -x$
$g'(y) = -2x$

Contradiction!

The equation $g'(y) = -2x$ is a contradiction. The left side, $g'(y)$, must be a function of only $y$ (or a constant). The right side, $-2x$, is a function of $x$. It is impossible for a function of $y$ to equal a non-constant function of $x$.

Therefore, no such function $g(y)$ exists, which means no potential function $f(x,y)$ exists. The field is not conservative.

Check Your Understanding

Problem. Is the vector field $\vec{F}(x,y) = \langle 3x^2y^2, 2x^3y \rangle$ conservative? If so, find a potential function $f$.

We are looking for $f(x,y)$ such that $f_x = 3x^2y^2$ and $f_y = 2x^3y$.

Integrate $f_x$ w.r.t. $x$:
$f(x,y) = \int 3x^2y^2 \,dx = (x^3)y^2 + g(y)$.
Differentiate $f$ w.r.t. $y$:
$f_y = \frac{\partial}{\partial y}(x^3y^2 + g(y)) = 2x^3y + g'(y)$.
Set equal to $Q = f_y$:
$2x^3y + g'(y) = 2x^3y$.
Solve for $g(y)$:
$g'(y) = 0$.
This is not a contradiction! It just means $g(y)$ is a constant.
$g(y) = K$.
Conclusion:
Yes, the field is conservative.
The potential function is $f(x,y) = x^3y^2 + K$. (We can choose $K=0$ and use $f(x,y) = x^3y^2$).

Topic 2: The 2D Test for Conservatism

Finding the potential function $f$ just to prove a field is conservative is a lot of work. Example 1B (the failure) gives us a clue for a faster test.

If $\vec{F} = \langle P, Q \rangle = \nabla f = \langle f_x, f_y \rangle$, then $P = f_x$ and $Q = f_y$.
By Clairaut's Theorem on mixed partials, $f_{xy} = f_{yx}$.
But $f_{xy} = \frac{\partial}{\partial y}(f_x) = \frac{\partial P}{\partial y}$ (or $P_y$).
And $f_{yx} = \frac{\partial}{\partial x}(f_y) = \frac{\partial Q}{\partial x}$ (or $Q_x$).
Therefore, if $\vec{F}$ is conservative, we must have $P_y = Q_x$.

Theorem: Test for a Conservative Field (2D)

Let $\vec{F} = \langle P(x,y), Q(x,y) \rangle$ be a vector field on a simply-connected domain $D$.

$\vec{F}$ is conservative if and only if: $$ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} $$

A simply-connected domain is one that has no "holes". The $xy$-plane is simply-connected. A disk is simply-connected. The $xy$-plane minus the origin is not simply-connected (it has a "hole" at $(0,0)$). This "hole" is the source of many counter-examples, but for most problems in this class, the domain is all of $\mathbb{R}^2$ and this condition holds.

Example 2A: Use the test on $\vec{F} = \langle 3 + 2xy, x^2 - 3y^2 \rangle$

Solution (Step-by-step):

Identify P and Q.
$P(x,y) = 3 + 2xy$
$Q(x,y) = x^2 - 3y^2$
Compute partials.
$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3 + 2xy) = 2x$.
$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 - 3y^2) = 2x$.
Compare.
$P_y = 2x$ and $Q_x = 2x$. They are equal.
Conclusion.
Since $P_y = Q_x$ and the domain is $\mathbb{R}^2$ (which is simply-connected), the vector field $\vec{F}$ is conservative.

Example 2B: Use the test on $\vec{F} = \langle x+y, x-y \rangle$

Solution (Step-by-step):

Identify P and Q.
$P(x,y) = x+y$
$Q(x,y) = x-y$
Compute partials.
$\frac{\partial P}{\partial y} = 1$.
$\frac{\partial Q}{\partial x} = 1$.
Compare.
$P_y = 1$ and $Q_x = 1$. They are equal.
Conclusion.
The vector field $\vec{F}$ is conservative.

Check Your Understanding

Problem. Use the 2D test to determine if $\vec{F} = \langle e^x \cos y, -e^x \sin y \rangle$ is conservative.

Identify P and Q.
$P = e^x \cos y$
$Q = -e^x \sin y$
Compute partials.
$\frac{\partial P}{\partial y} = e^x (-\sin y) = -e^x \sin y$.
$\frac{\partial Q}{\partial x} = (-e^x \sin y) = -e^x \sin y$.
Compare.
$P_y = Q_x$.
Conclusion.
Yes, the field is conservative.

Topic 3: A Bridge to Green's Theorem

We just learned that if a 2D field $\vec{F} = \langle P, Q \rangle$ is conservative, then $P_y = Q_x$ (which means $Q_x - P_y = 0$).

What does Green's Theorem (Sec 16.4) tell us about this?

Green's Theorem states that for any simple closed path $C$ enclosing a region $D$: $$ \oint_C P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA $$

Assuming $\vec{F} = \langle P(x,y) , Q(x,y) \rangle$, Let's apply this to a conservative vector field.
If $\vec{F}$ is conservative, we know $Q_x - P_y = 0$.
Therefore, Green's Theorem simplifies: $$ \oint_C P\,dx + Q\,dy = \iint_D (0)\,dA = 0 $$

Green's Theorem Proves a Key Result

Thanks to Green's Theorem, we have now *proven* that for any conservative vector field $\vec{F}$ on a simply-connected domain:

The line integral over any simple closed loop $C$ is zero. $$ \oint_C \vec{F} \cdot d\vec{r} = 0 $$

This is a major component of the Fundamental Theorem, and we proved it using our knowledge of Section 16.4!

This result—that closed-loop integrals are zero—is what leads us to our big shortcut.

Topic 4: The Fundamental Theorem and Path Independence

The fact that $\oint_C \vec{F} \cdot d\vec{r} = 0$ for conservative fields has a profound consequence.

Consider two different paths, $C_1$ and $C_2$, that both start at $A$ and end at $B$.
If we go from $A$ to $B$ along $C_1$, and then *backwards* from $B$ to $A$ along $C_2$ (call this $-C_2$), we've created a single closed loop $C = C_1 \cup (-C_2)$.
The integral over this loop must be zero: $\oint_C \vec{F} \cdot d\vec{r} = \int_{C_1} \vec{F} \cdot d\vec{r} + \int_{-C_2} \vec{F} \cdot d\vec{r} = 0$.
Since $\int_{-C_2} = -\int_{C_2}$, this means $\int_{C_1} \vec{F} \cdot d\vec{r} - \int_{C_2} \vec{F} \cdot d\vec{r} = 0$.
This proves: $\int_{C_1} \vec{F} \cdot d\vec{r} = \int_{C_2} \vec{F} \cdot d\vec{r}$.

Path Independence vs. Closed Loops

A line integral $\int_C \vec{F} \cdot d\vec{r}$ is path-independent on a domain $D$ if and only if $\oint_C \vec{F} \cdot d\vec{r} = 0$ for all closed paths $C$ in $D$.

Geometric Interpretation: The Altitude Analogy

Think of the potential function $f(x,y)$ as a mountain or a landscape, where the value $f$ is the altitude at point $(x,y)$.

The conservative field $\vec{F} = \nabla f$ is a "gradient map" that points in the direction of steepest ascent.
The line integral $\int_C \vec{F} \cdot d\vec{r}$ represents the total net change in altitude as you walk along the path $C$.
Path Independence: This means your total altitude change only depends on your Start Point (A) and End Point (B). It doesn't matter if you take a direct path or a long, winding switchback; your net altitude change, $f(B) - f(A)$, is identical.
Closed Loop = 0: This means if you take any hike that ends exactly where you started ($A=B$), your net change in altitude must be zero. This is geometrically obvious!

This brings all our ideas together. The *only* vector fields that have this path-independent "altitude" property are conservative vector fields.

Equivalence Theorem

For a vector field $\vec{F}$ on a simply-connected domain $D$, the following statements are all equivalent (if one is true, all are true):

$\vec{F}$ is conservative (i.e., $\vec{F} = \nabla f$, or $P_y = Q_x$ in 2D).
$\oint_C \vec{F} \cdot d\vec{r} = 0$ for every closed loop $C$ in $D$.
$\int_C \vec{F} \cdot d\vec{r}$ is path-independent in $D$.

If the integral only depends on the start and end points, there must be a simple way to evaluate it. This is the Fundamental Theorem.

Theorem: The Fundamental Theorem of Line Integrals

Let $C$ be a smooth curve with start point $A = \vec{r}(a)$ and end point $B = \vec{r}(b)$.

If $\vec{F}$ is a continuous conservative vector field with potential function $f$ (meaning $\vec{F} = \nabla f$), then:

$$ \int_C \vec{F} \cdot d\vec{r} = \int_C \nabla f \cdot d\vec{r} = f(B) - f(A) $$

In words: The line integral of a conservative field is just (Potential at End) - (Potential at Start).

Example 4A: Compute $\displaystyle \int_C \vec{F} \cdot d\vec{r}$

Where $\vec{F} = \langle 2x, 2y \rangle$ and $C$ is the top half of the unit circle $x^2+y^2=1$ from $(1,0)$ to $(-1,0)$.

Solution (Step-by-step):

Check if $\vec{F}$ is conservative.
From Example 1A, yes, it is. We found its potential function is $f(x,y) = x^2 + y^2$.
Identify Start and End points.
The path starts at $A = (1,0)$.
The path ends at $B = (-1,0)$.
Apply the Fundamental Theorem.
We do not need to parametrize! We just evaluate $f$ at the endpoints.
$\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A) = f(-1,0) - f(1,0)$.
Calculate.
$f(-1,0) = (-1)^2 + (0)^2 = 1$.
$f(1,0) = (1)^2 + (0)^2 = 1$.
Value = $1 - 1 = \boxed{0}$.

Compare: The old way would be $\vec{r}(t) = \langle \cos t, \sin t \rangle$ for $t=0$ to $t=\pi$. $d\vec{r} = \langle -\sin t, \cos t \rangle dt$. $\vec{F} = \langle 2\cos t, 2\sin t \rangle$. $\vec{F} \cdot d\vec{r} = (-2\cos t \sin t + 2\sin t \cos t)dt = 0 \,dt$. The integral is $\int_0^\pi 0 \,dt = 0$. The theorem gave us the same answer with almost no work.

Example 4B: Closed Loop Integral

Compute $\oint_C (y\,dx + x\,dy)$, where $C$ is the closed path from $A=(0,0)$ to $B=(1,1)$ along the parabola $y=x^2$, and then from $B=(1,1)$ back to $A=(0,0)$ along the line $y=x$.

Solution (Step-by-step):

Analyze the integral.
This is $\oint_C \vec{F} \cdot d\vec{r}$ with $\vec{F} = \langle y, x \rangle$.
The path $C$ is a closed loop, starting at $(0,0)$ and ending at $(0,0)$.
Check if $\vec{F}$ is conservative.
We use the 2D test (Topic 2):
$P=y \implies P_y = 1$.
$Q=x \implies Q_x = 1$.
Since $P_y = Q_x$, the field is conservative.
Apply the Equivalence Theorem.
Since $\vec{F}$ is conservative and $C$ is a closed loop, the Equivalence Theorem (and our bridge from Green's Theorem) tells us the integral must be $\boxed{0}$.
Pedagogical Note:
Think about the "old way." We would have to compute two separate line integrals:
$C_1$ (parabola): $\int_{C_1} (y\,dx + x\,dy)$
$C_2$ (line): $\int_{C_2} (y\,dx + x\,dy)$
...and then add them: $W = \int_{C_1} + \int_{C_2}$.
The Fundamental Theorem lets us get the answer $0$ with almost no work.

Check Your Understanding

Problem. Let $\vec{F} = \langle 3x^2y^2, 2x^3y \rangle$.
(a) Is $\vec{F}$ conservative?
(b) What is $\int_C \vec{F} \cdot d\vec{r}$ for any path $C$ from $(1, 2)$ to $(3, 0)$?

(a) Test for Conservatism:
$P = 3x^2y^2 \implies P_y = 6x^2y$.
$Q = 2x^3y \implies Q_x = 6x^2y$.
Yes, $P_y = Q_x$, so the field is conservative.
(b) Evaluate integral:
Since $\vec{F}$ is conservative, the integral is path-independent. We can use the FTLI.
First, find $f$: From the CYU in Topic 1, we know the potential function is $f(x,y) = x^3y^2$.
Next, use $f(B) - f(A)$:
Start: $A = (1, 2)$. End: $B = (3, 0)$.
$\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A) = f(3,0) - f(1,2)$.
$f(3,0) = (3)^3(0)^2 = 0$.
$f(1,2) = (1)^3(2)^2 = 4$.
Value = $0 - 4 = \boxed{-4}$.

Topic 5: Conservation of Energy

A major application of the Fundamental Theorem is in physics, where it leads directly to the Law of Conservation of Energy.

Let's consider a particle of mass $m$ moving along a path $C$, parametrized by $\vec{r}(t)$.
Its velocity is $\vec{v}(t) = \vec{r}'(t)$.
Its acceleration is $\vec{a}(t) = \vec{r}''(t)$.
By Newton's Second Law, the force $\vec{F}$ acting on the particle is $\vec{F}(\vec{r}(t)) = m\vec{a}(t) = m\vec{r}''(t)$.

The **Work** done by this force field from $t=a$ to $t=b$ is: $$ W = \int_C \vec{F} \cdot d\vec{r} = \int_a^b \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \,dt $$ Substitute $\vec{F} = m\vec{r}''$ and $\vec{v} = \vec{r}'$: $$ W = \int_a^b (m\vec{r}''(t)) \cdot \vec{r}'(t) \,dt = m \int_a^b \vec{r}'' \cdot \vec{r}' \,dt $$

Now, let's use a clever derivative trick. What is $\frac{d}{dt}(\vec{v} \cdot \vec{v})$?
By the dot product rule: $\frac{d}{dt}(\vec{v} \cdot \vec{v}) = \vec{v}' \cdot \vec{v} + \vec{v} \cdot \vec{v}' = 2(\vec{v} \cdot \vec{v}') = 2(\vec{r}' \cdot \vec{r}'')$.
So, $\vec{r}' \cdot \vec{r}'' = \frac{1}{2} \frac{d}{dt}(\vec{v} \cdot \vec{v}) = \frac{1}{2} \frac{d}{dt}|\vec{v}|^2$.

Substitute this back into the work integral: $$ W = m \int_a^b \left( \frac{1}{2} \frac{d}{dt}|\vec{v}(t)|^2 \right) \,dt = \frac{1}{2}m \int_a^b \frac{d}{dt}|\vec{v}(t)|^2 \,dt $$

By the Calc I FTC, this integral is: $$ W = \frac{1}{2}m \left[ |\vec{v}(t)|^2 \right]_a^b = \frac{1}{2}m \left( |\vec{v}(b)|^2 - |\vec{v}(a)|^2 \right) $$

We define $K = \frac{1}{2}m|\vec{v}|^2$ as the Kinetic Energy. Let $K(A)$ and $K(B)$ be the kinetic energies at the start and end points.

The Work-Energy Theorem

The work done by a force field $\vec{F}$ on a particle along a path $C$ is equal to the change in the particle's kinetic energy:

$$ W = \int_C \vec{F} \cdot d\vec{r} = K(B) - K(A) = \Delta K $$

Now, what if the force field $\vec{F}$ is conservative?
If $\vec{F}$ is conservative, then $\vec{F} = \nabla f$ for some potential $f$.
By the FTLI, $W = \int_C \vec{F} \cdot d\vec{r} = f(B) - f(A)$.

Let's combine our two results for Work:
$W = K(B) - K(A)$
$W = f(B) - f(A)$
Therefore: $K(B) - K(A) = f(B) - f(A)$.

In physics, it's conventional to define the Potential Energy $P$ as the negative of the potential function $f$. So $P = -f$. This means $\vec{F} = \nabla f = -\nabla P$.
With this definition, $f(B) - f(A) = -P(B) - (-P(A)) = P(A) - P(B)$.

So, $K(B) - K(A) = P(A) - P(B)$.
Rearranging gives: $K(B) + P(B) = K(A) + P(A)$.

The Law of Conservation of Energy

If a particle moves under the influence of a conservative force field $\vec{F}$, its total energy (Kinetic + Potential) is constant.

Let $E = K + P$ be the total energy. Then $E(B) = E(A)$.

$$ \frac{1}{2}m|\vec{v}(b)|^2 + P(B) = \frac{1}{2}m|\vec{v}(a)|^2 + P(A) $$

Example 5A: Work-Energy Theorem

A particle of mass $m=2$ moves along a path $C$. Its velocity at the start $A$ is $|\vec{v}(a)| = 3$. At the end $B$, its velocity is $|\vec{v}(b)| = 5$. What is the total work done on the particle by the force field $\vec{F}$ that moved it?

Solution (Step-by-step):

Identify the goal. We need to find the total Work $W$.
Identify the theorem. The Work-Energy Theorem relates work to the change in kinetic energy: $W = K(B) - K(A)$.
Calculate Kinetic Energies.
$K(A) = \frac{1}{2}m|\vec{v}(a)|^2 = \frac{1}{2}(2)(3)^2 = (1)(9) = 9$.
$K(B) = \frac{1}{2}m|\vec{v}(b)|^2 = \frac{1}{2}(2)(5)^2 = (1)(25) = 25$.
Calculate Work.
$W = K(B) - K(A) = 25 - 9 = \boxed{16}$.
(The work done on the particle is 16 joules, if using SI units).

Example 5B: Conservation of Energy

A particle of mass $m=4$ is at rest ($K(A)=0$) at point $A=(1,0)$. It is moved by the conservative force field $\vec{F} = \langle 2x, 2y \rangle$. What is the particle's speed when it reaches point $B=(0,2)$?

Solution (Step-by-step):

Identify the goal. We need to find the speed $|\vec{v}(b)|$ at point $B$.
Identify the theorem. The force field is conservative, so we can use Conservation of Energy. We can use the form $K(B) - K(A) = f(B) - f(A)$.
Find the potential function $f$.
From Example 1A, the potential function for $\vec{F} = \langle 2x, 2y \rangle$ is $f(x,y) = x^2 + y^2$.
Calculate values at Start Point $A=(1,0)$.
$f(A) = (1)^2 + (0)^2 = 1$.
The particle is "at rest", so $K(A) = 0$.
Calculate values at End Point $B=(0,2)$.
$f(B) = (0)^2 + (2)^2 = 4$.
$K(B) = \frac{1}{2}m|\vec{v}(b)|^2 = \frac{1}{2}(4)|\vec{v}(b)|^2 = 2|\vec{v}(b)|^2$.
Set up the Conservation equation and solve.
$K(B) - K(A) = f(B) - f(A)$
$(2|\vec{v}(b)|^2) - (0) = (4) - (1)$
$2|\vec{v}(b)|^2 = 3$
$|\vec{v}(b)|^2 = 3/2$
The speed is $\boxed{|\vec{v}(b)| = \sqrt{3/2}}$.

Check Your Understanding

Problem. A particle with mass $m=10$ moves in a conservative force field $\vec{F}$ with potential energy $P(x,y) = x^2 + 3y$. The particle is at $A=(1, 5)$ and has speed 2. It moves to $B=(3, 1)$. What is its speed at $B$?

Use Conservation of Energy: $K(B) + P(B) = K(A) + P(A)$.
Calculate values at Start $A=(1,5)$.
$P(A) = (1)^2 + 3(5) = 1 + 15 = 16$.
$K(A) = \frac{1}{2}m|\vec{v}(a)|^2 = \frac{1}{2}(10)(2)^2 = 5(4) = 20$.
Calculate values at End $B=(3,1)$.
$P(B) = (3)^2 + 3(1) = 9 + 3 = 12$.
$K(B) = \frac{1}{2}m|\vec{v}(b)|^2 = \frac{1}{2}(10)|\vec{v}(b)|^2 = 5|\vec{v}(b)|^2$.
Set up the equation and solve for $|\vec{v}(b)|$.
$(5|\vec{v}(b)|^2) + (12) = (20) + (16)$
$5|\vec{v}(b)|^2 + 12 = 36$
$5|\vec{v}(b)|^2 = 24$
$|\vec{v}(b)|^2 = 24/5 = 4.8$
$|\vec{v}(b)| = \boxed{\sqrt{24/5}}$ (or $\sqrt{4.8} \approx 2.19$).

Lecture Conclusion

What You Should Have Learned

The definition of a conservative vector field ($\vec{F} = \nabla f$) and its potential function $f$.
The 2D Test for conservatism: $P_y = Q_x$ on a simply-connected domain.
How Green's Theorem proves that $\oint_C \vec{F} \cdot d\vec{r} = 0$ for all conservative fields.
That Path Independence is equivalent to Zero Closed Loops.
The Fundamental Theorem of Line Integrals is the ultimate shortcut: $\int_C \nabla f \cdot d\vec{r} = f(\text{End}) - f(\text{Start})$.
This theorem is the basis for the Law of Conservation of Energy ($K(A)+P(A) = K(B)+P(B)$).

Final Practice Set

Practice #1 (Full 2D Problem)

Problem. Let $\vec{F} = \langle 4x^3y^2 - 2xy^3, 2x^4y - 3x^2y^2 \rangle$.
(a) Show that $\vec{F}$ is conservative using the 2D test.
(b) Find a potential function $f(x,y)$.
(c) Evaluate $\int_C \vec{F} \cdot d\vec{r}$ where $C$ is the curve $\vec{r}(t) = \langle t \cos(\pi t), t + \sin(\pi t) \rangle$ from $t=0$ to $t=1$.

(a) Test for Conservatism:
$P = 4x^3y^2 - 2xy^3 \implies P_y = 8x^3y - 6xy^2$.
$Q = 2x^4y - 3x^2y^2 \implies Q_x = 8x^3y - 6xy^2$.
Since $P_y = Q_x$, the field is conservative.
(b) Find $f$:
$f_x = 4x^3y^2 - 2xy^3 \implies f = \int (4x^3y^2 - 2xy^3) \,dx = x^4y^2 - x^2y^3 + g(y)$.
Differentiate w.r.t $y$: $f_y = 2x^4y - 3x^2y^2 + g'(y)$.
Set equal to $Q$: $2x^4y - 3x^2y^2 + g'(y) = 2x^4y - 3x^2y^2$.
$g'(y) = 0 \implies g(y) = K$.
Potential function: $f(x,y) = x^4y^2 - x^2y^3$.
(c) Evaluate Integral:
The path is horrible. Use the FTLI.
Start ($t=0$): $A = \vec{r}(0) = \langle 0 \cos(0), 0 + \sin(0) \rangle = \langle 0, 0 \rangle$.
End ($t=1$): $B = \vec{r}(1) = \langle 1 \cos(\pi), 1 + \sin(\pi) \rangle = \langle -1, 1 \rangle$.
$\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A) = f(-1, 1) - f(0, 0)$.
$f(-1, 1) = (-1)^4(1)^2 - (-1)^2(1)^3 = (1)(1) - (1)(1) = 0$.
$f(0, 0) = 0$.
Value = $0 - 0 = \boxed{0}$.

Practice #2 (3D Potential Function)

Problem. Let $\vec{F} = \langle e^z, 1, xe^z \rangle$.
(a) Find a potential function $f(x,y,z)$ (assume it is conservative).
(b) Evaluate $\int_C \vec{F} \cdot d\vec{r}$ where $C$ is the line segment from $(0,0,0)$ to $(1, 2, 3)$.

(a) Find $f$:
We need $f_x = e^z$, $f_y = 1$, $f_z = xe^z$.
$f_x = e^z \implies f = \int e^z \,dx = xe^z + g(y,z)$.
Differentiate w.r.t $y$: $f_y = 0 + g_y(y,z)$.
Set equal to $Q$: $g_y(y,z) = 1$.
Integrate $g_y$ w.r.t $y$: $g(y,z) = \int 1 \,dy = y + h(z)$.
Our $f$ is now: $f = xe^z + y + h(z)$.
Differentiate w.r.t $z$: $f_z = xe^z + 0 + h'(z)$.
Set equal to $R$: $xe^z + h'(z) = xe^z$.
$h'(z) = 0 \implies h(z) = K$.
Potential function: $f(x,y,z) = xe^z + y$.
(b) Evaluate Integral:
Use the FTLI.
Start $A=(0,0,0)$, End $B=(1,2,3)$.
$\int_C \vec{F} \cdot d\vec{r} = f(B) - f(A) = f(1,2,3) - f(0,0,0)$.
$f(1,2,3) = (1)e^3 + (2) = e^3 + 2$.
$f(0,0,0) = (0)e^0 + (0) = 0$.
Value = $\boxed{e^3 + 2}$.