Lesson 28: Fundamental Theorem of Calculus for Line Integrals (17.3)

Accessible transcription generated on 3/30/2026

Original Notes

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Lesson 28: Fundamental Theorem of Calculus for Line Integrals (17.3)

Example from last class: \(\vec{F} = \langle x, -y \rangle\)

\(C_1 = \vec{r}_1(t) = \langle 2t, t^2 \rangle, 0 \le t \le 1\)

A coordinate plane plot illustrating different paths between the origin (0,0) and the point (2,1). A green curved path, labeled C1, connects (0,0) and (2,1), arching upwards. A blue straight line path, labeled C2, also connects (0,0) and (2,1) directly. Two additional paths are sketched in red: one starts at (0,0), dips below the x-axis, and then curves back up to reach (2,1); another red path curves from above to the endpoint. Arrows on all paths indicate a direction of travel from (0,0) toward (2,1). The diagram visually represents multiple ways to travel between the same two points in a vector field.
Visual Description: A coordinate plane plot illustrating different paths between the origin (0,0) and the point (2,1). A green curved path, labeled C1, connects (0,0) and (2,1), arching upwards. A blue straight line path, labeled C2, also connects (0,0) and (2,1) directly. Two additional paths are sketched in red: one starts at (0,0), dips below the x-axis, and then curves back up to reach (2,1); another red path curves from above to the endpoint. Arrows on all paths indicate a direction of travel from (0,0) toward (2,1). The diagram visually represents multiple ways to travel between the same two points in a vector field.
\[ \int_{C_1} \vec{F} \cdot \vec{T} \, ds = \int_{C_1} \vec{F} \cdot d\vec{r} = \int_{0}^{1} \vec{F}(\vec{r}_1(t)) \cdot \vec{r}_1'(t) \, dt \] \[ = \int_{0}^{1} \langle 2t, -t^2 \rangle \cdot \langle 2, 2t \rangle \, dt \] \[ = \int_{0}^{1} (4t - 2t^3) \, dt = \frac{3}{2} \]

\(\underline{C_2}: \vec{r}_2(t) = \langle t, \frac{t}{2} \rangle, 0 \le t \le 2\)

\[ \int_{C_2} \vec{F} \cdot \vec{T} \, ds = \int_{0}^{2} \vec{F}(\vec{r}_2(t)) \cdot \vec{r}_2'(t) \, dt = \int_{0}^{2} \langle t, -\frac{t}{2} \rangle \cdot \langle 1, \frac{1}{2} \rangle \, dt \] \[ = \int_{0}^{2} \frac{3t}{4} \, dt = \left. \frac{3t^2}{8} \right|_0^2 = \frac{3}{2} \]

Today! When is \(\int_{C} \vec{F} \cdot \vec{T} \, ds\) NOT dependent on the curve?

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Recall!

\[ \int_{a}^{b} f'(t) dt = f(b) - f(a) \] This is the Fundamental Theorem of Calculus.
A vector field \(\vec{F}\) is called a Gradient field or Conservative if there exists a scalar function \(\phi\) such that: \[ \vec{F} = \vec{\nabla}\phi \] In this context, \(\phi\) is called the potential function.

Fundamental Theorem of Calculus for Line Integrals

A 2D coordinate system showing a path C between two points. The x and y axes are drawn as simple black lines. A green, winding curve labeled C starts at a point A labeled '(start)' in the bottom-left quadrant and ends at a point B labeled '(end)' in the top-right quadrant. The path is complex and crosses the axes multiple times, illustrating that the path taken does not matter for conservative fields. There is an arrow on the path C indicating the direction from A to B.
Visual Description: A 2D coordinate system showing a path C between two points. The x and y axes are drawn as simple black lines. A green, winding curve labeled C starts at a point A labeled '(start)' in the bottom-left quadrant and ends at a point B labeled '(end)' in the top-right quadrant. The path is complex and crosses the axes multiple times, illustrating that the path taken does not matter for conservative fields. There is an arrow on the path C indicating the direction from A to B.

Suppose \(\vec{F} = \vec{\nabla}\phi\), then:

\[ \int_{C} \vec{F} \cdot d\vec{r} = \int_{C} \vec{\nabla}\phi \cdot d\vec{r} = \phi(\text{end}) - \phi(\text{start}) \]
If \(\vec{F}\) is Conservative, then the line integral does not depend on the path (curve).

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Going back to the example from last class:

Is \(\vec{F} = \langle x, -y \rangle\) conservative?

Can we find \(\phi(x,y)\) such that \(\vec{F} = \vec{\nabla}\phi\)?

Can we find \(\phi(x,y)\) such that \(\phi_x = x\) & \(\phi_y = -y\)?

Guess: \(\phi(x,y) = \frac{x^2 - y^2}{2}\)

Verify:

\[ \frac{\partial\phi}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x^2 - y^2}{2} \right) = x, \quad \frac{\partial\phi}{\partial y} = \frac{\partial}{\partial y} \left( \frac{x^2 - y^2}{2} \right) = -y. \]

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Application of the Fundamental Theorem of Line Integrals

\(\vec{F} = \langle x, -y \rangle\) is conservative, \(\vec{F} = \vec{\nabla}\phi\) for \(\phi(x,y) = \frac{x^2 - y^2}{2}\)

to compute \(\int_C \vec{F} \cdot \vec{T} ds = \int_C \vec{F} \cdot d\vec{r}\)

we can apply FTCLI

Diagram illustrating path independence in a vector field on a 2D Cartesian plane. The origin (0,0) and a point (2,1) in the first quadrant are connected by three different paths. A straight blue line path with an arrow points directly from the origin to (2,1). A smooth green curve, labeled 'C', also runs from the origin to (2,1). A third path, drawn in red, is highly irregular and winding, passing through the second, third, and fourth quadrants before eventually terminating at (2,1). Arrows on all three paths indicate the direction of travel from the start point (0,0) to the end point (2,1). This visual demonstrates that for a conservative vector field, the line integral between two points depends only on the endpoints and is independent of the specific path taken.
Visual Description: Diagram illustrating path independence in a vector field on a 2D Cartesian plane. The origin (0,0) and a point (2,1) in the first quadrant are connected by three different paths. A straight blue line path with an arrow points directly from the origin to (2,1). A smooth green curve, labeled 'C', also runs from the origin to (2,1). A third path, drawn in red, is highly irregular and winding, passing through the second, third, and fourth quadrants before eventually terminating at (2,1). Arrows on all three paths indicate the direction of travel from the start point (0,0) to the end point (2,1). This visual demonstrates that for a conservative vector field, the line integral between two points depends only on the endpoints and is independent of the specific path taken.
\[ \begin{aligned} \int_C \vec{F} \cdot \vec{T} ds &= \phi(\text{end}) - \phi(\text{start}) \\ &= \phi(2, 1) - \phi(0, 0) \\ &= \left( \frac{2^2 - 1^2}{2} \right) - \left( \frac{0^2 - 0^2}{2} \right) \\ &= \frac{3}{2} \end{aligned} \]

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When is \(\vec{F} = \langle f, g \rangle\) Conservative?

If it is conservative, we can find \(\phi(x, y)\) such that:

\[ \langle f, g \rangle = \vec{F} = \nabla\phi = \langle \phi_x, \phi_y \rangle \]

This leads to the system of equations:

\[ \begin{cases} f = \phi_x \\ g = \phi_y \end{cases} \]

Recall: Mixed partials are the same:

\[ \phi_{xy} = \phi_{yx} \] \[ \frac{\partial}{\partial y}(\phi_x) = \frac{\partial}{\partial x}(\phi_y) \]

Substituting the definitions for \(f\) and \(g\):

\[ \frac{\partial f}{\partial y} = \frac{\partial g}{\partial x} \]

Theorem: \(\vec{F} = \langle f, g \rangle\) is conservative when \(f_y = g_x\).

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Examples: Testing for Conservative Vector Fields and Applicability of FTCLI

1) \(\vec{F} = \langle \underbrace{x}_{f}, \underbrace{y}_{g} \rangle\) \[ \left. \begin{aligned} f_y &= \frac{\partial}{\partial y}(x) = 0 \\ g_x &= \frac{\partial}{\partial x}(y) = 0 \end{aligned} \right\} \Rightarrow \vec{F} \text{ is conservative} \] CAN APPLY FTCLI to compute \(\int_C \vec{F} \cdot d\vec{r}\)
2) \(\vec{F} = \langle \underbrace{x^2}_{f}, \underbrace{-xy}_{g} \rangle\) \[ \left. \begin{aligned} f_y &= \frac{\partial}{\partial y}(x^2) = 0 \\ g_x &= \frac{\partial}{\partial x}(-xy) = -y \end{aligned} \right\} \Rightarrow \vec{F} \text{ is NOT conservative} \] CANNOT Apply FTCLI to compute \(\int_C \vec{F} \cdot d\vec{r}\)
3) \(\vec{F} = \langle \underbrace{ye^x + 1}_{f}, \underbrace{e^x + 2y}_{g} \rangle\) \[ \left. \begin{aligned} f_y &= e^x \\ g_x &= e^x \end{aligned} \right\} \Rightarrow \vec{F} \text{ is conservative} \] CAN APPLY FTCLI to compute \(\int_C \vec{F} \cdot d\vec{r}\)

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Example: \(\vec{F} = \langle x^2, -xy \rangle\) is NOT conservative

Consider the vector field \(\vec{F} = \langle x^2, -xy \rangle\). We will evaluate the line integral of this field along two different paths, \(C_1\) and \(C_2\), connecting the points \((0,1)\) and \((1,0)\).

Paths defined:

  • \(C_1: \vec{r}_1(t) = \langle \sin t, \cos t \rangle, \quad 0 \le t \le \frac{\pi}{2}\)
  • \(C_2: \vec{r}_2(t) = \langle t, -t+1 \rangle, \quad 0 \le t \le 1\)
A hand-drawn diagram on a grid showing two integration paths, C1 and C2, in the first quadrant of a 2D Cartesian plane. Both paths start at the point (0, 1) on the vertical y-axis and end at the point (1, 0) on the horizontal x-axis. Path C1 is depicted as a blue circular arc curving away from the origin, with a blue arrow showing the direction of movement from (0,1) to (1,0). Path C2 is a straight green line segment directly connecting (0, 1) and (1, 0), with a green arrow indicating the same direction of travel.
Visual Description: A hand-drawn diagram on a grid showing two integration paths, C1 and C2, in the first quadrant of a 2D Cartesian plane. Both paths start at the point (0, 1) on the vertical y-axis and end at the point (1, 0) on the horizontal x-axis. Path C1 is depicted as a blue circular arc curving away from the origin, with a blue arrow showing the direction of movement from (0,1) to (1,0). Path C2 is a straight green line segment directly connecting (0, 1) and (1, 0), with a green arrow indicating the same direction of travel.

Evaluating the integral along path \(C_1\):

\[C_1: \int_{C_1} \vec{F} \cdot d\vec{r} = \int_{0}^{\frac{\pi}{2}} \vec{F}(\vec{r}_1(t)) \cdot \vec{r}_1'(t) dt\] \[= \int_{0}^{\frac{\pi}{2}} \langle \sin^2 t, -\sin t \cos t \rangle \cdot \langle \cos t, -\sin t \rangle dt\] \[= \int_{0}^{\frac{\pi}{2}} ( \sin^2 t \cos t + \sin^2 t \cos t ) dt\] \[= \int_{0}^{\frac{\pi}{2}} 2 \sin^2 t \cos t dt = 2 \left[ \frac{\sin^3 t}{3} \right]_0^{\frac{\pi}{2}} = \frac{2}{3}\]

Evaluating the integral along path \(C_2\):

\[C_2: \int_{C_2} \vec{F} \cdot d\vec{r} = \int_{0}^{1} \langle t^2, -(t)(-t+1) \rangle \cdot \langle 1, -1 \rangle dt\] \[= \int_{0}^{1} \langle t^2, t^2 - t \rangle \cdot \langle 1, -1 \rangle dt\] \[= \int_{0}^{1} (t^2 - (t^2 - t)) dt\] \[= \int_{0}^{1} t dt = \left[ \frac{t^2}{2} \right]_0^1 = \frac{1}{2}\]

Since the line integral depends on the path taken (\(\frac{2}{3} \neq \frac{1}{2}\)), the vector field \(\vec{F}\) is not conservative.

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#### Finding a Potential Function Find \(\phi\) such that \(\vec{F} = \vec{\nabla}\phi\)
Given: \(\vec{F} = \langle ye^x + 1, e^x + 2y \rangle\). We want \(\phi\) such that \(\vec{F} = \vec{\nabla}\phi = \langle \phi_x, \phi_y \rangle\). To find \(\phi\), we solve the system of equations:
  1. \(\frac{\partial \phi}{\partial x} = ye^x + 1\)
  2. \(\frac{\partial \phi}{\partial y} = e^x + 2y\)
Step 1: Integrate equation (1) with respect to \(x\), treating \(y\) as a constant. \[ \frac{\partial \phi}{\partial x} = ye^x + 1 \xrightarrow{\text{integrate w.r.t } x, \text{ } y \text{ is const.}} \phi(x,y) = ye^x + x + K(y) \] Step 2: Substitute this expression for \(\phi(x,y)\) into equation (2). \[ e^x + 2y = \frac{\partial \phi}{\partial y} = e^x + 0 + K'(y) \] Solving for \(K'(y)\): \[ K'(y) = 2y \] Step 3: Integrate \(K'(y)\) with respect to \(y\). \[ K'(y) = 2y \xrightarrow{\text{integrate w.r.t } y} K(y) = y^2 + K \] Final Result: \[ \phi(x,y) = ye^x + x + y^2 + K \text{ satisfies } \vec{F} = \vec{\nabla}\phi. \]

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Example: Conservative Vector Fields and Line Integrals

Given the vector field \(\vec{F} = \langle ye^x + 1, e^x + 2y \rangle\), find the line integral \(\int_C \vec{F} \cdot \vec{T} ds\).

Assume \(\vec{F} = \vec{\nabla}\phi\), where the potential function is \(\phi(x,y) = ye^x + x + y^2 + k\).

Apply FTCLI:

A coordinate plane diagram showing a complex, winding path labeled 'C'. The path starts at a point labeled 'start' with coordinates (1,0) located on the x-axis. The green curve meanders through several quadrants before ending at a point labeled 'end' with coordinates (0,1) located on the y-axis.
Visual Description: A coordinate plane diagram showing a complex, winding path labeled 'C'. The path starts at a point labeled 'start' with coordinates (1,0) located on the x-axis. The green curve meanders through several quadrants before ending at a point labeled 'end' with coordinates (0,1) located on the y-axis.

According to the Fundamental Theorem of Calculus for Line Integrals (FTCLI), the line integral of a conservative vector field is independent of the path and depends only on the endpoints.

\[\int_C \vec{F} \cdot \vec{T} ds = \phi(\text{end}) - \phi(\text{start})\] \[= \phi(0,1) - \phi(1,0)\] \[= (1 + 1 + k) - (1 + k)\] \[= 1\]

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3D Vector Fields

A vector field \(\vec{F} = \langle f, g, h \rangle\) is conservative if there exists a scalar potential function \(\phi(x, y, z)\) such that: \[\langle f, g, h \rangle = \vec{F} = \vec{\nabla}\phi = \langle \phi_x, \phi_y, \phi_z \rangle\]
This implies the following component relationships: \[ \left. \begin{aligned} f &= \phi_x \\ g &= \phi_y \\ h &= \phi_z \end{aligned} \right\} \]

Apply! Mixed partials are Same

\[ \phi_{xy} = \phi_{yx} \rightsquigarrow f_y = g_x \] \[ \phi_{yz} = \phi_{zy} \rightsquigarrow g_z = h_y \] \[ \phi_{zx} = \phi_{xz} \rightsquigarrow h_x = f_z \]
The vector field \(\vec{F} = \langle f, g, h \rangle\) is conservative when: \[ f_y = g_x, \quad f_z = h_x \quad \text{and} \quad g_z = h_y \]

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Example

\( \vec{F} = \langle x^2 + e^z, 2z, x e^z + 2y + z^2 \rangle \), is \( \vec{F} \) conservative?

Let \( \vec{F} = \langle f, g, h \rangle \). To determine if the vector field is conservative, we calculate and compare the relevant partial derivatives:

\( f_y = 0 \)
\( g_x = 0 \)
\( g_z = 2 \)
\( h_y = 2 \)
\( f_z = e^z \)
\( h_x = e^z \)
\[ \left. \begin{aligned} f_y &= g_x \\ g_z &= h_y \\ f_z &= h_x \end{aligned} \right\} \]

\( \vec{F} \) is conservative

To find the potential function \( \phi \) such that \( \vec{F} = \vec{\nabla}\phi = \langle \phi_x, \phi_y, \phi_z \rangle \), we set up the following system:

Then solve:

① \( \phi_x = x^2 + e^z \)
② \( \phi_y = 2z \)
③ \( \phi_z = x e^z + 2y + z^2 \)

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Worked Example: Finding a Potential Function

(1) \(\phi_x = x^2 + e^z\)
(2) \(\phi_y = 2z\)
(3) \(\phi_z = x e^z + 2y + z^2\)

Integrate with respect to \(x\) (treating \(y, z\) as constants):

\[\phi(x, y, z) = \frac{x^3}{3} + x e^z + k_1(y, z)\]

Substituting in (2):

\[2z = \phi_y = \frac{\partial k_1(y, z)}{\partial y}\]

Integrate with respect to \(y\) (\(z\) is constant):

\[k_1(y, z) = 2yz + k_2(z)\]

Substituting this back into the expression for \(\phi\):

\[\phi(x, y, z) = \frac{x^3}{3} + x e^z + 2yz + k_2(z)\]

Substituting in (3):

\[x e^z + 2y + z^2 = \phi_z = x e^z + 2y + k_2'(z)\]

By canceling the common terms on both sides, we find:

\[k_2'(z) = z^2\]

Integrate with respect to \(z\):

\[k_2(z) = \frac{z^3}{3} + K\]

Combining all results gives the final potential function:

\[\phi(x, y, z) = \frac{x^3}{3} + x e^z + 2yz + \frac{z^3}{3} + K\]