Lesson 27: Line integrals of Vectorfields (17.2- part II)

Accessible transcription generated on 3/27/2026

Original Notes

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Lesson 27: Line integrals of Vectorfields (17.2- part II)

Review: Line integral of a function \(f\) over Curve \(C\): \(\vec{r}(t)\), \(a \le t \le b\)

\[ \int_{C} f \, ds = \int_{a}^{b} f(\vec{r}(t)) |\vec{r}'(t)| \, dt \quad \rightsquigarrow \quad \text{Mass of curve with density } f \] \[ \int_{C} 1 \, ds = \int_{a}^{b} |\vec{r}'(t)| \, dt \quad \rightsquigarrow \quad \text{Arc Length} \]

Today: integrate / accumulate a component of vectorfield along a curve

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Example

Evaluate \(\int_C xy^3 \, ds\)

\(C\) is the unit circle centered at the origin.

A coordinate plane diagram showing the unit circle C. The circle is centered at the origin (0,0) of a standard Cartesian coordinate system, where the x and y axes intersect. An arrowhead is drawn on the upper-right portion of the circle, indicating a counter-clockwise orientation for the parameterization of the curve.
Visual Description: A coordinate plane diagram showing the unit circle C. The circle is centered at the origin (0,0) of a standard Cartesian coordinate system, where the x and y axes intersect. An arrowhead is drawn on the upper-right portion of the circle, indicating a counter-clockwise orientation for the parameterization of the curve.

First, we parameterize the curve \(C\): \[C: \vec{r}(t) = \langle x(t), y(t) \rangle\] \[= \langle \cos t, \sin t \rangle, \quad 0 \leq t \leq 2\pi\]

Next, we find the derivative of the position vector and its magnitude: \[\vec{r}'(t) = \langle -\sin t, \cos t \rangle\] \[|\vec{r}'(t)| = 1\]

Now, we substitute these into the formula for the line integral with respect to arc length: \[\int_C xy^3 \, ds = \int_0^{2\pi} x(t) (y(t))^3 |\vec{r}'(t)| \, dt\] \[= \int_0^{2\pi} \cos t (\sin t)^3 \, dt\] \[= \left. \frac{(\sin t)^4}{4} \right|_0^{2\pi} = 0\]

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Scalar Projections

Geometric representation of a scalar projection. A vector \(\vec{u}\) with magnitude \(|u|\) is shown at an angle \(\theta\) relative to a horizontal vector \(\vec{v}\). A vertical dashed line is dropped from the terminal point of \(\vec{u}\) to meet the horizontal vector \(\vec{v}\) at a right angle. The horizontal segment along vector \(\vec{v}\) from the shared origin to the dashed line is labeled \(L\), representing the component of \(\vec{u}\) in the direction of \(\vec{v}\).
Visual Description: Geometric representation of a scalar projection. A vector \(\vec{u}\) with magnitude \(|u|\) is shown at an angle \(\theta\) relative to a horizontal vector \(\vec{v}\). A vertical dashed line is dropped from the terminal point of \(\vec{u}\) to meet the horizontal vector \(\vec{v}\) at a right angle. The horizontal segment along vector \(\vec{v}\) from the shared origin to the dashed line is labeled \(L\), representing the component of \(\vec{u}\) in the direction of \(\vec{v}\).
\[ L = \text{Scal}_{\vec{v}} \vec{u} \] \[ \frac{L}{|u|} = \cos \theta \implies L = |u| \cos \theta \] \[ \vec{u} \cdot \vec{v} = |u| |v| \cos \theta \] \[ \implies |u| \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|v|} \]

Today:

Compute Line integrals of \(\text{Scal}_{\vec{T}} \vec{F}\), \(\text{Scal}_{\vec{N}} \vec{F}\)

Illustration showing vector components along a curve \(C\). A green curve \(C\) originates at the bottom left and arcs toward the top right. At three discrete points along this path, three vectors are illustrated: a blue tangent vector \(\vec{T}\) pointing along the curve, a red normal vector \(\vec{N}\) pointing perpendicular to the curve, and a purple vector \(\vec{F}\) representing a vector field at that location. At the final point at the top right, the vectors are clearly labeled \(\vec{N}\) (red), \(\vec{F}\) (purple), and \(\vec{T}\) (blue).
Visual Description: Illustration showing vector components along a curve \(C\). A green curve \(C\) originates at the bottom left and arcs toward the top right. At three discrete points along this path, three vectors are illustrated: a blue tangent vector \(\vec{T}\) pointing along the curve, a red normal vector \(\vec{N}\) pointing perpendicular to the curve, and a purple vector \(\vec{F}\) representing a vector field at that location. At the final point at the top right, the vectors are clearly labeled \(\vec{N}\) (red), \(\vec{F}\) (purple), and \(\vec{T}\) (blue).

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Component of \(\vec{F}\) in \(\vec{T}\) direction

Consider a curve \(C: \vec{r}(t)\) for \(a \le t \le b\), where \(\vec{T}\) is the unit tangent vector.

\[\vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|}\]

A technical diagram illustrating the geometric relationship between a vector field and a curve. A green line representing a curve C arcs from the bottom left toward the top right. At a specific point on the curve, two vectors are drawn. A blue vector labeled T is shown tangent to the curve at that point. A purple vector labeled F originates from the same point on the curve and points away from it at an angle. This visual aid demonstrates how vector F can be projected onto the unit tangent vector T.
Visual Description: A technical diagram illustrating the geometric relationship between a vector field and a curve. A green line representing a curve C arcs from the bottom left toward the top right. At a specific point on the curve, two vectors are drawn. A blue vector labeled T is shown tangent to the curve at that point. A purple vector labeled F originates from the same point on the curve and points away from it at an angle. This visual aid demonstrates how vector F can be projected onto the unit tangent vector T.
The scalar component of \(\vec{F}\) in the direction of \(\vec{T}\) is: \[scal_{\vec{T}} \vec{F} = \frac{\vec{F} \cdot \vec{T}}{|\vec{T}|} = \vec{F} \cdot \vec{T}\] Where the resulting value is a scalar.

Line integral: \[\int_C \vec{F} \cdot \vec{T} \, ds\]

Physical Meaning:

If \(\vec{F} = \text{Force Field} \rightsquigarrow \text{Work} = \int_C \vec{F} \cdot \vec{T} \, ds\)

If \(\vec{F} = \text{Velocity} \rightsquigarrow \text{Flow} = \int_C \vec{F} \cdot \vec{T} \, ds\)

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Computing \(\int_C \vec{F} \cdot \vec{T} \, ds\)

\(C: r(t), \quad a \le t \le b\)

\[\vec{T}(t) = \frac{r'(t)}{|r'(t)|}\]

By substituting the definition of the unit tangent vector and the differential arc length \(ds = |r'(t)| \, dt\), the line integral is expressed as:

\[\int_C \vec{F} \cdot \vec{T} \, ds = \int_a^b \left( \vec{F}(r(t)) \cdot \frac{r'(t)}{|r'(t)|} \right) |r'(t)| \, dt\]

Simplifying the expression leads to the formula for computing the line integral of a vector field along a curve:

\[\int_C \vec{F} \cdot \vec{T} \, ds = \int_a^b \vec{F}(r(t)) \cdot r'(t) \, dt\]

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Other Notation!

\[ \int_{C} \vec{F} \cdot \vec{T} \, ds = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) \, dt \] \[ = \int_{C} \vec{F} \cdot d\vec{r} \]

Suppose \(\vec{F} = \langle f, g, h \rangle\), \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\)

\[ \int_{C} \vec{F} \cdot \vec{T} \, ds = \int_{a}^{b} (f x' + g y' + h z') \, dt \] \[ = \int_{a}^{b} f x' \, dt + \int_{a}^{b} g y' \, dt + \int_{a}^{b} h z' \, dt \] \[ = \int_{C} f \, dx + \int_{C} g \, dy + \int_{C} h \, dz = \int_{C} f \, dx + g \, dy + h \, dz \]

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Work Done in a Force Field

Eg: Find Work done in presence of Force Field \(\vec{F} = \langle x, -y \rangle\) moving along the Curve \(C\) given by \(\vec{r}(t) = \langle 2t, t^2 \rangle, 0 \le t \le 1\)

Vector field plot representing the force field \(\vec{F} = \langle x, -y \rangle\) on a Cartesian grid ranging from approximately -3 to 3 on both axes. The vectors illustrate the field's behavior: in the first quadrant, vectors point down and to the right; in the second, down and to the left; in the third, up and to the left; and in the fourth, up and to the right. Overlaid on this field is a green parabolic curve representing the path \(C\) defined by \(\vec{r}(t) = \langle 2t, t^2 \rangle\). The curve starts at the origin (0, 0) and ends at the point (2, 1). A green arrow on the path indicates the direction of travel along the curve from the origin toward (2, 1).
Visual Description: Vector field plot representing the force field \(\vec{F} = \langle x, -y \rangle\) on a Cartesian grid ranging from approximately -3 to 3 on both axes. The vectors illustrate the field's behavior: in the first quadrant, vectors point down and to the right; in the second, down and to the left; in the third, up and to the left; and in the fourth, up and to the right. Overlaid on this field is a green parabolic curve representing the path \(C\) defined by \(\vec{r}(t) = \langle 2t, t^2 \rangle\). The curve starts at the origin (0, 0) and ends at the point (2, 1). A green arrow on the path indicates the direction of travel along the curve from the origin toward (2, 1).

\[ \text{Work} = \int_{C} \vec{F} \cdot \vec{T} \, ds \]

\[ = \int_{0}^{1} \left[ \vec{F}(\vec{r}(t)) \cdot \vec{T}(t) \right] |\vec{r}'(t)| \, dt \]

Given:
\(\vec{r}' = \langle 2, 2t \rangle\), \(|\vec{r}'| = \sqrt{4 + 4t^2}\)

\[ \vec{T} = \frac{\vec{r}'}{|\vec{r}'|} = \frac{\langle 2, 2t \rangle}{\sqrt{4 + 4t^2}} \]

\[ \vec{F}(\vec{r}(t)) = \langle x(t), -y(t) \rangle = \langle 2t, -t^2 \rangle \]

Substituting into the work integral:

\[ = \int_{0}^{1} \left( \langle 2t, -t^2 \rangle \cdot \frac{\langle 2, 2t \rangle}{\sqrt{4 + 4t^2}} \right) \sqrt{4 + 4t^2} \, dt \]

\[ = \int_{0}^{1} (4t - 2t^3) \, dt = \left[ 2t^2 - \frac{t^4}{2} \right]_{0}^{1} = \frac{3}{2} \]

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Other ways of framing the same problem

\(\vec{F} = \langle x, -y \rangle, \quad C: \vec{r}_(t) = \langle 2t, t^2 \rangle, \quad 0 \leq t \leq 1\)

  1. Compute \(\int_{C} \vec{F} \cdot \vec{T} \, ds \quad \text{or} \quad \int_{C} \vec{F} \cdot d\vec{r}\)
  2. Find \(\int_{C} x \, dx - y \, dy \quad \text{or} \quad \int_{C} x \, dx - \int_{C} y \, dy\)

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Circulation

Suppose \(C\) is a closed curve (where the start and end are the same points).

A hand-drawn illustration of a closed curve labeled C. The curve is represented as an irregular green loop. Arrows are drawn on the curve indicating a counter-clockwise orientation, showing the path returns to its starting point.
Visual Description: A hand-drawn illustration of a closed curve labeled C. The curve is represented as an irregular green loop. Arrows are drawn on the curve indicating a counter-clockwise orientation, showing the path returns to its starting point.

A line integral over such a curve is called CIRCULATION.

Notation:

\[ \oint_{C} \vec{F} \cdot \vec{T} \, ds = \oint_{C} \vec{F} \cdot d\vec{r} \]

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Example: Evaluating a Line Integral over a Closed Curve

eg: Evaluate \(\oint_C \vec{F} \cdot \vec{T} ds\), where \(\vec{F} = \langle x, -y \rangle\) and \(C\) is the Closed Curve Below.

A vector field plot on a Cartesian coordinate plane illustrating a line integral over a closed triangular path C. The vector field \(\vec{F} = \langle x, -y \rangle\) is represented by a grid of blue arrows. In the first quadrant, where the path is located, the vectors generally point to the right and down. The closed path C is a triangle with vertices at the origin (0,0), (2,0), and (0,2). It is composed of three directed line segments: C1 is a horizontal path along the x-axis from (0,0) to (2,0); C2 is a diagonal path from (2,0) to (0,2); and C3 is a vertical path along the y-axis from (0,2) back to (0,0). The direction of travel indicated by arrows is counter-clockwise. The axes are labeled with integer units from -3 to 3.
Visual Description: A vector field plot on a Cartesian coordinate plane illustrating a line integral over a closed triangular path C. The vector field \(\vec{F} = \langle x, -y \rangle\) is represented by a grid of blue arrows. In the first quadrant, where the path is located, the vectors generally point to the right and down. The closed path C is a triangle with vertices at the origin (0,0), (2,0), and (0,2). It is composed of three directed line segments: C1 is a horizontal path along the x-axis from (0,0) to (2,0); C2 is a diagonal path from (2,0) to (0,2); and C3 is a vertical path along the y-axis from (0,2) back to (0,0). The direction of travel indicated by arrows is counter-clockwise. The axes are labeled with integer units from -3 to 3.

\[ \oint_C \vec{F} \cdot \vec{T} ds = \int_{C_1} \vec{F} \cdot \vec{T} ds + \int_{C_2} \vec{F} \cdot \vec{T} ds + \int_{C_3} \vec{F} \cdot \vec{T} ds \]

\(C_1\):

\[ \vec{r}_1(t) = \vec{r}_0 + t \vec{v} \] \[ = \langle 0,0 \rangle + t \langle 1,0 \rangle = \langle t, 0 \rangle \] \[ 0 \leq t \leq 2 \]

\(\vec{r}_1'(t) = \langle 1, 0 \rangle\)
\(\vec{F}(\vec{r}_1(t)) = \langle t, 0 \rangle\)

\[ \int_{C_1} \vec{F} \cdot \vec{T} dt = \int_0^2 \vec{F}(\vec{r}_1(t)) \cdot \vec{r}_1'(t) dt \] \[ = \int_0^2 t dt = 2 \]

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Line Integral Calculation for Path \(C_2\)

A vector field plot on a Cartesian coordinate system ranging from -3 to 3 on both axes. The vector field is represented by blue arrows that point away from the y-axis and towards the x-axis, corresponding to the field \(\vec{F}(x, y) = \langle x, -y \rangle\). Superimposed on the field is a directed triangular path with vertices at (0,0), (2,0), and (0,2). Segment \(C_1\) is on the x-axis directed from (0,0) to (2,0). Segment \(C_2\) is a diagonal line from (2,0) to (0,2). Segment \(C_3\) is on the y-axis directed from (0,2) to (0,0). The points (2,0) and (0,2) are explicitly labeled on the axes.
Visual Description: A vector field plot on a Cartesian coordinate system ranging from -3 to 3 on both axes. The vector field is represented by blue arrows that point away from the y-axis and towards the x-axis, corresponding to the field \(\vec{F}(x, y) = \langle x, -y \rangle\). Superimposed on the field is a directed triangular path with vertices at (0,0), (2,0), and (0,2). Segment \(C_1\) is on the x-axis directed from (0,0) to (2,0). Segment \(C_2\) is a diagonal line from (2,0) to (0,2). Segment \(C_3\) is on the y-axis directed from (0,2) to (0,0). The points (2,0) and (0,2) are explicitly labeled on the axes.

For the segment \(C_2\), we define the parametrization \(r_2(t)\) starting from point \((2, 0)\) and moving towards point \((0, 2)\) using the vector equation \(r_2(t) = r_0 + t\vec{v}\):

\[ \begin{aligned} r_2(t) &= \langle 2, 0 \rangle + t \langle -2, 2 \rangle \\ &= \langle -2t + 2, 2t \rangle, \quad 0 \leq t \leq 1 \end{aligned} \]

The derivative of the position vector is:

\[ r_2'(t) = \langle -2, 2 \rangle \]

Evaluating the vector field \(\vec{F}(x, y) = \langle x, -y \rangle\) along the path \(r_2(t)\):

\[ \vec{F}(r_2(t)) = \langle -2t + 2, -2t \rangle \]

The line integral over \(C_2\) is then calculated as follows:

\[ \begin{aligned} \int_{C_2} \vec{F} \cdot \vec{T} \, ds &= \int_{0}^{1} \vec{F}(r_2(t)) \cdot r_2'(t) \, dt \\ &= \int_{0}^{1} \langle -2t + 2, -2t \rangle \cdot \langle -2, 2 \rangle \, dt \\ &= \int_{0}^{1} (4t - 4 - 4t) \, dt \\ &= \int_{0}^{1} -4 \, dt \\ &= -4 \end{aligned} \]

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Example: Line Integral over a Closed Triangular Path

\(C_3\): \(\vec{r}_3(t) = \vec{r}_0 + t \vec{v}\)
\(\quad = \langle 0, 2 \rangle + t \langle 0, -1 \rangle\)
\(\quad = \langle 0, 2 - t \rangle\)
\(\quad 0 \le t \le 2\)

A vector field plot in the Cartesian plane with a superimposed triangular path. The vector field \(\vec{F}(x,y) = \langle x, -y \rangle\) is represented by a grid of blue arrows. In the first quadrant, the arrows point towards the x-axis and away from the y-axis (downwards and to the right). A right-angled triangle path \(C\) is highlighted in green with its vertices at the origin \((0,0)\), \((2,0)\), and \((0,2)\). The path consists of three segments: \(C_1\) along the positive x-axis from 0 to 2, \(C_2\) along the hypotenuse from \((2,0)\) to \((0,2)\), and \(C_3\) along the positive y-axis from 2 back to 0. Small green arrowheads on these segments indicate a counter-clockwise orientation. The coordinates \((0,2)\) and \((2,0)\) are explicitly labeled on the graph, which ranges from -3 to 3 on both axes.
Visual Description: A vector field plot in the Cartesian plane with a superimposed triangular path. The vector field \(\vec{F}(x,y) = \langle x, -y \rangle\) is represented by a grid of blue arrows. In the first quadrant, the arrows point towards the x-axis and away from the y-axis (downwards and to the right). A right-angled triangle path \(C\) is highlighted in green with its vertices at the origin \((0,0)\), \((2,0)\), and \((0,2)\). The path consists of three segments: \(C_1\) along the positive x-axis from 0 to 2, \(C_2\) along the hypotenuse from \((2,0)\) to \((0,2)\), and \(C_3\) along the positive y-axis from 2 back to 0. Small green arrowheads on these segments indicate a counter-clockwise orientation. The coordinates \((0,2)\) and \((2,0)\) are explicitly labeled on the graph, which ranges from -3 to 3 on both axes.

\(\vec{r}_3'(t) = \langle 0, -1 \rangle\)
\(\vec{F}(\vec{r}_3(t)) = \langle 0, t - 2 \rangle\)

\[ \int_{C_3} \vec{F} \cdot \vec{T} \, ds = \int_0^2 \vec{F}(\vec{r}_3(t)) \cdot \vec{r}_3'(t) \, dt = \int_0^2 (-t + 2) \, dt = \left[ -\frac{t^2}{2} + 2t \right]_0^2 = 2 \]

\[ \oint_C \vec{F} \cdot \vec{T} \, ds = \int_{C_1} \vec{F} \cdot \vec{T} \, ds + \int_{C_2} \vec{F} \cdot \vec{T} \, ds + \int_{C_3} \vec{F} \cdot \vec{T} \, ds = 2 - 4 + 2 = 0 \]

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Flux

\[ \int_C \vec{F} \cdot \vec{N} \, ds \]
Diagram illustrating the relationship between a curve and its normal vector for flux calculation. A green curve flows from bottom-left to top-right. At a point on the curve, a blue arrow represents the tangent vector pointing in the direction of the path. A solid red arrow, representing the unit normal vector N, points perpendicular to the tangent vector towards the outside (right). A dashed red arrow points in the opposite direction, towards the inside. This illustrates the convention for a unit vector perpendicular to the curve going outwards.
Visual Description: Diagram illustrating the relationship between a curve and its normal vector for flux calculation. A green curve flows from bottom-left to top-right. At a point on the curve, a blue arrow represents the tangent vector pointing in the direction of the path. A solid red arrow, representing the unit normal vector N, points perpendicular to the tangent vector towards the outside (right). A dashed red arrow points in the opposite direction, towards the inside. This illustrates the convention for a unit vector perpendicular to the curve going outwards.

Definition:

\[ \vec{T} = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \]

\(\vec{N}\) = unit vector perpendicular to \(T\) going outwards OR following Right hand Rule.

Suppose \(\vec{r}(t) = \langle x(t), y(t) \rangle\)

\[ T(t) = \frac{1}{|\vec{r}'(t)|} \langle x'(t), y'(t) \rangle \]

By convention:

\[ N(t) = \frac{1}{|\vec{r}'(t)|} \langle y'(t), -x'(t) \rangle \]

Flux Formula:

\[ \text{Flux} = \int_C \vec{F} \cdot \vec{N} \, ds = \int_a^b \vec{F}(\vec{r}(t)) \cdot \langle y', -x' \rangle \, dt \]

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Example not worked out in class

Example: Given \(\vec{F} = \langle 2x, 2y \rangle\), find the flux across the unit circle \(C: \vec{r}(t) = \langle \cos t, \sin t \rangle, 0 \le t \le 2\pi\).

We start by defining the parametric components of the curve \(C\): \[\vec{r}(t) = \langle x(t), y(t) \rangle = \langle \cos t, \sin t \rangle\] The derivative of the position vector is: \[\vec{r}'(t) = \langle x'(t), y'(t) \rangle = \langle -\sin t, \cos t \rangle\] The magnitude of the velocity vector is: \[|\vec{r}'(t)| = 1\] The outward unit normal vector \(\vec{N}(t)\) is given by: \[\vec{N}(t) = \frac{1}{|\vec{r}'(t)|} \langle y'(t), -x'(t) \rangle = \langle \cos t, \sin t \rangle\] Evaluating the vector field \(\vec{F}\) along the curve: \[\vec{F}(\vec{r}(t)) = \langle 2 \cos t, 2 \sin t \rangle\]

Diagram showing a unit circle on a Cartesian coordinate plane centered at the origin. The circle has an arrow indicating a counter-clockwise orientation. At a point in the first quadrant, two vectors originate: a blue vector labeled 'T' pointing tangent to the circle in the direction of the orientation, and a red vector labeled 'N' pointing radially outward, representing the outward unit normal vector. A red dashed line extends from the point on the circle back towards the origin, aligned with the normal vector.
Visual Description: Diagram showing a unit circle on a Cartesian coordinate plane centered at the origin. The circle has an arrow indicating a counter-clockwise orientation. At a point in the first quadrant, two vectors originate: a blue vector labeled 'T' pointing tangent to the circle in the direction of the orientation, and a red vector labeled 'N' pointing radially outward, representing the outward unit normal vector. A red dashed line extends from the point on the circle back towards the origin, aligned with the normal vector.

The flux across the curve is calculated as the line integral: \[\oint_C \vec{F} \cdot \vec{N} \, ds = \int_0^{2\pi} (\vec{F}(\vec{r}(t)) \cdot \vec{N}(t)) |\vec{r}'(t)| \, dt\] Substituting the expressions we found: \[= \int_0^{2\pi} (2 \cos^2 t + 2 \sin^2 t) \, dt = \int_0^{2\pi} 2 \, dt = 4\pi\]