Lesson 33: Surface Integrals - III (17.6)

Accessible transcription generated on 4/10/2026

Original Notes

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Lesson 33: Surface Integrals - III (17.6)

Review: Surface integrals of scalar functions

\(S: \vec{r}(u,v)\), \(u,v\) in Region \(R\) inside \(\mathbb{R}^2\)

\[\iint_S f \, dS = \iint_R f(\vec{r}(u,v)) |\vec{r}_u \times \vec{r}_v| \, dA\]

What does \(\vec{r}_u \times \vec{r}_v\) represent? Normal vector. Note that \(-(\vec{r}_u \times \vec{r}_v)\) is also a normal vector.

\(z = f(x,y) \rightsquigarrow \vec{r}(u,v) = \langle u, v, f(u,v) \rangle\)

  • \(\vec{r}_u \times \vec{r}_v = \langle -f_u, -f_v, 1 \rangle\)
  • \(|\vec{r}_u \times \vec{r}_v| = \sqrt{f_u^2 + f_v^2 + 1}\)

Example: \(4x + 3y + z = 12 \rightsquigarrow z = 12 - 4x - 3y = f(x,y)\)

\(\vec{r}(u,v) = \langle u, v, f(u,v) \rangle\)

\(\vec{r}_u \times \vec{r}_v = \langle 4, 3, 1 \rangle\)

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Review example:

Evaluate \[ \iint_S (x^2 + 3z)^{1/2} \, dS \] where \( S \) is the surface \( z = x^2 \approx f(x,y) \) with boundaries \( 0 \le x \le 2 \) and \( 0 \le y \le 4 \).
First, we parameterize the surface \( S \): \[ r(u, v) = \langle u, v, f(u, v) \rangle = \langle u, v, u^2 \rangle \] with limits \( 0 \le u \le 2 \) and \( 0 \le v \le 4 \). Next, we calculate the normal vector and its magnitude: \[ r_u \times r_v = \langle -2u, 0, 1 \rangle \] \[ |r_u \times r_v| = \sqrt{4u^2 + 1} \] Setting up the surface integral by substituting variables and the magnitude of the normal vector: \[ \iint_S (x^2 + 3z)^{1/2} \, dS = \int_0^4 \int_0^2 \underbrace{(u^2 + 3u^2)^{1/2}}_{(x^2 + 3z)^{1/2}} \underbrace{(\sqrt{4u^2 + 1})}_{|r_u \times r_v|} \, du \, dv \] Simplifying the integrand: \[ = \int_0^4 \int_0^2 2u \sqrt{4u^2 + 1} \, du \, dv \] ... evaluate by substitution.

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Integrating vector fields over a Surface

Flux integral or Surface integrals of vector fields

Notation: \[\iint_{S} \vec{F} \cdot d\vec{S} = \iint_{S} (\vec{F} \cdot \vec{n}) \, dS\]

Where the term \((\vec{F} \cdot \vec{n})\) is a scalar, and \(\vec{n} = \text{unit normal}\).

Expanding the integral into parametric form:

\[= \iint_{\substack{u, v \\ \text{Bounds}}} (\vec{F} \cdot \vec{n}) |\vec{r}_u \times \vec{r}_v| \, dA\]
A 3D diagram representing a parametric surface patch. It shows a blue curved quadrilateral surface with grid lines. At a specific point on the surface, two tangent vectors are shown: a green vector labeled \vec{r}_u and a magenta vector labeled \vec{r}_v, which follow the directions of the parametric grid lines. A third vector, the normal vector, is shown pointing outward from the surface, perpendicular to the plane formed by \vec{r}_u and \vec{r}_v. This illustrates the relationship between tangent vectors and the surface normal.
Visual Description: A 3D diagram representing a parametric surface patch. It shows a blue curved quadrilateral surface with grid lines. At a specific point on the surface, two tangent vectors are shown: a green vector labeled \vec{r}_u and a magenta vector labeled \vec{r}_v, which follow the directions of the parametric grid lines. A third vector, the normal vector, is shown pointing outward from the surface, perpendicular to the plane formed by \vec{r}_u and \vec{r}_v. This illustrates the relationship between tangent vectors and the surface normal.
unit normal \[\vec{n} = \frac{\vec{r}_u \times \vec{r}_v}{|\vec{r}_u \times \vec{r}_v|} \quad \text{or} \quad -\frac{(\vec{r}_u \times \vec{r}_v)}{|\vec{r}_u \times \vec{r}_v|}\]

Substituting the expression for the unit normal back into the flux integral, the magnitude terms \(|\vec{r}_u \times \vec{r}_v|\) cancel out, resulting in the computational formula:

\[\iint_{S} \vec{F} \cdot \vec{n} \, dS = \iint_{\substack{u, v \\ \text{Bounds}}} \vec{F} \cdot (\pm \vec{r}_u \times \vec{r}_v) \, dA\]

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Should I choose \(r_u \times r_v\) or \(-r_u \times r_v\)?

Convention! Choose normal pointing outward or upward.

OR you will be asked to choose a specific one.

The above choices define the "Positive" Normal Direction.

A diagram of a green sphere with a dashed midline indicating depth. A purple arrow points radially away from the sphere's surface and is labeled 'outward'. A blue arrow points from the surface toward the center and is labeled 'inward'. This illustrates the choice between outward and inward normal vectors on a closed surface.
Visual Description: A diagram of a green sphere with a dashed midline indicating depth. A purple arrow points radially away from the sphere's surface and is labeled 'outward'. A blue arrow points from the surface toward the center and is labeled 'inward'. This illustrates the choice between outward and inward normal vectors on a closed surface.
A diagram of a green vertical cylinder with purple and blue arrows. Purple arrows point horizontally outward from the cylinder's curved surface, representing outward normals. Blue arrows point horizontally inward towards the cylinder's vertical center line, representing inward normals.
Visual Description: A diagram of a green vertical cylinder with purple and blue arrows. Purple arrows point horizontally outward from the cylinder's curved surface, representing outward normals. Blue arrows point horizontally inward towards the cylinder's vertical center line, representing inward normals.
A diagram showing a curved green surface, similar to a saddle shape, in a 3-dimensional coordinate system. A purple arrow points vertically up from a point on the surface and is labeled 'upward (+ve z coordinate)'. A blue arrow points vertically down from the same point and is labeled 'downward (-ve z coordinate)'. This illustrates the convention for upward and downward normal vectors on an open surface.
Visual Description: A diagram showing a curved green surface, similar to a saddle shape, in a 3-dimensional coordinate system. A purple arrow points vertically up from a point on the surface and is labeled 'upward (+ve z coordinate)'. A blue arrow points vertically down from the same point and is labeled 'downward (-ve z coordinate)'. This illustrates the convention for upward and downward normal vectors on an open surface.

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#### Example

\(S: z = x^2, \quad 0 \le x \le 2, \quad 0 \le y \le 4\)

A 3D coordinate plot illustrating a parabolic cylinder surface in the first octant. The coordinate system consists of a vertical z-axis, a horizontal y-axis extending to the right, and an x-axis extending toward the bottom-left. The surface is defined by the equation z = x^2 and is constrained by the domain 0 ≤ x ≤ 2 and 0 ≤ y ≤ 4. The diagram shows the surface's shape using four green parabolic arcs drawn in vertical planes parallel to the xz-plane, spaced at intervals along the y-axis. Each arc starts at the y-axis (where x=0 and z=0) and curves upward as the x-value increases. Dashed lines delineate the rectangular footprint of the surface on the xy-plane and connect the upper endpoints of the parabolic arcs to indicate the full three-dimensional extent of the cylinder segment.
Visual Description: A 3D coordinate plot illustrating a parabolic cylinder surface in the first octant. The coordinate system consists of a vertical z-axis, a horizontal y-axis extending to the right, and an x-axis extending toward the bottom-left. The surface is defined by the equation z = x^2 and is constrained by the domain 0 ≤ x ≤ 2 and 0 ≤ y ≤ 4. The diagram shows the surface's shape using four green parabolic arcs drawn in vertical planes parallel to the xz-plane, spaced at intervals along the y-axis. Each arc starts at the y-axis (where x=0 and z=0) and curves upward as the x-value increases. Dashed lines delineate the rectangular footprint of the surface on the xy-plane and connect the upper endpoints of the parabolic arcs to indicate the full three-dimensional extent of the cylinder segment.

The surface can be parameterized as: \[r(u, v) = \langle u, v, u^2 \rangle\] where \(0 \le u \le 2, \quad 0 \le v \le 4\)

The normal vector is found using the cross product of the partial derivatives: \[r_u \times r_v = \langle -2u, 0, 1 \rangle\]

Depending on the required orientation, the normal vector is:
upward Normal is \(\langle -2u, 0, 1 \rangle\)
downward Normal is \(\langle 2u, 0, -1 \rangle\)

facing +ve x axis is \(\langle 2u, 0, -1 \rangle\)
facing -ve x axis is \(\langle -2u, 0, 1 \rangle\)

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Example: Flux Integral over a Surface

Eg: Evaluate flux integral for \(\vec{F} = \langle 1, z, -x \rangle\) over the surface \(S: z = x^2\), \(0 \le x \le 2\), \(0 \le y \le 4\) with normal pointing downward.

Solution:

First, we parametrize the surface \(S\):

\[ r(u, v) = \langle u, v, u^2 \rangle \] \[ r_u \times r_v = \langle -2u, 0, 1 \rangle \]

Since the problem specifies a downward-pointing normal vector, we use:

\[ -(r_u \times r_v) = \langle 2u, 0, -1 \rangle \]

The flux integral is defined as:

\[ \iint_S \vec{F} \cdot \vec{n} \, dS = \iint_{\substack{u,v \\ \text{Bounds}}} \vec{F}(r(u,v)) \cdot (-(r_u \times r_v)) \, du \, dv \]

Substituting the field \(\vec{F}(r(u,v)) = \langle 1, u^2, -u \rangle\) and the normal vector into the integral with the given bounds:

\[ = \int_0^4 \int_0^2 \langle 1, u^2, -u \rangle \cdot \langle +2u, 0, -1 \rangle \, du \, dv \] \[ = \int_0^4 \int_0^2 3u \, du \, dv = \left. \frac{3u^2}{2} \right|_0^2 \times \left. v \right|_0^4 = 24 \]

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Worked Example: Flux Integral

Evaluate the flux integral for \(\vec{F} = \langle x, y, z \rangle\) over the surface \(S: 4x + 3y + z = 12\) in the 1st octant with the normal pointing towards the positive \(x\)-axis.

Evaluate: \[\iint_{S} \vec{F} \cdot \vec{n} \, dS\]

For the surface \(S\), we can parametrize it as: \[S: \vec{r}(u,v) = \langle u, v, 12 - 4u - 3v \rangle, \quad u \ge 0, \quad v \ge 0, \quad 4u + 3v \le 12\]

A 2D coordinate graph illustrating the region of integration R in the uv-plane. The horizontal axis is labeled u and the vertical axis is labeled v. A straight line with a negative slope represents the equation 4u + 3v = 12. The triangular region in the first quadrant bounded by this line and the axes is shaded with diagonal lines and labeled R, representing the set of points where 4u + 3v is less than or equal to 12.
Visual Description: A 2D coordinate graph illustrating the region of integration R in the uv-plane. The horizontal axis is labeled u and the vertical axis is labeled v. A straight line with a negative slope represents the equation 4u + 3v = 12. The triangular region in the first quadrant bounded by this line and the axes is shaded with diagonal lines and labeled R, representing the set of points where 4u + 3v is less than or equal to 12.

Calculating the cross product of the partial derivatives: \[\vec{r}_u \times \vec{r}_v = \langle 4, 3, 1 \rangle\] \[-(\vec{r}_u \times \vec{r}_v) = \langle -4, -3, -1 \rangle\]

Since the problem specifies the normal must point towards the positive \(x\)-axis, we choose the vector with the positive \(x\)-component: \(\langle 4, 3, 1 \rangle\).

Setting up the integral over the region \(R\): \[\iint_{S} \vec{F} \cdot \vec{n} \, dS = \iint_{R} \vec{F}(\vec{r}(u,v)) \cdot (\vec{r}_u \times \vec{r}_v) \, dA\] \[= \iint_{R} \langle u, v, 12 - 4u - 3v \rangle \cdot \langle 4, 3, 1 \rangle \, dA\] \[= \iint_{R} (4u + 3v + 12 - 4u - 3v) \, dA\] \[= \iint_{R} 12 \, dA\]

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A coordinate graph in the uv-plane showing a triangular region R. The horizontal axis is labeled 'u' and the vertical axis is labeled 'v'. A straight line connects the v-intercept at 4 to the u-intercept at 3, labeled with the equation 4u + 3v = 12. The region R is the interior of the triangle formed by this line and the axes, and it is shaded with green diagonal lines. Green arrows point horizontally from the v-axis (u=0) to the line 4u + 3v = 12, indicating the range of u for a fixed v. A horizontal pink line segment is drawn at a fixed v value to represent the width of the region as (12 - 3v)/4.
Visual Description: A coordinate graph in the uv-plane showing a triangular region R. The horizontal axis is labeled 'u' and the vertical axis is labeled 'v'. A straight line connects the v-intercept at 4 to the u-intercept at 3, labeled with the equation 4u + 3v = 12. The region R is the interior of the triangle formed by this line and the axes, and it is shaded with green diagonal lines. Green arrows point horizontally from the v-axis (u=0) to the line 4u + 3v = 12, indicating the range of u for a fixed v. A horizontal pink line segment is drawn at a fixed v value to represent the width of the region as (12 - 3v)/4.
\[ \iint_R 12 \, dA = 12 \iint_R 1 \, dA \] \[ = 12 \ast \text{Area of } R \] \[ = 12 \ast \frac{1}{2}(3 \times 4) \] \[ = 72 \]

Alternatively

Fix \(v\) with \(0 \le u \le \frac{12-3v}{4}\)

\[ \iint_R 12 \, dA = \int_{0}^{4} \int_{0}^{\frac{12-3v}{4}} 12 \, du \, dv \] \[ = \int_{0}^{4} \frac{12}{4} (12 - 3v) \, dv \] \[ = 3 \left[ 12v - \frac{3v^2}{2} \right]_{0}^{4} = 72 \]

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eg: Evaluate \(\iint_S \vec{F} \cdot d\vec{S}\), where \(\vec{F} = \langle x^7, z^3, y^3 \rangle\)

The surface \(S\) is defined by \(y^2 + z^2 = 4\), for \(0 \leq x \leq 7\).
The normal is positive outwards.

A 3D coordinate system diagram illustrating the cylindrical surface S. The axes shown include a vertical axis and a horizontal axis to the right, forming the yz-plane, and a diagonal axis extending towards the bottom-left, representing the x-axis. A cylinder is drawn centered along this x-axis, starting from the origin (x=0) and extending to x=7. The circular ends of the cylinder have a radius of 2, based on the equation y^2 + z^2 = 4. The diagram uses solid lines for the visible exterior of the cylinder and the front circular face, while the obscured part of the circular base at the origin is indicated with a dashed line.
Visual Description: A 3D coordinate system diagram illustrating the cylindrical surface S. The axes shown include a vertical axis and a horizontal axis to the right, forming the yz-plane, and a diagonal axis extending towards the bottom-left, representing the x-axis. A cylinder is drawn centered along this x-axis, starting from the origin (x=0) and extending to x=7. The circular ends of the cylinder have a radius of 2, based on the equation y^2 + z^2 = 4. The diagram uses solid lines for the visible exterior of the cylinder and the front circular face, while the obscured part of the circular base at the origin is indicated with a dashed line.

Parametrization:

\(x = u\), \(0 \leq u \leq 7\)
\(y^2 + z^2 = 4\) is a circle of Radius 2.
\[ \left. \begin{aligned} y &= 2 \cos v \\ z &= 2 \sin v \end{aligned} \right\} 0 \leq v \leq 2\pi \]

\(r(u, v) = \langle u, 2 \cos v, 2 \sin v \rangle\)
\(r_u = \langle 1, 0, 0 \rangle\)
\(r_v = \langle 0, -2 \sin v, 2 \cos v \rangle\)
\(r_u \times r_v = \langle 0, -2 \cos v, -2 \sin v \rangle\)
\(-(r_u \times r_v) = \langle 0, 2 \cos v, 2 \sin v \rangle\)

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Determining Surface Orientation and Flux Integral

A coordinate diagram in the y-z plane showing a circle centered at the origin. A vertical axis is labeled 'z' and a horizontal axis is labeled 'y'. There is a green circle drawn on these axes. Two blue arrows represent a vector field pointing from the circle's boundary towards the center: one arrow points horizontally from the positive y-axis toward the origin, and another points vertically from the positive z-axis toward the origin. This visualizes that the calculated normal vector field points inwards.
Visual Description: A coordinate diagram in the y-z plane showing a circle centered at the origin. A vertical axis is labeled 'z' and a horizontal axis is labeled 'y'. There is a green circle drawn on these axes. Two blue arrows represent a vector field pointing from the circle's boundary towards the center: one arrow points horizontally from the positive y-axis toward the origin, and another points vertically from the positive z-axis toward the origin. This visualizes that the calculated normal vector field points inwards.

To determine the orientation of the surface, we calculate the cross product of the partial derivatives of the parameterization \(\mathbf{r}(u,v)\).

\[\mathbf{r}_u \times \mathbf{r}_v = \langle 0, -2\cos v, -2\sin v \rangle\]

We test specific values of the parameter \(v\) to see the direction of the normal vectors:

  • For \(v = 0 \rightarrow \langle 0, -2, 0 \rangle\)
  • For \(v = \frac{\pi}{2} \rightarrow \langle 0, 0, -2 \rangle\)

As shown in the diagram, these vectors point toward the origin, meaning \(\mathbf{r}_u \times \mathbf{r}_v\) is facing inwards.

To orient the surface correctly (typically outwards for flux calculations), we choose the negative of the calculated cross product:

\[-(\mathbf{r}_u \times \mathbf{r}_v) = \langle 0, 2\cos v, 2\sin v \rangle\]

We then set up the surface integral for the flux of the vector field \(\mathbf{F}\) across surface \(S\):

\[\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \int_0^{2\pi} \int_0^7 \mathbf{F}(\mathbf{r}(u,v)) \cdot (-(\mathbf{r}_u \times \mathbf{r}_v)) \, du \, dv\]

Substituting the components of \(\mathbf{F}\) and our chosen normal vector into the integral:

\[= \int_0^{2\pi} \int_0^7 \langle u^7, 8\sin^3 v, 8\cos^3 v \rangle \cdot \langle 0, 2\cos v, 2\sin v \rangle \, du \, dv\]

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Integral Calculation Continued

\[ = \int_{0}^{2\pi} \int_{0}^{7} (16 \sin^3 v \cos v + 16 \cos^3 v \sin v) \, du \, dv \] \[ = \int_{0}^{2\pi} \int_{0}^{7} 16 \sin v \cos v \, du \, dv \] \[ = \int_{0}^{2\pi} (7 \times 16) \sin v \cos v \, dv \] \[ = 7 \times 16 \times \left. \frac{(\sin v)^2}{2} \right|_{0}^{2\pi} = 0 \]